Laws of Motion - Class 11 Physics - Chapter 5 - Notes, NCERT Solutions & Extra Questions
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Notes - Laws of Motion | Class 11 NCERT | Physics
Laws of Motion Class 11 Notes: Comprehensive Guide with Examples
Understanding the laws of motion is crucial for anyone studying physics, especially at the Class 11 level. These laws, formulated by Sir Isaac Newton, form the foundation of classical mechanics and describe the relationship between the motion of an object and the forces acting on it. Let’s dive into the concepts with some detailed explanations and examples.
Introduction to Laws of Motion
Overview of Newton's Laws
Newton's laws of motion consist of three fundamental principles that elucidate the dynamics of motion. These laws not only apply to everyday activities but also to the grand phenomena observed in the universe.
Understanding Newton’s First Law of Motion
Concept of Inertia
The first law states that: "Every object continues in its state of rest, or in uniform motion in a straight line, unless acted upon by an external force."
This principle is known as the law of inertia. It implies that objects resist changes in their state of motion.
Galileo’s Contribution to Inertia
Galileo's experiments with inclined planes were pioneering, showing that objects would continue to move indefinitely on a frictionless plane, laying the groundwork for Newton’s first law.
Newton’s Second Law of Motion
Force and Acceleration
The second law is articulated as: "The rate of change of momentum of an object is directly proportional to the applied force and occurs in the direction of the force."
This can be expressed with the formula: $$ \mathbf{F} = m \mathbf{a} $$ where $ \mathbf{F} $ is the net force acting on the object, ( m ) is the mass, and $ \mathbf{a} $ is the acceleration.
Momentum and its Relation to Force
Momentum $( \mathbf{p} )$ is defined as the product of mass and velocity $( \mathbf{p} = m \mathbf{v} )$. According to the second law, force is the change in momentum over time.
Newton’s Third Law of Motion
Action and Reaction Forces
The third law states: "To every action, there is always an equal and opposite reaction."
This means that for every force exerted by one body on another, there is a force of equal magnitude but in the opposite direction exerted back on the first body.
Law of Conservation of Momentum
Derivation from Newton’s Laws
The law of conservation of momentum is derived from Newton’s second and third laws. In an isolated system, the total momentum remains constant if no external forces act on it.
Common Forces in Mechanics
Gravitational Force
Gravitational force is universal and acts between any two masses. On Earth, it provides the weight that acts downward on objects.
Contact Forces
Contact forces include the normal reaction force, tension, and friction. These arise from the interaction between objects in contact.
Friction and its Types
Static Friction vs. Kinetic Friction
Static Friction: Opposes the initiation of motion.
Kinetic Friction: Acts against the motion once it has started.
These forces prevent or slow down the relative motion between two surfaces in contact.
Circular Motion and Centripetal Force
Dynamics of Circular Motion
In circular motion, an object moving at a speed $\mathbf{v} $ along a path of radius $ \mathbf{R} $ experiences a centripetal force directed towards the centre of the circle: $$ \mathbf{f_{c}} = \frac{m v^2}{R} $$
Problem-Solving in Mechanics
Free-Body Diagrams
Free-body diagrams are essential for visualising the forces acting on an object. Here’s a basic approach to drawing them:
graph TD;
A[Object] -- Gravity --> B[Earth]
A -- Normal force --> C[Table]
A -- Applied force --> D[Push]
A -- Friction --> E[Friction Force]
Inclined Planes and Friction
To solve problems involving inclined planes, resolve the forces parallel and perpendicular to the plane's surface. Use the static and kinetic friction equations to determine if the object will move or remain at rest.
Applications and Examples
Real-world Applications of Newton's Laws
Sports: Understanding the motion of balls.
Vehicles: Calculations involving acceleration and braking.
Daily Activities: Walking, using tools, and more.
Impulse and Real-life Scenarios
Impulse $( \mathbf{J} = \mathbf{F} \Delta t )$ changes the momentum of objects and is exemplified in sports like cricket and baseball.
