Motion in a Plane - Class 11 Physics - Chapter 4 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Motion in a Plane | NCERT | Physics | Class 11
Airplanes $A$ and $B$ are in the same vertical plane and flying with constant velocity. At time $t = 0$ sec, an observer in $A$ finds $B$ at a distance of $500$ m and moving perpendicular to the line of motion of $A$. If at $t = t_{0}$ sec, $A$ just escapes being hit by $B$, then $t_{0}$ in seconds is:
To solve this problem, we consider the velocities of airplanes $A$ and $B$ and their directions of motion.
Given that:
Airplane $B$ is observed to be moving perpendicular to Airplane $A$'s line of motion.
Consider the plane of motion with x-axis along $A$'s trajectory and y-axis perpendicular to it. Assuming airplane $A$ is stationary for relative motion analysis, let's compute:
Relative velocity of $B$ with respect to $A$ along x-axis: $ (v_{BA}){ox} = (v_B){ox} - (v_A){ox} = 0 $
We know that $(v_A){ox} = 100\sqrt{3}$ m/s (assuming the escape speed given indirectly).
Assuming $(v_B)_{ox} = v \cos 30^\circ$, we equate: $ v \cos 30^\circ = 100 \sqrt{3} \Rightarrow v = 200 , \text{m/s} $ because $\cos 30^\circ = \frac{\sqrt{3}}{2}$.Perpendicular distance $d$ to be covered by $B$ to reach the path of $A$ is given as 500 m. Thus, the time needed for $B$ to travel this distance at the velocity perpendicular to $A$ is calculated using: $ t = \frac{d}{v_{oy}} $ where $v_{oy}$ represents velocity of $B$ perpendicular to $A$, which is $v \sin 30^\circ = 200 \times \frac{1}{2} = 100 , \text{m/s}$. Thus, $ t = \frac{500}{100} = 5 , \text{sec} $
Conclusion:The time $t_0$ at which airplane $A$ just avoids collision with airplane $B$ is $t_0 = 5$ seconds. This calculation uses the components of velocity and geometry of the motion to determine the narrow escape of $A$.
A player kicks a football at an angle of $45^{\circ}$ with a velocity of $30 \mathrm{~ms}^{-1}$. A second player 120 m away along the direction of the kick starts running to receive the ball at that instant. Find the speed with which the second player should run to reach the ball before it hits the ground ($\mathrm{g}=10 \mathrm{~ms}^{-2}$).
A) $5 \sqrt{2} \mathrm{~m/s}$ B) $4 \sqrt{2} \mathrm{~m/s}$ C) $5 \mathrm{~m/s}$ D) $10 \sqrt{2} \mathrm{~m/s}$
The correct option is A) $5 \sqrt{2} \mathrm{~m/s}$.
To find the speed at which the second player should run, we first determine the range $R$ of the football when kicked with an initial speed $u = 30 \mathrm{~ms}^{-1}$ at an angle $\theta = 45^\circ$, with gravitational acceleration $g = 10 \mathrm{~ms}^{-2}$. The range of the projectile is calculated by: $$ R = \frac{u^2 \sin 2\theta}{g} $$ Given $\theta = 45^\circ$, $\sin 2\theta = \sin 90^\circ = 1$. Plugging in the values, we get: $$ R = \frac{30^2 \cdot 1}{10} = 90 \mathrm{~m} $$
The range of the football is 90 m.
Since the second player is 120 m away from the point the football was kicked, he needs to travel $120 - 90 = 30 \mathrm{~m}$ to meet the ball.
The total time $T_f$ it takes for the ball to reach the ground, also known as the time of flight, is given by: $$ T_f = \frac{2u \sin \theta}{g} $$
For $\theta = 45^\circ$, $\sin \theta = \sin 45^\circ = \frac{\sqrt{2}}{2}$. The time of flight becomes: $$ T_f = \frac{2 \times 30 \times \frac{\sqrt{2}}{2}}{10} = 3\sqrt{2} \mathrm{~s} $$
The speed $v$ at which the second player must run is then determined by the distance to be covered divided by the time available: $$ v = \frac{x}{T_f} = \frac{30}{3\sqrt{2}} = 10/\sqrt{2} = 5\sqrt{2} \mathrm{~m/s} $$
The speed with which the second player should run is $5\sqrt{2} \mathrm{~m/s}$.
A ring of mass $2\pi , \mathrm{kg}$ and radius $0.25 , \mathrm{m}$ is making $300 , \mathrm{rpm}$ about an axis perpendicular to its plane. The tension (in newtons) developed in the ring is, approximately
A) 50
B) 100
C) $\mathbf{175}$
D) 246
The correct answer is D) 246.
To solve for the tension developed in the ring, we consider the Free Body Diagram (FBD) of a small segment of the ring, PQ, that subtends an angle $2d\theta$. The radial components of the tension provide the necessary centripetal force to keep this segment in motion, while the tangential components of the tension cancel each other out.
By balancing the forces, we have:
$$ 2T \sin(d\theta) = m_{PQ} R \omega^2 $$
Using the small angle approximation, $\sin(d\theta) \approx d\theta$, and noting the mass of the segment $m_{PQ}$ as a fraction of the total mass $M$ of the ring, we get:
$$ m_{PQ} = \frac{2d\theta}{2\pi} \times 2\pi = 2d\theta $$
Substituting the values and solving for $T$, the equation becomes:
$$ 2T d\theta = (2d\theta) R (2\pi n)^2 $$
Here, $n = 300 , \text{rpm} = 5 , \text{rps}$. Plugging the values into the equation and simplifying, we calculate $T$ as:
$$ T = 4\pi^2 n^2 R = 4\pi^2 \times 5^2 \times 0.25 \approx 246 , \text{N} $$
Therefore, the tension developed in the ring is approximately 246 N.
An aeroplane is flying horizontally with a velocity of $600 \mathrm{~km} / \mathrm{h}$ at a height of $1960 \mathrm{~m}$. When it is vertically above a point $A$ (point $A$ on the ground), a bomb is released from it. The bomb strikes the ground at point $B$. The horizontal distance $AB$ is:
A. $1.2 \mathrm{~km}$
B. $0.33 \mathrm{~km}$
C. $3.33 \mathrm{~km}$
D. $33 \mathrm{~km}$
The correct option is C$$ 3.33 \text{ km} $$
To calculate the horizontal distance $AB$ that the bomb travels, we use the flight time to the ground, computed by considering the initial motion as purely vertically under gravity.
Horizontal velocity of the bomb $$ u = 600 \text{ km/h} $$
Convert the velocity to meters per second for consistency in units, $$ u = 600 \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 600 \times \frac{5}{18} \text{ m/s} $$
Height from which the bomb is dropped $$ h = 1960 \text{ m} $$
The acceleration due to gravity $$ g = 9.8 \text{ m/s}^2 $$
The time $t$ for the bomb to hit the ground can be calculated from the kinematic equation for freely falling objects (no initial vertical velocity): $$ t = \sqrt{\frac{2h}{g}} $$
Substitute the values: $$ t = \sqrt{\frac{2 \times 1960}{9.8}} = 20 \text{ s} $$
Use this time to find horizontal displacement $AB$: $$ AB = u \times t = \left(600 \times \frac{5}{18}\right) \text{ m/s} \times 20 \text{ s} = 3333.33 \text{ m} $$
Finally, converting this into kilometers: $$ AB = 3.33 \text{ km} $$
Thus, the bomb lands 3.33 km away from point A, corresponding to option C.
A balloon moves up vertically such that if a stone is projected with a horizontal velocity u relative to the balloon, the stone always hits the ground at a fixed point at a distance $\frac{2u^{2}}{g}$ horizontally away from it. Find the height of the balloon as a function of time.
To solve the problem, we need to determine the height of the balloon as it changes over time, given the condition related to the stone's motion.
Step 1: Establish the Time of Flight of the Stone
The time $ t $ it takes for the stone to reach the ground can be derived from the kinematics equation for an object under gravity, assuming no initial vertical velocity: $$ t = \sqrt{\frac{2h}{g}} $$ where $ h $ is the height of the balloon and $ g $ is the acceleration due to gravity.
Step 2: Define the Horizontal Range
From the problem statement, the stone hits the ground at a fixed distance: $$ R = \frac{2u^2}{g} $$ However, we know the horizontal range ($ R $) can also be expressed as: $$ R = u t $$ Given $ t $ from the first equation, substituting it in gives: $$ \frac{2u^2}{g} = u \sqrt{\frac{2h}{g}} $$
Step 3: Solve for the Height (h)
Simplifying the equation: $$ \frac{2u}{g} = \sqrt{\frac{2h}{g}} $$ Squaring both sides: $$ \frac{4u^2}{g^2} = \frac{2h}{g} $$ Rearranging for $ h $ yields: $$ h = \frac{2u^2}{g} $$
Key Observation and Conclusion
Interestingly, this finds $ h $, the initial height of the balloon, as a constant independent of time. However, to reconcile this with the temporal aspect mentioned in the question, we must review the expression derived for horizontal range: $$ \frac{2u^2}{g} = ut $$ Rearrange this for $ h $ in terms of time: $$ h = ut $$ indicating a linear relationship between the height of the balloon and time, assuming the initial height to be zero and with $ u $ being the vertical rate of rise of the balloon relative to the ground.
Thus, the height of the balloon as a function of time $ t $ is given by: $$ h(t) = ut $$ where $ h(t) $ is the height at time $ t $, and $ u $ is the vertical velocity of the balloon rising upwards.
A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with a speed of $0.5 \mathrm{~m/s}$. What is the height of the plane of the circle from the vertex of the funnel? (Take $\mathrm{g} = 10 \mathrm{~m/s}^{2}$)
A) $0.25 \mathrm{~cm}$
B) $2 \mathrm{~cm}$
C) $4 \mathrm{~cm}$
D) $2.5 \mathrm{~cm}$
To find the height of the plane of the circle from the vertex of the funnel, we need to understand the dynamics and forces acting on the particle.
The particle stays on a circular path within the funnel due to the balance between gravitational force and normal force from the surface. For a particle of mass $m$, moving with velocity $v$ on the path of radius $r$, the equations can be set using gravitational force ($\mathrm{mg}$) and normal force ($N$) components along and perpendicular to the slanted surface of the funnel.
From the balance of forces, we have:
Vertical force balance (along the slope): $$ \mathrm{mg} = N \sin \theta $$
Horizontal force balance (perpendicular to the slope, providing centripetal force): $$ \frac{m v^2}{r} = N \cos \theta $$
Dividing these two equations (to eliminate $N$), $$ \frac{\mathrm{mg}}{\frac{mv^2}{r}} = \frac{N \sin \theta}{N \cos \theta} $$ This simplifies to, $$ \tan \theta = \frac{rg}{v^2} $$ where $\theta$ is the angle of the slant of the funnel with respect to the vertical. Furthermore, if $h$ is the height of the circular path from the vertex, in terms of the slant height $r$, this angle $\theta$ also satisfies, $$ \tan \theta = \frac{r}{h} $$ Equating the two expressions for $\tan \theta$ gives us: $$ \frac{r}{h} = \frac{rg}{v^2} $$ From which, upon simplifying, we find: $$ h = \frac{v^2}{g} $$
Given $v = 0.5 , \mathrm{m/s}$ and $g = 10 , \mathrm{m/s}^2$, substituting these values returns: $$ h = \frac{(0.5)^2}{10} = 0.025 , \mathrm{m} = \textbf{2.5 cm} $$
Therefore, the correct answer is Option D: $2.5 , \mathrm{cm}$.
An airplane moving horizontally with a speed of $720 \mathrm{~km/h}$ drops a food packet, while flying at a height of $396.9 \mathrm{~m}$. The time taken by the food packet to reach the ground and its horizontal range is (Take $\mathrm{g}=9.8 \mathrm{~m/s}^{2}$):
A) $3 \mathrm{~sec}$ and $2000 \mathrm{~m}$
B) $5 \mathrm{~sec}$ and $500 \mathrm{~m}$
C) $8 \mathrm{~sec}$ and $1500 \mathrm{~m}$
D) $9 \mathrm{~sec}$ and $1800 \mathrm{~m}$
The correct answer is Option D: $9 \mathrm{sec}$
and $1800 \mathrm{~m}$
.
To solve for the time of fall ($t$) and the horizontal range ($R$):
Calculate the Time of Fall: The time $t$ it takes for the packet to reach the ground is calculated using the formula for free fall: $$ t = \sqrt{\frac{2h}{g}} $$ where $h = 396.9 , \mathrm{m}$ (height) and $g = 9.8 , \mathrm{m/s}^2$ (acceleration due to gravity). Plugging in the values: $$ t = \sqrt{\frac{2 \times 396.9}{9.8}} = 9 , \mathrm{sec} $$
Calculate the Horizontal Range: The horizontal velocity $u$ of the airplane is given as $720 , \mathrm{km/h}$, which needs to be converted into $\mathrm{m/s}$: $$ u = 720 \frac{\mathrm{km}}{\mathrm{hr}} \times \frac{1000 ,\mathrm{m}}{1 , \mathrm{km}} \times \frac{1 , \mathrm{hr}}{3600 , \mathrm{s}} = 200 , \mathrm{m/s} $$ Given this velocity and the time calculated previously, the horizontal range $R$ can be calculated as: $$ R = u \times t = 200 , \mathrm{m/s} \times 9 , \mathrm{sec} = 1800 , \mathrm{m} $$
Hence, the food packet takes $9$ seconds to hit the ground and travels a horizontal range of $1800$ meters.
A clock is in the $x$-$y$ plane. Find the angular velocity of the second hand of a clock (in rad/s).
(A) $\frac{\pi}{30}(\hat{i})$
(B) $\frac{\pi}{30}(-\hat{j})$
(C) $\frac{\pi}{60}(\hat{k})$
(D) $\frac{\pi}{30}(-\hat{k})$
The correct choice is (D) $\frac{\pi}{30}(-\hat{k})$.
The angular velocity $\omega$ of the second hand on a clock can be calculated using the formula: $$ \omega = \frac{2\pi}{T} $$ where $T$ = 60 seconds (the time it takes for the second hand to complete one full rotation, i.e., one minute). Thus, substituting into the formula provides: $$ \omega = \frac{2\pi}{60} = \frac{\pi}{30} $$
Since the rotation of the second hand is in the clockwise direction, and by convention, clockwise rotation in the $x$-$y$ plane corresponds to a negative direction along the $z$-axis (i.e., $-\hat{k}$), we represent angular velocity as: $$ \vec{\omega} = \frac{\pi}{30}(-\hat{k}) $$
Therefore, the answer is (D) $\frac{\pi}{30}(-\hat{k})$.
A man of mass $50 \mathrm{ kg}$ is running on a plank of mass $150 \mathrm{ kg}$ with a speed of $8 \mathrm{ m/s}$ relative to the plank as shown in the figure (both were initially at rest and velocity of man with respect to ground anyhow remains constant). The plank is placed on a smooth horizontal surface. The man, while running, whistles with frequency $f_{0}$. A detector placed on the plank detects the frequency.
The man jumps off with the same velocity (w.r.t. to ground) from point D and slides on the smooth horizontal surface. (Assume the coefficient of friction between the man and the horizontal is zero). The speed of sound in still medium is $330 \mathrm{ m/s}$. Answer the following questions based on the above situations.
$\frac{332}{324} f_{0}$
$\frac{330}{322} f_{0}$
$\frac{328}{336} f_{0}$
$\frac{328}{338} f_{0}$
To solve this problem, we need to determine the shift in frequency detected by a detector placed on the plank when the man, who is running on the plank and whistling, jumps off the plank. The phenomena at play here involve the conservation of momentum and the Doppler effect for sound waves.
Step-by-Step :
Initial Setup and Known Values:
Mass of the man, $ M_m = 50 \ \text{kg}$
Mass of the plank, $ M_p = 150 \ \text{kg}$
Velocity of the man relative to the plank, $ v_{mp} = 8 \ \text{m/s}$ (towards the left)
Speed of sound in still air, $ v_{sound} = 330 \ \text{m/s}$
Frequency of man’s whistle, $ f_0$
Velocity Calculations:
Since the system is initially at rest and there is no friction between the plank and the horizontal surface, we apply the conservation of momentum.
Let $ v_p $ be the velocity of the plank with respect to the ground, and $ v_m $ be the velocity of the man with respect to the ground.
Initial momentum of the system = Final momentum of the system
Therefore, $ M_m \cdot v_m = M_p \cdot v_p $
Given the velocity of the man with respect to the plank is 8 m/s: $$ v_m = v_p + 8 $$
Applying conservation of momentum: $$ 50 \cdot v_m = 150 \cdot v_p $$
Substitute $ v_m = v_p + 8 $: $$ 50 \cdot (v_p + 8) = 150 \cdot v_p $$ $$ 50v_p + 400 = 150v_p $$ $$ 100v_p = 400 $$ $$ v_p = 4 \ \text{m/s} $$
Therefore, $ v_m = v_p + 8 = 4 + 8 = 12 \ \text{m/s}$.
Doppler Effect Application:
When the man jumps off the plank, we need to determine the frequency detected by the detector on the plank.
For Doppler effect, the detected frequency $ f' $ is given by: $$ f' = f_0 \left( \frac{v_{sound} + v_{d}}{v_{sound} + v_{s}} \right) $$
Here ( v_d = 4 \ \text{m/s} ) (plank's velocity relative to the ground) and ( v_s = 12 \ \text{m/s} ) (man’s velocity relative to the ground).
Since the detector (plank) is moving towards the source (man is moving away): $$ f' = f_0 \left( \frac{330 + 4}{330 + 12} \right) $$ $$ f' = f_0 \left( \frac{334}{342} \right) $$ Simplify the fraction: $$ f' = f_0 \left( \frac{167}{171} \right) $$
Final Answer:
The frequency detected by the detector on the plank when the man jumps off is:
$$ \boxed{\frac{328}{336} f_0} $$
Two small metal balls of different mass $m_{1}$ and $m_{2}$ are connected by strings of equal length to a fixed point. When the balls are given charges, the angles that the two strings make with the vertical are 30° and 60°, respectively. The ratio $m_{1} / m_{2}$ is close to:
A) 1.7
B) 3.0
C) 0.58
D) 2.0
The correct option is A) 1.7.
Given:
Two small metal balls of different masses $m_1$ and $ m_2$
The balls are connected by strings of equal length to a fixed point
When the balls are charged, they make angles of 30° and 60° with the vertical, respectively
We need to find the ratio $\frac{m_1}{m_2} $.
From the equilibrium condition for the forces, we have:
For ball with mass $ m_2 $: $$ m_2 g \sin 60^\circ = F \cos 45^\circ$$
For ball with mass $ m_1 $: $$ m_1 g \sin 30^\circ = F \cos 45^\circ$$
By dividing these two equations, we get: $$ \frac{m_1 g \sin 30^\circ}{m_2 g \sin 60^\circ} = \frac{F \cos 45^\circ}{F \cos 45^\circ} $$
Simplifying further: $$ \frac{m_1 \sin 30^\circ}{m_2 \sin 60^\circ} = 1 $$
Now, substituting the trigonometric values: $$ \sin 30^\circ = \frac{1}{2} $$ $$\sin 60^\circ = \frac{\sqrt{3}}{2} $$
So, $$ \frac{m_1 \cdot \frac{1}{2}}{m_2 \cdot \frac{\sqrt{3}}{2}} = 1 $$ $$\frac{m_1}{m_2} = \sqrt{3} $$
Approximating $ \sqrt{3} \approx 1.732 $, we get the ratio: $$ \frac{m_1}{m_2} \approx 1.7 $$
Thus, the ratio $ \frac{m_1}{m_2} $ is approximately 1.7.
Four persons A, B, C and D initially at the corners of a square of side length d. If every person starts moving with the same speed $v$ such that each one faces the other always, the persons will meet after a time:
A $\frac{d}{v}$
B $\frac{\sqrt{2} d}{v}$
C $\frac{d}{2v}$
D $\frac{d}{\sqrt{2}v}$
To determine the time it takes for four persons ( A, B, C, ) and ( D ) to meet after starting at the corners of a square with side length ( d ), all moving at speed ( v ) and always facing each other, follow these steps:
Understand the initial setup:
Four persons start at the corners of a square of side length ( d ).
Each person moves towards the center of the square.
Motion towards the center:
Since they are consistently facing each other, each person's trajectory is a direct path towards the center of the square.
Calculate the distance:
The distance from any corner of the square to the center can be found by constructing a right triangle from the square's geometry:
The half-diagonal of the square forms the hypotenuse of this right triangle. If the side length of the square is ( d ), the distance from the corner to the center is: $$ \text{Distance} = \frac{d}{\sqrt{2}} $$
Determine the time:
Since all persons are moving at the same speed $ v $, they each take the same time to travel this distance.
Using the formula for time, $ t = \frac{\text{Distance}}{\text{Speed}} $: $$ t = \frac{\frac{d}{\sqrt{2}}}{v} = \frac{d}{\sqrt{2}v} $$
Therefore, the time required for all four persons to meet at the center of the square is $ \frac{d}{\sqrt{2}v} $.
Answer: The correct option is D $ \frac{d}{\sqrt{2}v} $.
An inclined plane is located at an angle of $a=53^{\circ}$ to the horizontal. There is a hole at point $\mathrm{B}$ in the inclined plane as shown in the figure. A particle is projected along the plane with speed $v_{0}$ at an angle $\beta=37^{\circ}$ to the horizontal in such a way so that it gets into the hole. Find the speed $v_{0}$ (in $\mathrm{m} / \mathrm{s}$), if $h=1 \mathrm{~m}, \mathrm{I}=8 \mathrm{~m}$ and $\sqrt{5} \approx 2.25$. (Neglect any type of friction)
The projectile's path is given by the equation:
$$ y = (\tan \theta) x + \frac{g' x^2}{2 u^2 \cos^2 \theta} $$
Here, the values are assigned as:
$y = h$
$x = l$
$\theta = \beta$
$u = v_0$
Additionally, $g'$ represents the component of gravitational acceleration $g$ acting along the inclined plane. In this case, $g' = g \sin a$.
Given these values, we substitute them into the equation:
$$ h = l \tan \beta - \frac{g \sin a \cdot l^2}{2 v_0^2 \cos^2 \beta} $$
First, solving for $v_0$, we rearrange the above equation to isolate $v_0$:
$$ v_0 = \sqrt{\frac{g \sin a \cdot l^2}{2 \cos^2 \beta (l \tan \beta - h)}} $$
Substitute the given values:
$a = 53^\circ \Rightarrow \sin 53^\circ \approx \frac{4}{5}$
$\beta = 37^\circ \Rightarrow \cos 37^\circ \approx \frac{4}{5}, \tan 37^\circ \approx \frac{3}{4}$
$g = 10 , \text{m/s}^2$
$h = 1 , \text{m}$
$l = 8 , \text{m}$
Substituting these values:
$$ v_0 = \sqrt{\frac{10 \times \frac{4}{5} \times 8^2}{2 \cos^2 37^\circ \left(8 \tan 37^\circ - 1\right)}} $$
Simplify the values:
$$ v_0 = \sqrt{\frac{10 \times \frac{4}{5} \times 64}{2 \times \left(\frac{4}{5}\right)^2 \left(8 \times \frac{3}{4} - 1\right)}} $$
Calculate further:
$$ v_0 = \sqrt{\frac{10 \times \frac{4}{5} \times 64}{2 \times \frac{16}{25} \left(6 - 1\right)}} $$
$$ v_0 = \sqrt{\frac{512}{\frac{32}{25} \times 5}} $$
$$ v_0 = \sqrt{\frac{512 \times 25}{160}} $$
$$ v_0 = \sqrt{\frac{12800}{160}} $$
$$ v_0 = \sqrt{80} $$
$$ v_0 \approx 9 , \text{m/s} $$
Thus, the speed $v_0$ required for the particle to reach the hole is 9 m/s.