Understanding the laws of motion is fundamental not only for academic success but also for comprehending the mechanics of daily life and the natural world. These principles are the cornerstone of classical mechanics, offering insights into the intricate balance of forces and motion.
This comprehensive guide provides a solid foundation for anyone delving into the fascinating world of physics. By mastering these concepts, you will be well-equipped to tackle more advanced topics and real-world applications.
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NCERT Solutions - Laws of Motion | NCERT | Physics | Class 11
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass $10 \mathrm{~g}$ floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of $30 \mathrm{~km} / \mathrm{h}$ on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
To determine the magnitude and direction of the net force acting on each object, we need to analyze whether there is any acceleration or if the objects are in a state of equilibrium. According to Newton's first law, if an object is moving with constant velocity or is stationary, the net external force acting on it must be zero (i.e., it is either at rest or in uniform motion).
(a) A drop of rain falling down with a constant speed
Since the rain drop is falling with a constant speed (implying no acceleration), the net force acting on it must be zero. This means the gravitational force acting downward is exactly balanced by the air resistance acting upward.
Magnitude of net force: $0 , \text{N}$
Direction of net force: Not applicable (since the net force is zero)
(b) A cork of mass $10 , \mathrm{g}$ floating on water
A cork floating on water is in a state of equilibrium, with the buoyant force from the water balancing the gravitational force acting downward.
Magnitude of net force: $0 , \text{N}$
Direction of net force: Not applicable (since the net force is zero)
(c) A kite skillfully held stationary in the sky
A kite held stationary in the sky must be in equilibrium. The forces acting (gravity, tension in the string, and aerodynamic lift) balance each other out.
Magnitude of net force: $0 , \text{N}$
Direction of net force: Not applicable (since the net force is zero)
(d) A car moving with a constant velocity of $30 , \mathrm{km} / \mathrm{h}$ on a rough road
Since the car is moving with a constant velocity, it is not accelerating. This implies that all the forces acting on the car (engine force, friction, air resistance, etc.) are balanced.
Magnitude of net force: $0 , \text{N}$
Direction of net force: Not applicable (since the net force is zero)
(e) A high-speed electron in space far from all material objects, and free of electric and magnetic fields
In this case, the electron experiences no external forces. Therefore, it remains in a state of constant velocity as described by Newton's first law.
Magnitude of net force: $0 , \text{N}$
Direction of net force: Not applicable (since the net force is zero)
In summary, for all these cases, the net external force is zero due to the conditions implying no acceleration or equilibrium states.
A pebble of mass $0.05 \mathrm{~kg}$ is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,
(a) during its upward motion,
(b) during its downward motion,
(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $45^{\circ}$ with the horizontal direction?
Ignore air resistance.
Calculations Result
$$ W = 0.05 , \text{kg} \times 9.8 , \text{m/s}^2 = 0.49 , \text{N} $$
Answers:
(a) During its upward motion:
Direction of Net Force: Downward
Magnitude: $ 0.49 , \text{N} $
(b) During its downward motion:
Direction of Net Force: Downward
Magnitude: $ 0.49 , \text{N} $
(c) At the highest point where it is momentarily at rest:
Direction of Net Force: Downward
Magnitude: $ 0.49 , \text{N} $
Conclusion:
The direction and magnitude of the force do not change if the pebble is thrown at an angle of $ 45^\circ $with the horizontal direction, because gravity acts uniformly in the vertical direction regardless of the horizontal motion.
Give the magnitude and direction of the net force acting on a stone of mass $0.1 \mathrm{~kg}$,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of $36 \mathrm{~km} / \mathrm{h}$,
(c ) just after it is dropped from the window of a train accelerating with $1 \mathrm{~m} \mathrm{~s}^{-2}$,
(d) lying on the floor of a train which is accelerating with $1 \mathrm{~m} \mathrm{~s}^{-2}$, the stone being at rest relative to the train.
Neglect air resistance throughout.