The displacement ($x$) of a particle as a function of time ($t$) is given by:
$$ X = a \sin (bt + c) $$
Where $a$, $b$, and $c$ are constants of motion. Choose the correct statement(s) from the following:
A. The motion repeats itself in a time interval $\frac{2 \pi}{b}$.
B. The energy of the particle remains constant.
C. The velocity of the particle is zero at $x = \pm \alpha$.
D. The acceleration of the particle is zero at $x = \pm \alpha$.
The correct answer is Option A: The motion repeats itself in a time interval $\frac{2\pi}{b}$.
Reasoning:
The displacement of the particle as given by the equation $X = a \sin(bt + c)$ implies that the particle is undergoing simple harmonic motion. In simple harmonic motion, the displacement function is cyclic and repeats after a period.
To determine when the motion repeats itself, observe how the function behaves over time: $$ X(t) = a \sin(bt + c) $$ and when $t$ is increased by $\frac{2\pi}{b}$, $$ X\left(t + \frac{2\pi}{b}\right) = a \sin\left( b(t + \frac{2\pi}{b}) + c \right) = a \sin(bt + c + 2\pi) $$ Since $\sin(\theta + 2\pi) = \sin(\theta)$, we get: $$ X\left(t + \frac{2\pi}{b}\right) = a \sin(bt + c) = X(t) $$ Thus, the displacement $X$ at time $t + \frac{2\pi}{b}$ is the same as at time $t$, confirming that the motion indeed repeats every $\frac{2\pi}{b}$ seconds.
Option B: Since kinetic and potential energies transform into each other but the total mechanical energy remains constant.
Option C and D: Although velocity becomes zero at the extreme points of the simple harmonic motion (i.e., $x = \pm \alpha$ where $\alpha$ is amplitude $a$), the statements regarding acceleration are incorrect. Acceleration is in fact maximum at the extremes of the motion, directly proportional to the displacement from the mean position (i.e., $x = \pm a$) due to the restoring force being maximum at these points.
A man of mass $M$ having a bag of mass $m$ slips from the roof of a tall building of height $H$ and starts falling vertically (figure). When at a height $h$ from the ground, he notices that the ground below him is pretty hard, but there is a pond at a horizontal distance $x$ from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water.
(A) $\frac{2 m x \sqrt{g}}{M(\sqrt{2 H}-\sqrt{2(H-h)})}$
B $\frac{2 m x \sqrt{g}}{M(\sqrt{2 H}-\sqrt{2(H-h)}}$
C $\frac{M x \sqrt{g}}{m(\sqrt{2 H}-\sqrt{2(H-h)}}$
D $\frac{m x}{m(\sqrt{2 g H}-\sqrt{2 g(H-h)}}$
To determine the minimum horizontal velocity imparted to the bag so that the man lands in the water, we start by using the principle of conservation of linear momentum in the horizontal direction. There is no external force acting horizontally, so the total horizontal momentum remains conserved.
Given:
Mass of the man = $M$
Mass of the bag = $m$
Height of the building = $H$
Height from which the man notices the pond = $h$
Horizontal distance to the pond = $x$
Let:
The horizontal velocity imparted to the bag with respect to the ground be $V_b$ (towards the left).
The horizontal velocity of the man be $V_m$ (towards the right).
Applying the conservation of linear momentum in the horizontal direction:
$$ -mV_b + MV_m = 0 $$
Solving for $V_b$:
$$ V_b = \frac{M}{m} V_m $$
Now, we need to find the minimum horizontal velocity required at point P (height $h$ above the ground) to reach the pond located at $x$ units horizontally.
The time taken to fall from height $h$ to the ground is determined by solving the kinematic equation:
$$ \begin{aligned} &v^2 = u^2 + 2g(H-h) \ &v = \sqrt{2g(H-h)} \end{aligned} $$
Here, $v$ is the velocity when the man reaches the ground. Using the kinematic equation for displacement $s = ut + \frac{1}{2}at^2$, we set $u=0$:
$$ \begin{aligned} s &= h \ h &= \sqrt{2g(H-h)} \cdot t + \frac{1}{2}gt^2 \end{aligned} $$
Solving for time $t$:
$$ \frac{1}{2}g t^2 + \sqrt{2g(H-h)}t - h = 0 $$
Using the quadratic formula:
$$ \begin{aligned} t &= \frac{-\sqrt{2g(H-h)} \pm \sqrt{(\sqrt{2g(H-h)})^2 + 2gh}}{g} \ &= \frac{\sqrt{2g(H-h)} \pm \sqrt{2gH}}{g} \end{aligned} $$
Since time cannot be negative, we take the positive root:
$$ t = \frac{\sqrt{2gH} - \sqrt{2g(H-h)}}{g} $$
Using the relationship for speed, $V_m$:
$$ V_m = \frac{x}{t} = \frac{x g}{\sqrt{2gH} - \sqrt{2g(H-h)}} $$
Substituting this in the earlier equation for $V_b$:
$$ \begin{aligned} V_b &= \frac{M}{m} \cdot \frac{x g}{\sqrt{2gH} - \sqrt{2g(H-h)}} \ &= \frac{M x \sqrt{g}}{m\left(\sqrt{2H} - \sqrt{2(H-h)}\right)} \end{aligned}$$
Thus, the minimum horizontal velocity imparted to the bag so that the man lands in the water is:
$$ \boxed{\frac{M x \sqrt{g}}{m\left(\sqrt{2H} - \sqrt{2(H-h)}\right)}}$$
The correct option is (C).
The distance travelled by a body during the last second of its total flight is d when the body is projected vertically up with a certain velocity. If the velocity of projection is doubled, the distance travelled by the body during the last second of its total flight is:
A. $2 \mathrm{~d}$
B. $d$
C. $2d+\frac{g}{2}$
D. $2d-\frac{g}{2}$
To determine the distance traveled by a body during the last second of its total flight when the velocity of projection is doubled, we can use the following approach:
Distance Traveled during the Nth Second:
The distance traveled in the $n$th second is given by the formula: $$ S_n = u + \frac{a}{2} \left(2n - 1\right) $$
Analyzing the Initial Scenario:
When the body is projected upward with velocity $u$, the distance traveled during the last second of its total flight is given as $d$.
For the time of flight, if we assume the initial velocity $u$, the upward and downward journey will have total time $T = \frac{2u}{g}$.
When Velocity is Doubled:
If the initial velocity is doubled to $2u$, the new time of flight becomes $T' = \frac{2(2u)}{g} = \frac{4u}{g}$.
Distance Traveled During the Last Second with New Initial Velocity:
For the nth case when the initial velocity is $2u$, the distance traveled during the last second ($n = \frac{4u}{g}$) is given by: $$ S_{last} = 2u + \frac{a}{2} \left[ 2 \left(\frac{4u}{g}\right) - 1 \right] $$
Simplify the Expression:
When $u$ is doubled: $$ S_{last} = 2u + \frac{g}{2} \left[ 2 \left(\frac{4u}{g}\right) - 1 \right] $$
Expanding and simplifying, we get: $$ S_{last} = 2u + \frac{g}{2} \left( \frac{8u}{g} - 1 \right) $$
Further simplification results in: $$ S_{last} = 2u + 4u - \frac{g}{2} $$
Relating to Initial Distance d:
Given the original distance traveled during the last second is $d$, for the doubled velocity, the expression becomes: $$ S_{last} = 2d - \frac{g}{2} $$
Thus, the correct answer is: Option D:$$ \boxed{2d - \frac{g}{2}} $$
Two vectors - 3 units and 5 units - are acting at 60 degrees to each other. What is the magnitude and direction of the resultant?
When two vectors act at an angle to one another, the resultant vector can be determined using the law of cosines. Given the vectors of magnitudes 3 units and 5 units acting at an angle of 60 degrees, we can use the following formula to find the magnitude of the resultant vector:
$$ R = \sqrt{A^2 + B^2 + 2AB\cos\theta} $$
Here:
( A = 3 ) units
( B = 5 ) units
( \theta = 60^\circ )
Substituting these values into the equation:
$$ R = \sqrt{3^2 + 5^2 + 2 \cdot 3 \cdot 5 \cdot \cos 60^\circ} $$
We know ( \cos 60^\circ = 0.5 ), so:
$$ R = \sqrt{9 + 25 + 2 \cdot 3 \cdot 5 \cdot 0.5} $$ $$ R = \sqrt{9 + 25 + 15} $$ $$ R = \sqrt{49} $$ $$ R = 7 \text{ units} $$
Thus, the magnitude of the resultant vector is 7 units.
To find the direction of the resultant vector, we utilize the formula for the angle ( \phi ) that the resultant makes with the vector of magnitude 3 units:
$$ \tan \phi = \frac{B \sin \theta}{A + B \cos \theta} $$
Substituting the given values:
$$ \tan \phi = \frac{5 \sin 60^\circ}{3 + 5 \cos 60^\circ} $$ $$ \tan \phi = \frac{5 \cdot \sqrt{3}/2}{3 + 5 \cdot 0.5} $$ $$ \tan \phi = \frac{5 \sqrt{3}/2}{3 + 2.5} $$ $$ \tan \phi = \frac{5 \sqrt{3}/2}{5.5} $$ $$ \tan \phi = \frac{5 \sqrt{3}}{11} $$
Therefore, the angle ( \phi ) can be found by taking the arctan:
$$ \phi = \tan^{-1} \left(\frac{5 \sqrt{3}}{11}\right) $$
The exact value of ( \phi ) will provide the direction of the resultant vector.
In summary:
Magnitude of the resultant vector: 7 units
Direction: Angle ( \phi ) calculated from the given formula
At a certain height, a shell at rest explodes into two equal fragments. One of the fragments receives a horizontal velocity $u$. The time interval after which the velocity vectors will be inclined at 120° to each other is
A $\frac{u}{\sqrt{3} g}$
B $\frac{\sqrt{3} u}{g}$
C $\frac{2u}{\sqrt{3} g}$
D $\frac{u}{2\sqrt{3} g}$
To solve the problem where a shell at rest explodes into two equal fragments and one fragment receives a horizontal velocity $ u $, we need to find the time interval after which the velocity vectors of the fragments will be inclined at $ 120^\circ $ to each other.
Here’s a step-by-step explanation:
Initial Conditions:
The shell is initially at rest.
Upon explosion, fragment A gets a horizontal velocity $ \mathbf{u} = u\mathbf{i} $.
Fragment B must have an initial velocity $-u\mathbf{i}$ (opposite direction) to conserve momentum.
Velocity Components After Explosion:
The fragments also experience acceleration due to gravity in the ( y )-direction.
For fragment A: $$ \mathbf{v}_1 = u\mathbf{i} + g t \mathbf{j} $$
For fragment B: $$ \mathbf{v}_2 = -u\mathbf{i} + g t \mathbf{j} $$
Calculating the Angle Between Vectors:
The angle $ \theta $ between the two velocity vectors is given by the dot product formula: $$ \cos \theta = \frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{|\mathbf{v}_1| |\mathbf{v}_2|} $$
For $ \theta = 120^\circ $, $ \cos 120^\circ = -\frac{1}{2} $.
Dot Product of Velocities:
Calculate the dot product of $ \mathbf{v}_1 $ and $ \mathbf{v}_2 $: $$ \mathbf{v}_1 \cdot \mathbf{v}_2 = (u\mathbf{i} + g t \mathbf{j}) \cdot (-u\mathbf{i} + g t \mathbf{j}) $$ $$ = u(-u) + gt(gt) $$ $$ = -u^2 + g^2 t^2 $$
Magnitude of the Velocities:
The magnitudes are: $$ |\mathbf{v}_1| = \sqrt{u^2 + (gt)^2} $$ $$ |\mathbf{v}_2| = \sqrt{u^2 + (gt)^2} $$
Putting it all Together:
Using the cos formula: $$ \cos 120^\circ = -\frac{1}{2} = \frac{-u^2 + g^2 t^2}{u^2 + g^2 t^2} $$
Solving for ( t ): $$ -\frac{1}{2} = \frac{-u^2 + g^2 t^2}{u^2 + g^2 t^2} $$ $$ -\frac{1}{2}(u^2 + g^2 t^2) = -u^2 + g^2 t^2 $$ $$ -\frac{1}{2} u^2 - \frac{1}{2} g^2 t^2 = -u^2 + g^2 t^2 $$ $$ \frac{1}{2} u^2 = \frac{3}{2} g^2 t^2 $$ $$ u^2 = 3 g^2 t^2 $$ $$ t^2 = \frac{u^2}{3 g^2} $$ $$ t = \frac{u}{\sqrt{3} g} $$
Thus, the time interval after which the velocity vectors will be inclined at $ 120^\circ $ to each other is: $$ \boxed{\frac{u}{\sqrt{3} g}} $$
Therefore, the correct answer is (A) $\frac{u}{\sqrt{3} g}$.
The displacement $x$ of a particle varies with time according to the relation $x=\frac{a}{b}(1 - e^{-bt})$. Then select the false alternative
A. At $t = 1 / b$ the displacment of the particle in nearly $(2/3)(a/b)$
B. The velocity and acceleration of the particle at t = 0 and a and -ab respectively
C. The particle cannot reach a point at a distance $x'$ from its starting point if $x = \frac{a}{b}$
D. The particle will not come back to its starting point as $ t \rightarrow \infin$
The given problem states that the displacement $ x $ of a particle varies with time according to the relation:
$$ x = \frac{a}{b}(1 - e^{-bt}) $$
We need to evaluate the following statements to determine which one is false:
At $ t = \frac{1}{b} $, the displacement of the particle is nearly $\frac{2}{3} \left(\frac{a}{b}\right)$.
The velocity and acceleration of the particle at $t = 0 $ are ( a ) and $ -ab $, respectively.
The particle cannot reach a point at a distance $ \frac{a}{b} $ from the starting point.
The particle will not come back to its starting point when $ t$ tends to infinity.
Analysis of Each Statement
1. Displacement at $t = \frac{1}{b}$:
Given the formula:
$$ x = \frac{a}{b}(1 - e^{-bt}) $$
Substitute ($ t = \frac{1}{b} $:
$$ x = \frac{a}{b} \left( 1 - e^{-b \cdot \frac{1}{b}} \right)$$ $$ x = \frac{a}{b} \left( 1 - e^{-1} \right) $$
Since $ e^{-1} $ is approximately 0.3679, we have:
$$ x \approx \frac{a}{b} (1 - 0.3679) $$ $$ x \approx \frac{a}{b} (0.6321)$$ $$ x \approx 0.6321 \left(\frac{a}{b}\right) $$
This is roughly equal to $\frac{2}{3} \left(\frac{a}{b}\right)$. Therefore, the first statement is true.
2. Velocity and Acceleration at $ t = 0 $:
Velocity $( v )$ at $ t = 0 $
Velocity is the first derivative of displacement:
$$ v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{a}{b}(1 - e^{-bt}) \right) $$ $$ v = \frac{a}{b} \cdot b \cdot e^{-bt} $$ $$ v = a e^{-bt} $$
At $t = 0$:
$$v = a e^{0} = a $$
So, the velocity at $ t = 0 $ is ( a ).
Acceleration $( a )$at $t = 0 $
Acceleration is the derivative of velocity:
$$ a = \frac{dv}{dt} = \frac{d}{dt} (a e^{-bt}) $
$ a = a \cdot (-b) \cdot e^{-bt} ] [ a = -ab e^{-bt} $
At $ t = 0 $
$$ a = -ab e^{0} = -ab $$
So, the acceleration at $t = 0 $ is $ -ab $.
Therefore, the second statement is true.
3. Can the particle reach a distance $ \frac{a}{b} $:
As $t $approaches infinity:
$$ x = \frac{a}{b} \left( 1 - e^{-bt} \right) $$
For $ t \to \infty $:
$$ x = \frac{a}{b} \left( 1 - e^{-\infty} \right) $$
$$ x = \frac{a}{b} (1 - 0) ] [ x = \frac{a}{b} $$
Thus, the particle can reach a distance of $\frac{a}{b} $. Therefore, the third statement is true.
4. Will the particle return to the starting point $( x = 0 )$ as $ t \to \infty $:
Using the given displacement formula:
$$[ x = \frac{a}{b} (1 - e^{-bt}) $$
As $ t \to \infty $:
$$ x \to \frac{a}{b} (1 - 0) = \frac{a}{b} $$
The particle does not return to $x = 0 $; instead, it approaches $\frac{a}{b} $. Therefore, the fourth statement is false.
Conclusion
The false statement is:
The particle will not come back to its starting point when $ t $ tends to infinity.
So, the final answer is:
Final Answer: D
A ball is projected from the bottom of a tower and is found to go above the tower and is caught by the thrower at the bottom of the tower after a time interval $t_{1}$. An observer at the top of the tower see the same ball go up above him and then come back at this level in a time interval $t_{2}$. The height of the tower is:
A. $\frac{1}{2} g t_{1} t_{2}$
B. $\frac{g t_{1} t_{2}}{8}$
C. $\frac{g}{8}\left(t_{1}^{2}-t_{2}^{2}\right)$
D. $\frac{g}{2}\left(t_{1}-t_{2}\right)^{2}$
To determine the height of the tower based on the given conditions, we proceed as follows:
Time Analysis:
The total time the ball takes to travel up and down is denoted as $ t_1 $.
The time observed by the person at the top of the tower, as the ball goes above and returns to the top, is denoted as $ t_2 $.
Time to Reach Maximum Height:
The time taken for the ball to reach its maximum height from the bottom is $ \frac{t_1}{2} $.
The ball's initial vertical velocity ($ u $) can be calculated using the equation of motion: $ v = u + at $.
At maximum height, $ v = 0 $, $ a = -g $, and $ t = \frac{t_1}{2} $. Thus, we have: $$ 0 = u - g \frac{t_1}{2} $$ Solving for $ u $, we get: $$ u = \frac{g t_1}{2} $$
Using the Observer's Perspective:
The time taken by the ball to reach maximum height from the observer at the top is $ \frac{t_2}{2} $.
Time Difference:
The time taken by the ball to travel from the bottom to the top (height of the tower) is $ \frac{t_1 - t_2}{2} $.
Height Calculation:
Using the kinematic equation ( s = ut + \frac{1}{2} a t^2 ) where $ s = h $, $ u = \frac{g t_1}{2} $, $ t = \frac{t_1 - t_2}{2} $, and $ a = -g $, we get: $$ h = \left( \frac{g t_1}{2} \right) \left( \frac{t_1 - t_2}{2} \right) + \frac{1}{2} ( -g ) \left( \frac{t_1 - t_2}{2} \right)^2 $$
Simplifying the Equation:
Substitute the values and simplify the expression: $$ h = \frac{g t_1 (t_1 - t_2)}{4} - \frac{g (t_1 - t_2)^2}{8} $$
Further simplification leads to: $$ h = \frac{g}{8} \left( t_1^2 - t_2^2 \right) $$
Therefore, the height of the tower is given by: $$ \boxed{\frac{g}{8} \left( t_1^2 - t_2^2 \right)} $$
Thus, the correct answer is:
Option C: $\frac{g}{8}\left(t_{1}^{2}-t_{2}^{2}\right)$
At a certain height, a shell at rest explodes into two equal fragments. One of the fragments receives a horizontal velocity $u$. The time interval after which the velocity vectors will be inclined at $120^\circ$ to each other is
A. $\frac{u}{\sqrt{3}g}$
B. $\frac{\sqrt{3}u}{g}$
C. $\frac{2u}{\sqrt{3}g}$
D. $\frac{u}{2\sqrt{3} g}$
To solve the problem of determining the time interval after which the velocity vectors of two fragments will be inclined at an angle of $120^{\circ}$ to each other, we can use the principles of conservation of momentum and kinematics.
Initial Setup:
A shell at rest explodes into two equal fragments.
One fragment receives a horizontal velocity $u$.
Let fragment 1 receive velocity $\vec{v}_1 = u\hat{i}$.
Since there are no external forces, momentum is conserved. Hence, the other fragment must receive an equal and opposite momentum: $\vec{v}_2 = -u\hat{i}$.
Velocity Components After Time $t$:The fragments gain velocity due to gravitational acceleration $g$. The vertical component of the velocity for both fragments will be $gt$ after time $t$:
Velocity of fragment 1 after time $t$: $$ \vec{v}_1 = u\hat{i} + gt\hat{j} $$
Velocity of fragment 2 after time $t$: $$ \vec{v}_2 = -u\hat{i} + gt\hat{j} $$
Finding the Time Interval Using the Angle Between the Velocity Vectors:We know that the angle between the two velocity vectors should be $120^\circ$. Therefore, the cosine of the angle between $\vec{v}_1$ and $\vec{v}_2$ is: $$ \cos(120^\circ) = -\frac{1}{2} $$
Using the formula for the dot product: $$ \vec{v}_1 \cdot \vec{v}_2 = |\vec{v}_1||\vec{v}_2|\cos(\theta) $$
Plugging in the values, we get: $$ (u\hat{i} + gt\hat{j}) \cdot (-u\hat{i} + gt\hat{j}) = |\vec{v}_1| |\vec{v}_2| \cos(120^\circ) $$ Simplifying the dot product: $$ -u^2 + g^2t^2 = |\vec{v}_1||\vec{v}_2| \left( -\frac{1}{2} \right) $$
Calculating Magnitudes:The magnitudes of $\vec{v}_1$ and $\vec{v}_2$ are: $$ |\vec{v}_1| = \sqrt{u^2 + g^2 t^2} $$ $$ |\vec{v}_2| = \sqrt{u^2 + g^2 t^2} $$
Substitute and Solve for $t$:Substituting back, we get: $$ -u^2 + g^2 t^2 = \left( \sqrt{u^2 + g^2 t^2} \right) \left( \sqrt{u^2 + g^2 t^2} \right) \left( -\frac{1}{2} \right) $$ Simplifying further, we get: $$ -u^2 + g^2 t^2 = -\frac{1}{2}(u^2 + g^2 t^2) $$ Rearranging the terms: $$ -u^2 + g^2 t^2 = -\frac{1}{2} u^2 - \frac{1}{2} g^2 t^2 $$ Multiplying both sides by $2$: $$ -2u^2 + 2g^2 t^2 = -u^2 - g^2 t^2 $$ Bringing like terms together: $$ u^2 = 3g^2 t^2 $$
Solving for $t$:$$ t^2 = \frac{u^2}{3g^2} $$ $$ t = \frac{u}{\sqrt{3}g} $$
Thus, the time interval after which the velocity vectors will be inclined at $120^\circ$ to each other is:
$$ \boxed{\frac{u}{\sqrt{3} g}} $$
For a body traveling with uniform acceleration, its final velocity is $v = \sqrt{180 - 7x}$, where $x$ is the distance traveled by the body. Then the acceleration is:
A. $-8 , \mathrm{m/s}^{2}$
B. $-3.5 , \mathrm{m/s}^{2}$
C. $-7 , \mathrm{m/s}^{2}$
D. $180 , \mathrm{m/s}^{2}$
To determine the acceleration of a body moving with uniform acceleration given the final velocity formula $v = \sqrt{180 - 7x}$, follow these steps:
Square both sides of the velocity equation to remove the square root: $$ v^2 = 180 - 7x $$
Differentiate both sides of the equation with respect to $x$. Using the chain rule, note that $\frac{d}{dx}(v^2) = 2v \frac{dv}{dx}$: $$ \frac{d}{dx}(v^2) = \frac{d}{dx}(180 - 7x) $$ This simplifies to: $$ 2v \frac{dv}{dx} = -7 $$
Solving for $\frac{dv}{dx}$, we get: $$ v \frac{dv}{dx} = \frac{-7}{2} $$
Relate this to acceleration. Since acceleration $a$ is defined as $a = v \frac{dv}{dx}$: $$ a = \frac{-7}{2} $$
Therefore, the acceleration is: $$ a = -3.5 , \mathrm{m/s}^2 $$
The correct option that matches this calculation is:
B. $-3.5 , \mathrm{m/s}^2$
This provides the acceleration of the body, indicating a deceleration due to the negative sign.