The magnitude of the net force acting on the stone when lying on the floor of an accelerating train is approximately:
$$ F_{\text{net}} = 0.985 , \text{N} $$
To summarize the results for each part:
(a) Dropped from the window of a stationary train:
Magnitude: $ 0.98 , \text{N} $
Direction: Downward
(b) Dropped from the window of a train running at constant velocity:
Magnitude: $ 0.98 , \text{N} $
Direction: Downward
(c) Dropped from the window of a train accelerating:
Magnitude: $ 0.98 , \text{N} $
Direction: Downward
(d) Lying on the floor of a train accelerating:
Magnitude: $ \approx 0.985 , \text{N} $
Direction: The net direction combines the downward and opposing horizontal force. The resultant force would be at an angle to the vertical.
One end of a string of length $l$ is connected to a particle of mass $m$ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$ the net force on the particle (directed towards the centre) is :
(i) $T$, (ii) $T-\frac{m v^{2}}{l}$, (iii) $T+\frac{m v^{2}}{l}$, (iv) 0
$T$ is the tension in the string. [Choose the correct alternative].
To determine the net force on the particle, we must understand the forces acting on it. The particle is moving in a circle at constant speed, so it experiences a centripetal force directed towards the center of the circle. This centripetal force is provided by the tension in the string.
The centripetal force $ F_c $ is given by: $$ F_c = \frac{m v^2}{l} $$
Since the only force providing the centripetal force is the tension ( T ) in the string, we have: $$ T = F_c = \frac{m v^2}{l} $$
Therefore, the correct alternative is: (i) ( T )
So, the net force on the particle (directed towards the centre) is ( T ).
A constant retarding force of $50 \mathrm{~N}$ is applied to a body of mass $20 \mathrm{~kg}$ moving initially with a speed of $15 \mathrm{~m} \mathrm{~s}^{-1}$. How long does the body take to stop?
The time it takes for the body to stop is:
$$ t = 6 , \text{s} $$
So, the body takes $ \boldsymbol{6} , \text{s}$ to stop.
A constant force acting on a body of mass $3.0 \mathrm{~kg}$ changes its speed from $2.0 \mathrm{~m} \mathrm{~s}^{-1}$ to $3.5 \mathrm{~m} \mathrm{~s}^{-1}$ in $25 \mathrm{~s}$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
The magnitude of the force $\mathbf{F}$ is:
$$ \mathbf{F} = 0.18 , \mathrm{N} $$
Direction:Since the direction of the force is the same as the direction of the acceleration, and the acceleration causes an increase in speed while maintaining the initial direction, the force must act in the same direction as the motion of the body.
Therefore, the magnitude of the force is 0.18 N in the direction of the body's motion.
A body of mass $5 \mathrm{~kg}$ is acted upon by two perpendicular forces $8 \mathrm{~N}$ and $6 \mathrm{~N}$. Give the magnitude and direction of the acceleration of the body.
Resultant Force Calculation
The magnitude of the resultant force is: $$ F_{\text{res}} = 10 , \text{N} $$
Acceleration Calculation
Using Newton's second law, $F = m \cdot a$, where $m$ is the mass of the body (5 kg):
$$ a = \frac{F_{\text{res}}}{m} = \frac{10 , \text{N}}{5 , \text{kg}} = 2 , \text{m/s}^2 $$
Direction Calculation
The direction of the resultant force can be given by the angle $\theta$ with respect to the 8 N force. Using the inverse tangent function:
$$ \theta = \tan^{-1} \left(\frac{6}{8}\right) = 36.87^\circ $$
Summary
Magnitude of acceleration: $$2 , \text{m/s}^2$$
Direction of acceleration: $$36.87^\circ$$ with respect to the 8 N force.
The driver of a three-wheeler moving with a speed of $36 \mathrm{~km} / \mathrm{h}$ sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is $400 \mathrm{~kg}$ and the mass of the driver is $65 \mathrm{~kg}$.