A body of mass $1 \mathrm{~kg}$ is projected at an angle $30^{\circ}$ with horizontal on a level ground at a speed $50 \mathrm{~m} / \mathrm{s}$. The magnitude of change in momentum of the body during its flight is $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$:
A $50 \mathrm{~kg} \mathrm{~ms}^{-1}$
B $100 \mathrm{~kg} \mathrm{~ms}^{-1}$
C $25 \mathrm{~kg} \mathrm{~ms}^{-1}$
D Zero
The correct option is A ( 50 , \text{kg} , \text{m/s} ).
To understand the change in momentum, consider the initial and final velocities of the body in both the horizontal and vertical directions.
Initial velocity components: $$ v_{x} = u \cos \theta = 50 \cos 30^\circ = 50 \times \frac{\sqrt{3}}{2} $$ $$ v_{y} = u \sin \theta = 50 \sin 30^\circ = 50 \times \frac{1}{2} = 25 , \text{m/s} $$
Due to the projectile motion on a level ground and symmetry, the final horizontal velocity component remains the same while the vertical component changes direction (downward) but with the same magnitude:
Final velocity components: $$ v_{x} = u \cos \theta = 50 \cos 30^\circ $$ $$ v_{y} = -50 \sin 30^\circ = -25 , \text{m/s} $$
The initial and final momentum vectors are thus: $$ \vec{p_{i}} = m u \cos \theta \hat{i} + m u \sin \theta \hat{j} $$ $$ \vec{p_{f}} = m u \cos \theta \hat{i} - m u \sin \theta \hat{j} $$
Calculating the magnitude of the change in momentum: $$ |\Delta \vec{p}| = |\vec{p_{f}} - \vec{p_{i}}| = 2 m u \sin \theta $$ $$ |\Delta \vec{p}| = 2 \times 1 \times 50 \times \frac{1}{2} = 50 , \text{kg} , \text{m/s} $$
Alternate :
In the horizontal direction: $$ \Delta p_{x} = 0 $$
In the vertical direction: $$ \Delta p_{y} = m u \sin \theta - (-m u \sin \theta) = 2 m u \sin \theta $$ $$ = 2 \times 1 \times 50 \times \frac{1}{2} = 50 , \text{kg} , \text{m/s} $$
Therefore, the change in momentum during its flight is 50 kg m/s.
A plane progressive wave is represented by the equation $Y = 0.1 \sin (200 \pi t - \frac{20 \pi}{17})$ where $y$ is displacement in meters, $t$ in seconds, and $x$ is distance from a fixed origin in meters. The frequency, wavelength, and speed of the wave respectively are:
A) 100 Hz, 1.7 m, 170 m/s
B) 150 Hz, 2.4 m, 200 m/s
C) 80 Hz, 1.1 m, 90 m/s
D) 120 Hz, 1.25 m, 207 m/s
The correct option is A: 100 Hz, 1.7 m, 170 m/s.
To find the frequency, wavelength, and speed, we compare the given wave equation:
[ Y = 0.1 \sin \left( 200 \pi t - \frac{20 \pi}{17} x \right) ]
with the standard form of a wave equation:
[ Y = A \sin (\omega t - kx) ]
From the given equation, we identify:
Angular frequency ($\omega$): [ \omega = 200 \pi ] Using the relationship $\omega = 2 \pi f$ (where $f$ is the frequency), we solve for $f$: [ 200 \pi = 2 \pi f ] [ f = \frac{200 \pi}{2 \pi} = 100 \text{ Hz} ]
Wave number ($k$): [ k = \frac{20 \pi}{17} ] Wavelength ($\lambda$) is related to $k$ by: [ \lambda = \frac{2 \pi}{k} ] Substituting the value of $k$: [ \lambda = \frac{2 \pi}{\frac{20 \pi}{17}} = \frac{2 \pi \times 17}{20 \pi} = \frac{34 \pi}{20 \pi} = 1.7 \text{ m} ]
Wave speed ($v$): Wave speed is given by the relation: [ v = \frac{\omega}{k} ] Substituting the values of $\omega$ and $k$: [ v = \frac{200 \pi}{\frac{20 \pi}{17}} ] Simplifying: [ v = 200 \pi \times \frac{17}{20 \pi} = 10 \times 17 = 170 \text{ m/s} ]
Thus, the frequency, wavelength, and speed of the wave are 100 Hz, 1.7 m, and 170 m/s, respectively.
A man wants to reach point B on the opposite bank of a river flowing at a speed as shown in the figure. What minimum speed relative to water should the man have so that he can reach point B? In which direction should he swim?
A $2 \mathrm{u}$
B $u / 2$
C $\frac{u}{\sqrt{2}}$
D. $\frac{3u}{\sqrt{2}}$
The correct option is C: $$ \frac{u}{\sqrt{2}} $$
Let ( v ) be the speed of the boatman in still water. The resultant velocity of ( v ) and the river's speed ( u ) should be directed along ( AB ).
The components of the absolute velocity of the boatman ( \vec{v}_b ) along the ( x )- and ( y )-directions are: $$ v_x = u - v \sin \theta \quad \text{and} \quad v_y = v \cos \theta $$
Given that ( \theta = 45^\circ ), we apply: $$ \tan 45^\circ = \frac{v_y}{v_x} $$ which simplifies to: $$ 1 = \frac{v \cos \theta}{u - v \sin \theta} $$ Rearranging terms, we get: $$ v = \frac{u}{\sin \theta + \cos \theta} = \frac{u}{\sqrt{2} \sin (\theta + 45^\circ)} $$
To minimize ( v ), set: $$ \theta + 45^\circ = 90^\circ \Rightarrow \theta = 45^\circ $$ and the minimal speed is: $$ v_{\text{min}} = \frac{u}{\sqrt{2}} $$
Thus, the minimum speed the man should have relative to the water to reach point ( B ) is ( \frac{u}{\sqrt{2}} ).
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Ask Chatterbot AINCERT Solutions - Motion in a Plane | NCERT | Physics | Class 11
State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Scalars | Description |
---|---|
Volume | A quantity with magnitude only. |
Mass | A quantity with magnitude only. |
Speed | A quantity with magnitude only. |
Density | A quantity with magnitude only. |
Number of moles | A quantity with magnitude only. |
Angular frequency | A quantity with magnitude only. |
Vectors | Description |
---|---|
Acceleration | A quantity with both magnitude and direction. |
Velocity | A quantity with both magnitude and direction. |
Displacement | A quantity with both magnitude and direction. |
Angular velocity | A quantity with both magnitude and direction. |
Pick out the two scalar quantities in the following list :
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
The two scalar quantities in the provided list are:
Work - Work is a scalar quantity as it only has magnitude and no direction.
Current - Electric current is also a scalar quantity as it has magnitude and direction is given as the flow of positive charges.
The rest of the quantities listed are vectors.
Pick out the only vector quantity in the following list :
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.
The only vector quantity in the list is impulse.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful :
(a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.
(a) Adding any two scalars:
This operation is meaningful. Scalars can always be added using ordinary algebra. For instance, adding two masses or two distances is perfectly valid.
(b) Adding a scalar to a vector of the same dimensions:
This operation is not meaningful. Scalars and vectors are different quantities. While a scalar has only magnitude, a vector has both magnitude and direction. Thus, adding them isn't mathematically valid.
(c) Multiplying any vector by any scalar:
This operation is meaningful. Multiplying a vector by a scalar scales the vector's magnitude without affecting its direction (unless the scalar is negative, in which case the direction reverses).
(d) Multiplying any two scalars:
This operation is meaningful. Scalars can always be multiplied by ordinary algebra. For example, multiplying speed by time gives distance.
(e) Adding any two vectors:
This operation is meaningful. Vectors can be added using vector addition rules, such as the triangle law or parallelogram law of addition.
(f) Adding a component of a vector to the same vector:
This operation is not meaningful. A component of a vector is simply one part of the total vector along a given axis. Adding a vector component to the entire vector itself does not conform to any valid vector operation rules.
To highlight the important aspects:
Adding scalars and multiplying scalars are valid operations.
Multiplying a vector by a scalar and adding vectors are valid operations.
Adding a scalar to a vector and adding a vector component to the vector are not valid operations.
Read each statement below carefully and state with reasons, if it is true or false :
(a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector.
(a) The magnitude of a vector is always a scalar
True. The magnitude of a vector represents its size or length and does not have direction, which classifies it as a scalar quantity.
(b) Each component of a vector is always a scalar
True. The components of a vector along each axis (e.g., (A_x), (A_y), and (A_z)) are scalars because they represent the magnitude of the vector in each direction without any direction themselves.
(c) The total path length is always equal to the magnitude of the displacement vector of a particle
False. The total path length (or distance traveled) is generally greater than or equal to the magnitude of the displacement vector unless the path is a straight line in one direction. The displacement vector only measures the shortest path between the initial and final points.
(d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time
True. The average speed is defined as the total path length divided by the time taken, whereas the magnitude of average velocity is the magnitude of the displacement divided by the time. Since the total path length is always greater than or equal to the magnitude of the displacement, the average speed is greater than or equal to the magnitude of average velocity.
(e) Three vectors not lying in a plane can never add up to give a null vector
True. If three vectors do not lie in the same plane (i.e., they are not coplanar), they cannot form a closed shape (triangle) and hence cannot add up to form a null vector (zero vector). The resultant of these vectors will not be zero.
Establish the following vector inequalities geometrically or otherwise :
(a) $|\mathbf{a}+\mathbf{b}| \leq|\mathbf{a}|+|\mathbf{b}|$
(b) $|\mathbf{a}+\mathbf{b}| \geq|| \mathbf{a}|-| \mathbf{b}||$
(c) $|\mathbf{a}-\mathbf{b}| \leq|\mathbf{a}|+|\mathbf{b}|$
(d) $|\mathbf{a}-\mathbf{b}| \geq|| \mathbf{a}|-| \mathbf{b}||$
When does the equality sign above apply?
(a) $ |\mathbf{a} + \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}| $
This inequality states that the magnitude of the sum of two vectors is less than or equal to the sum of their magnitudes.
Interpretation: This is a direct application of the Triangle Inequality in vector addition. Geometrically, it signifies that the length of one side of a triangle (formed by vectors) is less than or equal to the sum of the lengths of the other two sides.
Equality Condition: The equality holds when vectors $ \mathbf{a} $ and $ \mathbf{b} $ are collinear and point in the same direction.
(b) $ |\mathbf{a} + \mathbf{b}| \geq ||\mathbf{a}| - |\mathbf{b}|| $
This inequality states that the magnitude of the sum of two vectors is greater than or equal to the absolute difference of their magnitudes.
Interpretation: This can be derived using the properties of magnitudes and considering the cross product and dot product of vectors. It implies that the shortest distance between the tip of vector $\mathbf{a} $ and vector $ \mathbf{b}$ (when placed at the same tail) is at least the absolute difference of their magnitudes.
Equality Condition: The equality holds when vectors $ \mathbf{a} $ and $\mathbf{b} $ are collinear and point in the opposite directions.
(c) $ |\mathbf{a} - \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}| $
This inequality is similar to (a), where the difference of two vectors is considered.
Interpretation: This also follows the Triangle Inequality. The geometric interpretation here is that the length of one side of a triangle formed by vectors $ \mathbf{a} $ and $-\mathbf{b}$ is less than or equal to the sum of the lengths of $ \mathbf{a} $ and $ \mathbf{b}$.
Equality Condition: The equality condition for this inequality remains similar to that of (a); it holds when vectors$ \mathbf{a}$ and $ \mathbf{b}$ are collinear and point in the same direction.
(d) $ |\mathbf{a} - \mathbf{b}| \geq ||\mathbf{a}| - |\mathbf{b}|| $
This inequality is similar to (b), considering the difference of two vectors.
Interpretation: This can be inferred from considering vector subtraction and properties of magnitudes. The shortest distance represented by $ \mathbf{a} - \mathbf{b} $ will be greater than or equal to the absolute difference in their magnitudes.
Equality Condition: The equality holds when vectors $ \mathbf{a} $ and $ \mathbf{b}$ are collinear and point in the opposite directions.
Summary of Equality Conditions
For inequalities (a) and (c), the equality holds when vectors are collinear and point in the same direction.
For inequalities (b) and (d), the equality holds when vectors are collinear and point in opposite directions.
These explanations nicely encapsulate the geometric interpretations and equality conditions for these vector inequalities.
Given $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}$, which of the following statements are correct :
(a) $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and $\mathbf{d}$ must each be a null vector,
(b) The magnitude of $(\mathbf{a}+\mathbf{c})$ equals the magnitude of $(b+d)$,
(c) The magnitude of a can never be greater than the sum of the magnitudes of $\mathbf{b}, \mathbf{c}$, and $\mathbf{d}$,
(d) $\mathbf{b}+\mathbf{c}$ must lie in the plane of $\mathbf{a}$ and $\mathbf{d}$ if $\mathbf{a}$ and $\mathbf{d}$ are not collinear, and in the line of $\mathbf{a}$ and $\mathbf{d}$, if they are collinear ?
Let's examine each statement one by one based on the given equation $\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}$:
(a) $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and $\mathbf{d}$ must each be a null vector:
This statement is incorrect. The equation $\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}$only implies that these vectors sum to zero. It does not necessitate that each vector must be null.
(b) The magnitude of $(\mathbf{a} + \mathbf{c})$ equals the magnitude of $(\mathbf{b} + \mathbf{d})$:
This statement is also incorrect. The magnitudes of $(\mathbf{a} + \mathbf{c})$ and $(\mathbf{b} + \mathbf{d})$ do not necessarily have to be equal since the vectors could have different directions and magnitudes.
(c) The magnitude of $\mathbf{a}$ can never be greater than the sum of the magnitudes of $\mathbf{b}, \mathbf{c}$, and $\mathbf{d}$:
This statement is correct. By the triangle inequality, the magnitude of $\mathbf{a}$ cannot exceed the sum of the magnitudes of $\mathbf{b}, \mathbf{c}$, and $\mathbf{d}$.
(d) $\mathbf{b} + \mathbf{c}$ must lie in the plane of $\mathbf{a}$ and $\mathbf{d}$ if $\mathbf{a})$and $\mathbf{d}$ are not collinear, and in the line of $\mathbf{a}$ and $\mathbf{d}$, if they are collinear:
This statement is correct. Since $\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0}$, rearranging gives $\mathbf{b} + \mathbf{c} = -(\mathbf{a} + \mathbf{d})$. Therefore, $\mathbf{b} + \mathbf{c}$ lies in the plane of $\mathbf{a} $ and $\mathbf{d}$. If $\mathbf{a}$ and $\mathbf{d}$ are collinear, then $\mathbf{b} + \mathbf{c}$ must also be collinear with $\mathbf{a}$ and $\mathbf{d}$.
Summary:
Incorrect: (a), (b)
Correct: (c), (d)
Three girls skating on a circular ice ground of radius $200 \mathrm{~m}$ start from a point $P$ on the edge of the ground and reach a point $\mathrm{Q}$ diametrically opposite to $P$ following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate?
The displacement vector $\mathbf{PQ}$ for each of the three girls is the straight line joining points $P$ and $Q$. Since $Q$ is diametrically opposite to $P$ on the circular ground, the displacement is the diameter of the circle.
Let's calculate the magnitude of the displacement vector:
The radius of the circle is $\mathbf{R} = 200 \text{ m}$.
The diameter $\mathbf{D}$ of the circle is $2 \mathbf{R}$.
Therefore, the magnitude of the displacement vector for each girl is:
$$ |\mathbf{PQ}| = \mathbf{D} = 2 \times 200 \text{ m} = 400 \text{ m} $$
Path Length Comparison
Girl A (path A): The path is a curved path to the diametrically opposite point. The length of path A is greater than 400 m.
Girl B (path B): The path is also a curved path to the diametrically opposite point. The length of path B is greater than 400 m.
Girl C: The path is a straight line directly to point Q, which is the shortest possible path. The length of path C is exactly 400 m.
Conclusion
The magnitude of the displacement vector for each of the three girls is 400 m.
The actual length of the path skate is equal to the displacement vector for Girl C.
A cyclist starts from the centre $O$ of a circular park of radius $1 \mathrm{~km}$, reaches the edge $P$ of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?
(a) Net Displacement
Net displacement refers to the straight line distance between the initial and final positions of the cyclist.
The cyclist starts from the center O and returns to the center O.
Therefore, the net displacement is 0 km.
(b) Average Velocity
Average velocity is the total displacement divided by the total time taken.
Since the net displacement is 0 km, the average velocity is (\frac{0 \text{ km}}{10 \text{ min}} = 0 \text{ km/min}).
Hence, the cyclist's average velocity is 0 km/min.
(c) Average Speed
Average speed is the total distance traveled divided by the total time taken.
The total journey can be broken into three segments:
From O to P (radius of the circle): $r = 1 \text{ km}$
From P to Q (half the circumference of the circle):$\frac{1}{2} \times 2 \pi r = \pi \text{ km}$
From Q back to O (radius of the circle): $r = 1 \text{ km}$
Total distance traveled, $d_{\text{total}} = r + \pi r + r = 1 \text{ km} + \pi \text{ km} + 1 \text{ km} = (2 + \pi) \text{ km}$
The total time taken is 10 minutes, which is $\frac{10}{60} = \frac{1}{6} \text{ hours}$.
Average speed, $v_{\text{avg}} = \frac{d_{\text{total}}}{\text{total time}} = \frac{2 + \pi}{\frac{1}{6}} \text{ km/hr} = 6(2 + \pi) \text{ km/hr}$.
Using the approximate value (\pi \approx 3.14), we get: $$ v_{\text{avg}} = 6(2 + 3.14) \text{ km/hr} = 6 \times 5.14 \text{ km/hr} = 30.84 \text{ km/hr} $$
The average speed of the cyclist is 30.84 km/hr.
Summary
Net Displacement: 0 km
Average Velocity: 0 km/min
Average Speed: 30.84 km/hr
On an open ground, a motorist follows a track that turns to his left by an angle of $60^{\circ}$ after every $500 \mathrm{~m}$. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
To solve this problem, we'll keep track of the motorist's position after each turn using vector addition, considering the changes in direction and distances travelled. We will then compare the displacement of the motorist from the starting point to the path length covered.
1. Displacement after the third turn
Each turn involves moving $500 , \text{m}$ and turning $60^{\circ}$ to the left.
After the first turn: $\mathbf{r}_1 = 500 , \text{m} , \hat{i}$
After the second turn: $\mathbf{r}_2 = 500 , \text{m} , (\cos 60^{\circ} , \hat{i} + \sin 60^{\circ} , \hat{j})$
After the third turn: $\mathbf{r}_3 = 500 , \text{m} , (\cos 120^{\circ} , \hat{i} + \sin 120^{\circ} , \hat{j})$
Now summing these vectors:
$$ \mathbf{R_3} = 500\hat{i} + 500 (\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) + 500 (\cos 120^{\circ} \hat{i} + \sin 120^{\circ} \hat{j})$$
Using the values $\cos 60^{\circ} = \frac{1}{2}$, $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$, $\cos 120^{\circ} = -\frac{1}{2}$, and $\sin 120^{\circ} = \frac{\sqrt{3}}{2}$:
$$ \mathbf{R_3} = 500\hat{i} + 500 \left( \frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \right) + 500 \left( -\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \right) $$
Simplifying:
$$ \mathbf{R_3} = 500\hat{i} + 250\hat{i} + 250 \sqrt{3}\hat{j} - 250\hat{i} + 250 \sqrt{3}\hat{j} $$
$$ \mathbf{R_3} = 500\hat{i} + 250\sqrt{3}\hat{j} + 250\sqrt{3}\hat{j} $$
$$ \mathbf{R_3} = 500\hat{i} + 500\sqrt{3}\hat{j} $$
The magnitude of the displacement is:
$$ |\mathbf{R_3}| = \sqrt{(500)^2 + (500\sqrt{3})^2} $$
$$ |\mathbf{R_3}| = 500 \sqrt{1 + 3} $$
$$ |\mathbf{R_3}| = 500 \times 2 $$
$$ |\mathbf{R_3}| = 1000 , \text{m} $$
The total path length covered by the motorist after the third turn is:
$$ \text{Path length} = 3 \times 500 = 1500 , \text{m} $$
Comparison:
$\text{Displacement} = 1000 , \text{m}$
$\text{Path length} = 1500 , \text{m}$
2. Displacement after the sixth turn
After the sixth turn, the motorist completes a hexagon, which will bring him back to the starting point due to symmetry in a hexagonal path.
Comparison:
$\text{Displacement} = 0 , \text{m}$
$\text{Path length} = 6 \times 500 = 3000 , \text{m}$
3. Displacement after the eighth turn
We need to follow a similar vector approach considering he has already completed a hexagon after 6 turns:
Positions after 6 turns bring him back to origin $( \mathbf{r}_6 = 0 )$
7th turn: $ +500\hat{i} $
8th turn: $ +500 (\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) $
Summing these vectors:
$$ \mathbf{R_8} = 500\hat{i} + 500 \left(\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j} \right) $$
Simplifying:
$$ \mathbf{R_8} = 500\hat{i} + 250\hat{i} + 250\sqrt{3}\hat{j} $$
$$ \mathbf{R_8} = 750\hat{i} + 250\sqrt{3}\hat{j} $$
The magnitude of the displacement is:
$$ |\mathbf{R_8}| = \sqrt{(750)^2 + (250\sqrt{3})^2} $$
$$ |\mathbf{R_8}| = 750 \sqrt{1 + \left(\frac{\sqrt{3}}{3}\right)^2} $$
$$ |\mathbf{R_8}| = 750 \sqrt{1 + 1} $$
$$ |\mathbf{R_8}| = 750 \times \sqrt{2} $$
$$ |\mathbf{R_8}| = 883 , \text{m} $$
The total path length covered by the motorist after the eighth turn is:
$$ \text{Path length} = 8 \times 500 = 4000 , \text{m} $$
Comparison:
$\text{Displacement} = 883 , \text{m}$
$\text{Path length} = 4000 , \text{m}$
By comparing the magnitude of the displacements with the total path lengths:
After the third turn: Displacement $= 1000 , \text{m}$, Path length $= 1500 , \text{m}$
After the sixth turn: Displacement $= 0 , \text{m}$, Path length $= 3000 , \text{m}$
After the eighth turn: Displacement $= 883 , \text{m}$, Path length $= 4000 , \text{m}$
In all cases the displacement is less than the total path length except at the sixth turn where the motorist returns to the starting point (displacement is zero).
A passenger arriving in a new town wishes to go from the station to a hotel located $10 \mathrm{~km}$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23 \mathrm{~km}$ long and reaches the hotel in $28 \mathrm{~min}$. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?
(a) Average Speed
Using the calculation performed:
$$ \text{Average speed} = \frac{23 , \text{km}}{\frac{28}{60} , \text{hours}} = \frac{23}{\frac{28}{60}} = 49.29 , \text{km/h} $$
(b) Magnitude of Average Velocity
Using the calculation performed:
$$ \text{Magnitude of average velocity} = \frac{10 , \text{km}}{\frac{28}{60} , \text{hours}} = \frac{10}{\frac{28}{60}} = 21.43 , \text{km/h} $$
Are the two equal?
No, the average speed and magnitude of average velocity are not equal. The average speed is 49.29 km/h, whereas the magnitude of the average velocity is 21.43 km/h.
This difference arises because the total distance traveled (23 km) is greater than the displacement (10 km), indicating the circuitous path taken by the cabman.
The ceiling of a long hall is $25 \mathrm{~m}$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40 \mathrm{~m} \mathrm{~s}^{-1}$ can go without hitting the ceiling of the hall ?
The angle $\theta$ at which the maximum height is $25 \ \mathrm{m}$ is approximately $0.586 \ \text{radians}$.
Now, we need to find the value of $\cos \theta$: $$ \cos(0.586) \approx 0.833$$
We can use $\sin 2\theta$:
$$ \sin 2\theta = 2 \sin \theta \cos \theta $$
$$ \sin 2\theta = 2 \cdot 0.553 \cdot 0.833 \approx 0.923$$
Next, we calculate the horizontal range (R): $$ R = \frac{(v_{0}^2 \sin 2\theta)}{g} $$ $$ R = \frac{(40)^2 \cdot 0.923}{9.8} $$ $$R \approx \frac{1600 \cdot 0.923}{9.8} $$ $$ R \approx \frac{1476.8}{9.8}$$ $$ R \approx 150.69 \ \mathrm{m}$$
Therefore, the maximum horizontal distance that the ball can go without hitting the ceiling of the hall is approximately 150.69 meters.
A cricketer can throw a ball to a maximum horizontal distance of $100 \mathrm{~m}$. How much high above the ground can the cricketer throw the same ball ?
To solve for $v_0 $, let's break down the calculations step-by-step:
Use the equation for range $ R $: $$ R = \frac{v_0^2 \sin 2\theta}{g} $$
Since $ \theta = 45^\circ $: $$ \sin 2\theta = \sin 90^\circ = 1 $$
Plug in the values $ R = 100 , \text{m}, g = 9.8 , \text{m/s}^2 $: $$ 100 = \frac{v_0^2 \cdot 1}{9.8} $$ $$v_0^2 = 100 \cdot 9.8 ] [ v_0^2 = 980 $$ $$ v_0 = \sqrt{980} ] [ v_0 \approx 31.3 , \text{m/s} $$
Next, use the equation for maximum height ( h ): $$ h = \frac{v_0^2 \sin^2 \theta}{2g} $$ $$ \sin \theta = \sin 45^\circ = \frac{\sqrt{2}}{2}$$ $$ \sin^2 \theta = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{2}{4} = \frac{1}{2} $$
Substitute ( v_0 \approx 31.3 , \text{m/s} ) into the height equation: $$ h = \frac{(31.3)^2 \cdot \frac{1}{2}}{2 \cdot 9.8} $$ $$h = \frac{980 \cdot \frac{1}{2}}{19.6} $$ $$ h = \frac{490}{19.6} $$ $$ h \approx 25 , \text{m} $$
Therefore, the maximum height above the ground that the cricketer can throw the ball is approximately 25 meters.
A stone tied to the end of a string $80 \mathrm{~cm}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in $25 \mathrm{~s}$, what is the magnitude and direction of acceleration of the stone ?
1. Angular Speed $\omega$:
$$ \omega = \frac{2\pi \times 14}{25} \approx 3.52 \ \mathrm{rad/s} $$
2. Centripetal Acceleration $ a_c $:
$$ a_c = \omega^2 \times R $$ $$ a_c \approx (3.52)^2 \times 0.8 $$ $$ a_c \approx 9.90 \ \mathrm{m/s^2} $$
Direction of Acceleration:
The direction of the centripetal acceleration is towards the center of the circle.
Therefore, the magnitude of the acceleration of the stone is approximately 9.90 $\mathrm{m/s^2} $ and its direction is towards the center of the circle.
An aircraft executes a horizontal loop of radius $1.00 \mathrm{~km}$ with a steady speed of 900 $\mathrm{km} / \mathrm{h}$. Compare its centripetal acceleration with the acceleration due to gravity.
The centripetal acceleration of the aircraft, calculated using the formula $ a_c = \frac{v^2}{r} $, is 62.5 m/s².
Comparing this with the acceleration due to gravity $( g = 9.8 ; \text{m/s}^2 )$:
$$ \frac{a_c}{g} = \frac{62.5 ; \text{m/s}^2}{9.8 ; \text{m/s}^2} \approx 6.37 $$
Thus, the centripetal acceleration is approximately 6.37 times greater than the acceleration due to gravity.
Read each statement below carefully and state, with reasons, if it is true or false :
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre
Answer: True
Reason: For uniform circular motion, the acceleration experienced by the particle is the centripetal acceleration, which always points towards the centre of the circle. This acceleration is directed along the radius of the circle towards the centre. Therefore, the statement is true.
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
Answer: True
Reason: As described in the chapter, the velocity vector of a particle at any point on its path is tangent to its trajectory at that point. This is true for all types of motion, including circular motion and projectile motion. Therefore, the statement is true.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector
Answer: True
Reason: In uniform circular motion, the direction of the centripetal acceleration changes continuously pointing towards the center, but its magnitude remains constant. Over one complete cycle, the vector sum of all the centripetal accelerations cancels out due to symmetry, resulting in an average acceleration of zero. Hence, the acceleration vector averaged over one cycle is a null vector. Therefore, the statement is true.
Summary:
(a) True
(b) True
(c) True
The position of a particle is given by
$\mathbf{r}=3.0 t \hat{\mathbf{i}}-2.0 t^{2} \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}} \mathrm{m}$
where $t$ is in seconds and the coefficients have the proper units for $\mathbf{r}$ to be in metres.
(a) Find the $\mathbf{v}$ and $\mathbf{a}$ of the particle? (b) What is the magnitude and direction of velocity of the particle at $t=2.0 \mathrm{~s}$ ?
(a) Find $\mathbf{v}(t)$ and $\mathbf{a}(t)$ for the particle
Given the position vector: $$ \mathbf{r} = 3.0 t \hat{\mathbf{i}} - 2.0 t^{2} \hat{\mathbf{j}} + 4.0 \hat{\mathbf{k}} $$
Velocity is the derivative of the position vector with respect to time: $$ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} $$ $$ \mathbf{v}(t) = \frac{d}{dt} (3.0 t \hat{\mathbf{i}} - 2.0 t^{2} \hat{\mathbf{j}} + 4.0 \hat{\mathbf{k}}) $$ $$ \mathbf{v}(t) = 3.0 \hat{\mathbf{i}} - 4.0 t \hat{\mathbf{j}} $$
Acceleration is the derivative of the velocity vector with respect to time: $$ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} $$ $$ \mathbf{a}(t) = \frac{d}{dt} (3.0 \hat{\mathbf{i}} - 4.0 t \hat{\mathbf{j}})$$ $$ \mathbf{a}(t) = -4.0 \hat{\mathbf{j}} $$ $$ \mathbf{a}(t) \text{ is a constant vector.} $$
So,
$\mathbf{v}(t) = 3.0 \hat{\mathbf{i}} - 4.0 t \hat{\mathbf{j}}$
$\mathbf{a}(t) = -4.0 \hat{\mathbf{j}} $
(b) Find the magnitude and direction of velocity at $t = 2.0 , \text{s}$
First, calculate the velocity vector at $t = 2.0 , \text{s}$: $$ \mathbf{v}(2.0) = 3.0 \hat{\mathbf{i}} - 4.0 (2.0) \hat{\mathbf{j}} $$ $$ \mathbf{v}(2.0) = 3.0 \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}} $$
Magnitude of the velocity: $$ v = \sqrt{(v_x^2 + v_y^2)}$$ $$ v = \sqrt{(3.0^2 + (-8.0)^2)}$$ $$ v = \sqrt{(9 + 64)}$$ $$ v = \sqrt{73} $$ $$ v \approx 8.54 , \text{m/s}$$
Direction of the velocity: $$ \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) $$ $$ \theta = \tan^{-1}\left(\frac{-8.0}{3.0}\right) $$ $$ \theta \approx \tan^{-1}(-2.67)$$ $$ \theta \approx -70.3^\circ $$
Thus, at $t = 2.0 , \text{s}$:
The magnitude of the velocity is $\approx 8.54 , \text{m/s}$.
The direction of the velocity is $\approx -70.3^\circ$ with respect to the $+x$-axis.
A particle starts from the origin at $t=0 \mathrm{~s}$ with a velocity of $10.0 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$ and moves in the $x-y$ plane with a constant acceleration of $(8.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}}) \mathrm{m} \mathrm{s}^{-2}$. (a) At what time is the $x$-coordinate of the particle $16 \mathrm{~m}$ ? What is the $y$-coordinate of the particle at that time? (b) What is the speed of the particle at the time ?
(a) Time when the x-coordinate is 16 m:
From the given problem, we know the initial conditions and the equations of motion for a particle under constant acceleration.
The initial position vector $\mathbf{r}_{0} = 0$.
The initial velocity vector $\mathbf{v}_{0} = 10.0 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$.
The acceleration vector $\mathbf{a} = (8.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}}) \mathrm{m} /\mathrm{s}^{-2}$.
The equations of motion in the (x)- and (y)- directions are: $$ x = x_{0} + v_{0x} t + \frac{1}{2} a_{x} t^{2} $$ $$ y = y_{0} + v_{0y} t + \frac{1}{2} a_{y} t^{2} $$
Given that $x_{0} = 0$, $v_{0x} = 0$, $a_{x} = 8 \mathrm{m} /\mathrm{s}^{2}$:
$$ x = 0 + 0 \cdot t + \frac{1}{2} (8) t^{2}$$ $$ x = 4 t^{2} $$ Setting (x = 16 \mathrm{m}):
$$16 = 4 t^{2} $$ $$ t^{2} = 4 $$ $$ t = 2 \mathrm{s} \ (since \ t \geq 0) $$
Next, we find the (y)-coordinate at $t = 2 \mathrm{s}$: Given $y_{0} = 0$, $v_{0y} = 10 \mathrm{m} /\mathrm{s}$, $a_{y} = 2 \mathrm{m} /\mathrm{s}^{2}$:
$$ y = 10 t + \frac{1}{2} (2) t^{2} $$ $$ y = 10(2) + \frac{1}{2}(2)(2)^{2} $$ $$ y = 20 + 4 $$ $$y = 24 \mathrm{m} $$
Thus, at $t = 2 \mathrm{s}$, the (y)-coordinate is $24 \mathrm{m}$.
(b) Speed of the particle at $t=2 \mathrm{s}$:
To find the speed, we first find the velocity components at $t=2 \mathrm{s}$:
$$ v_{x} = v_{0x} + a_{x} t $$ $$ v_{x} = 0 + (8)(2) $$ $$ v_{x} = 16 \mathrm{m} /\mathrm{s}$$
$$ v_{y} = v_{0y} + a_{y} t $$ $$ v_{y} = 10 + (2)(2) $$ $$ v_{y} = 14 \mathrm{m} /\mathrm{s} $$
Now, the speed (v) is the magnitude of the velocity vector: $$ v = \sqrt{v_{x}^{2} + v_{y}^{2}}$$ $$v = \sqrt{16^{2} + 14^{2}} $$ $$ v = \sqrt{256 + 196} $$ $$ v = \sqrt{452} $$ $$ v \approx 21.3 \mathrm{m} /\mathrm{s}$$
Summary:
(a) The (x)-coordinate of the particle is $16 \mathrm{m}$ at $t = 2 \mathrm{s}$, and the corresponding (y)-coordinate is $24 \mathrm{m}$.
(b) The speed of the particle at $t = 2 \mathrm{s}$ is approximately $21.3 \mathrm{m} /\mathrm{s}$.
$\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are unit vectors along $x$-and $y$-axis respectively. What is the magnitude and direction of the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$, and $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ ? What are the components of a vector $\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ along the directions of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ ? [You may use graphical method]
Magnitude and Direction of the Vectors
For the vector $\hat{\mathbf{i}} + \hat{\mathbf{j}}$:
Magnitude: $\sqrt{2}$
Direction: 45° to the horizontal axis
For the vector $\hat{\mathbf{i}} - \hat{\mathbf{j}}$:
Magnitude: $\sqrt{2}$
Direction: -45° to the horizontal axis
Components of Vector $\mathbf{A} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}}$
To find the components of $\mathbf{A}$ along the directions of $\hat{\mathbf{i}} + \hat{\mathbf{j}}$ and $\hat{\mathbf{i}} - \hat{\mathbf{j}}$, we use projections.
Along $\hat{\mathbf{i}} + \hat{\mathbf{j}}$:
The unit vector in the direction of $\hat{\mathbf{i}} + \hat{\mathbf{j}}$ is: $$ \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} $$
The projection of $\mathbf{A}$ onto $\hat{\mathbf{i}} + \hat{\mathbf{j}}$ is given by: $$ (\mathbf{A} \cdot \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}) \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} $$
Calculating the dot product: $$ \mathbf{A} \cdot (\hat{\mathbf{i}} + \hat{\mathbf{j}}) = (2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}) \cdot (\hat{\mathbf{i}} + \hat{\mathbf{j}}) = 2(1) + 3(1) = 5 $$
Thus, the component of $\mathbf{A}$ along $\hat{\mathbf{i}} + \hat{\mathbf{j}}$ is: $$ \frac{5}{\sqrt{2}} $$
Along $\hat{\mathbf{i}} - \hat{\mathbf{j}}$:
The unit vector in the direction of $\hat{\mathbf{i}} - \hat{\mathbf{j}}$ is: $$ \frac{\hat{\mathbf{i}} - \hat{\mathbf{j}}}{\sqrt{2}} $$
The projection of $\mathbf{A}$ onto $\hat{\mathbf{i}} - \hat{\mathbf{j}}$ is given by: $$ (\mathbf{A} \cdot \frac{\hat{\mathbf{i}} - \hat{\mathbf{j}}}{\sqrt{2}}) \frac{\hat{\mathbf{i}} - \hat{\mathbf{j}}}{\sqrt{2}} $$
Calculating the dot product: $$ \mathbf{A} \cdot (\hat{\mathbf{i}} - \hat{\mathbf{j}}) = (2\hat{\mathbf{i}} + 3\hat{\mathbf{j}}) \cdot (\hat{\mathbf{i}} - \hat{\mathbf{j}}) = 2(1) - 3(1) = -1 $$
Thus, the component of $\mathbf{A}$ along $\hat{\mathbf{i}} - \hat{\mathbf{j}}$ is: $$ \frac{-1}{\sqrt{2}} $$
Summary
The magnitude of $\hat{\mathbf{i}} + \hat{\mathbf{j}}$ is $\sqrt{2}$ and its direction is 45° to the horizontal.
The magnitude of $\hat{\mathbf{i}} - \hat{\mathbf{j}}$ is $\sqrt{2}$ and its direction is -45° to the horizontal.
The component of $\mathbf{A}$ along $\hat{\mathbf{i}} + \hat{\mathbf{j}}$ is $\frac{5}{\sqrt{2}}$.
The component of $\mathbf{A}$ along $\hat{\mathbf{i}} - \hat{\mathbf{j}}$ is $\frac{-1}{\sqrt{2}}$.
For any arbitrary motion in space, which of the following relations are true :
(a) $\mathbf{v}_{\text {average }}=(1 / 2)\left(\mathbf{v}\left(t_{1}\right)+\mathbf{v}\left(t_{2}\right)\right)$
(b) $\mathbf{v}_{\text {average }}=\left[\mathbf{r}\left(t_{2}\right)-\mathbf{r}\left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$
(c) $\mathbf{v}(t)=$ average $=\mathbf{v}(0)+\mathbf{a} t$
(d) $\mathbf{r}(t)=\mathbf{r}(0)+\mathbf{v}(0) t+(1 / 2) \mathbf{a} t^{2}$
(e) $\mathbf{a}_{\text {average }}=\left[\mathbf{v}\left(t_{2}\right)-\mathbf{v}\left(t_{1}\right)\right] /\left(t_{2}-t_{1}\right)$
(The average' stands for average of the quantity over the time interval $t_{1}$ to $t_{2}$ )
Let's analyze each of the given relations to determine whether they are true for arbitrary motion in space.
(a) $$ \mathbf{v}_{\text {average }} = \frac{1}{2} \left(\mathbf{v}\left(t_{1}\right) + \mathbf{v}\left(t_{2}\right)\right) $$
This relation is true only if the acceleration is constant. For arbitrary motion, where acceleration may vary, this equality does not generally hold.
False
(b) $$ \mathbf{v}_{\text {average }} = \frac{\left[\mathbf{r}(t_{2}) - \mathbf{r}(t_{1})\right]}{(t_{2} - t_{1})} $$
This relation correctly defines the average velocity as the displacement divided by the time interval, and it is always true for arbitrary motion in space.
True
(c) $$ \mathbf{v}(t) = \mathbf{v}(0) + \mathbf{a} t $$
This is the equation of motion under constant acceleration. For arbitrary motion in space, where acceleration may change, this relation does not generally hold.
False
(d) $$ \mathbf{r}(t) = \mathbf{r}(0) + \mathbf{v}(0) t + \frac{1}{2} \mathbf{a} t^{2} $$
This is the equation of motion under constant acceleration. For arbitrary motion in space, where acceleration may change, this relation does not generally hold.
False
(e) $$ \mathbf{a}_{\text {average }} = \frac{\left[\mathbf{v}(t_{2}) - \mathbf{v}(t_{1})\right]}{(t_{2} - t_{1})} $$
This relation correctly defines the average acceleration as the change in velocity divided by the time interval, and it is always true for arbitrary motion in space.
True
Summary:
(a) False
(b) True
(c) False
(d) False
(e) True
Read each statement below carefully and state, with reasons and examples, if it is true or false :
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.
(a) Is conserved in a process:
False. A scalar quantity does not necessarily have to be conserved in a process. For example, temperature is a scalar quantity, but it is not conserved in a process (it can increase or decrease).
(b) Can never take negative values:
False. Scalar quantities can take negative values. For example, temperature (in Celsius) and electric potential can be negative.
(c) Must be dimensionless:
False. Scalar quantities can have dimensions. For example, mass (kg), distance (m), and time (s) are scalar quantities that are not dimensionless.
(d) Does not vary from one point to another in space:
False. Scalar quantities can vary from one point to another in space. For example, temperature in a region of space can vary from point to point.
(e) Has the same value for observers with different orientations of axes:
True. Scalars are invariant under transformations of the coordinate system. For example, the mass of an object remains the same regardless of the orientation of the axes.
In summary:
(a) False
(b) False
(c) False
(d) False
(e) True
An aircraft is flying at a height of $3400 \mathrm{~m}$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10.0 \mathrm{~s}$ a part is $30^{\circ}$, wat is the speed of the aircraft?
First, using the height and the angle, the horizontal distance ((D)) covered by the aircraft is computed as:
$$ D = \frac{3400 , \text{m}}{2 \tan (15^\circ)} = 6344 , \text{meters} $$
Then, with this distance, the speed ((v)) is:
$$ v = \frac{D}{t} = \frac{6344 , \text{meters}}{10 , \text{seconds}} = 634.4 , \text{m/s} $$
So, the speed of the aircraft is 634.4 m/s.
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Ask Chatterbot AINotes - Motion in a Plane | Class 11 NCERT | Physics
Comprehensive Notes on Motion in a Plane for Class 11: Key Concepts and Examples
Introduction to Motion in a Plane
Basics of Motion in a Plane
Motion in a plane involves moves that occur in two dimensions. Understanding this concept is a cornerstone in physics as it extends beyond linear motion and helps describe phenomena in the real world, like projectile motion and circular motion.
Scalars and Vectors
Understanding Scalars
A scalar quantity is defined by magnitude alone. Key examples include distance, speed, mass, and temperature. These quantities do not have a direction.
Understanding Vectors
Unlike scalars, vectors possess both magnitude and direction. Examples of vector quantities are displacement, velocity, and acceleration.
Representation of Vectors
Vectors are typically represented by arrows where the length indicates the magnitude, and the direction of the arrow indicates the direction of the vector.
Position and Displacement Vectors
Position Vectors
The position vector of an object gives its location with respect to a chosen origin.
Displacement Vectors
Displacement is a vector that represents the change in position of an object. It is the straight line connecting the initial and final positions, regardless of the path taken.
Vector Operations
Addition and Subtraction of Vectors
Graphical Method: The head-to-tail method and parallelogram method.
Analytical Method: Vectors can be added or subtracted by breaking them into their components.
Multiplication of Vectors by Real Numbers
Positive Multiplication: The direction remains the same.
Negative Multiplication: The direction reverses.
Resolution of Vectors
Concept of Resolution
Resolution involves breaking a vector into its components along commonly used axes.
Unit Vectors
Unit vectors have a magnitude of one and are used to indicate direction. Common unit vectors are $ \hat{i} $, $ \hat{j} $, and $ \hat{k} $ for the x, y, and z axes respectively.
Motion in Two Dimensions
Component Form of Vectors
Vectors in two dimensions can be expressed as $\mathbf{A} = A_x \hat{i} + A_y \hat{j} $.
Velocity in a Plane
Average Velocity: The total displacement divided by the total time.
Instantaneous Velocity: The derivative of the position vector with respect to time.
Acceleration in a Plane
Average Acceleration: The change in velocity divided by the time taken.
Instantaneous Acceleration: The derivative of the velocity vector with respect to time.
Projectile Motion
Basic Principles
The motion of a projectile is a combination of two independent motions: horizontal motion with constant velocity and vertical motion with constant acceleration due to gravity.
Characteristics of Projectile Motion
Path: Parabolic trajectory
Time of Flight: The total time the projectile is in the air.
Maximum Height: The highest point reached by the projectile.
Range: The horizontal distance covered by the projectile.
Equations of Motion
$ x = v_{0x} t = (v_0 \cos \theta_0) t $
$y = (v_0 \sin \theta_0) t - \frac{1}{2} g t^2 $
Example Calculation
Time of Maximum Height: $t_m = \frac{v_0 \sin \theta_0}{g} $
Range: $ = \frac{v_{0}^{2} \sin 2 \theta_{0}}{g} $
Uniform Circular Motion
Definition and Examples
Uniform circular motion refers to motion in a circular path at a constant speed.
Centripetal Acceleration
The acceleration directed towards the centre of the circle is called centripetal acceleration and is given by: $$ a_c = \frac{v^2}{R} $$
graph TD;
A[Circular Path] --> B[Centre of Circle];
A --> C[[ Tangential Velocity]] ;
B --> D[[Centripetal Acceleration]];
C --> D;
Angular Speed
Angular speed ( \omega ) is the rate of change of angular displacement and is related to the linear speed by: $$ \omega = \frac{v}{R} $$ $$ a_c = \omega^2 R $$
Summary and Key Takeaways
Recap of Main Topics
This guide covers scalar and vector quantities, vector operations, motion in two dimensions, projectile motion, and uniform circular motion.
Important Equations and Concepts
Displacement Vector: $ \Delta \mathbf{r} = \mathbf{r'} - \mathbf{r} $
Projectile Motion Path: $y = ( \tan \theta_{0}) x - \frac{gx^2}{2(v_0 \cos \theta_0)^2} $
Centripetal Acceleration: $ a_c = \frac{v^2}{R} $
By understanding these concepts, students can effectively approach problems involving motion in a plane and excel in their physics studies.
Frequently Asked Questions (FAQs)
Common Questions on Motion in a Plane
What is the difference between scalar and vector quantities?
How do you calculate the trajectory of a projectile?
What is the significance of centripetal acceleration in circular motion?
Tips for Studying and Understanding the Concepts
Practice Breaking Vectors into Components: This helps in simplifying complex problems.
Visualisation: Use diagrams and graphs to understand the direction and magnitude of vectors.
Real-Life Examples: Relate the concepts to real-world scenarios like sports, driving, and space missions.
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