The average retarding force on the three-wheeler is $ \mathbf{1160 , N} $ (newtons). Since it is a retarding force, the direction is opposite to the direction of motion.
A rocket with a lift-off mass $20,000 \mathrm{~kg}$ is blasted upwards with an initial acceleration of $5.0 \mathrm{~m} \mathrm{~s}^{-2}$. Calculate the initial thrust (force) of the blast.
The initial thrust (force) of the blast is:
$$ \mathbf{F} = 20000 \times (5 + 9.8) = 296000 , \text{N} $$
So, the initial thrust is 296,000 N.
A body of mass $0.40 \mathrm{~kg}$ moving initially with a constant speed of $10 \mathrm{~m} \mathrm{~s}^{-1}$ to the north is subject to a constant force of $8.0 \mathrm{~N}$ directed towards the south for $30 \mathrm{~s}$. Take the instant the force is applied to be $t=0$, the position of the body at that time to be $x=0$, and predict its position at $\mathrm{t}=-5 \mathrm{~s}, 25 \mathrm{~s}, 100 \mathrm{~s}$.
Step 1: Calculate the acceleration
Given:
Force, $F = 8 \text{ N}$ (directed south)
Mass, $m = 0.40 \text{ kg}$
Using Newton's second law, $$ F = ma $$ $$ a = \frac{F}{m} $$
Substitute the values: $$ a = \frac{8}{0.40} = 20 \text{ m/s}^2 $$ (south)
Step 2: Calculate the velocity at $t = 30 \text{ s}$
Given:
Initial velocity, $u = 10 \text{ m/s}$ (north)
Acceleration, $a = 20 \text{ m/s}^2$ (south)
Time, $t = 30 \text{ s}$
The velocity after time $t$ is given by: $$ v = u + at $$
Since south is the negative direction: $$ v = 10 - 20 \cdot 30 $$ $$ v = 10 - 600 $$ $$ v = -590 \text{ m/s} $$ (negative sign means the direction is south)
Step 3: Calculate the displacement at $t = 30 \text{ s}$
Using the equation of motion: $$ s = ut + \frac{1}{2}at^2 $$
Substitute the values: $$ s = 10 \cdot 30 + \frac{1}{2} \cdot (-20) \cdot (30)^2 $$ $$ s = 300 - 0.5 \cdot 20 \cdot 900 $$ $$ s = 300 - 9000 $$ $$ s = -8700 \text{ m} $$ (negative sign means the direction is south)
Now we can find the positions at different times:
$ \text{initial conditions at } t = 0: \text{ position } x = 0 \text{ m} $
At $t = 30 \text{ s}$: $x = -8700 \text{ m}$ (south)
Prediction for $t = -5 \text{ s}$:
Since $t = -5 \text{ s}$ is before the force is applied, the body moves with the initial constant speed of $10 \text{ m/s} \text{ north}$: $$ s = ut $$ $$ s = 10 \cdot (-5) = -50 \text{ m} $$ (negative sign means north) Thus, $x = -50 \text{ m}$ north.
Prediction for $t = 25 \text{ s}$:
Since $t = 25 \text{ s}$ is within the force application period, we use the same method: $$ s = ut + \frac{1}{2}at^2 $$ $$ s = 10 \cdot 25 + \frac{1}{2} \cdot (-20) \cdot (25)^2 $$ $$ s = 250 - 0.5 \cdot 20 \cdot 625 $$ $$ s = 250 - 6250 $$ $$ s = -6000 \text{ m} $$ (south)
Prediction for $t = 100 \text{ s}$:
For $t = 100 \text{ s}$, it consists of 30 seconds under force and then 70 seconds with constant velocity south: $$ \text{position at } t = 30 \text{ s} = -8700 \text{ m} $$
From $t = 30 \text{ s}$ to $t = 100 \text{ s}$:
Velocity after force application, $v = -590 \text{ m/s}$ (south)
Time duration after force application, $t' = 100 - 30 = 70 \text{ s}$
Total displacement in $70 \text{ s}$ under constant velocity: $$ s' = vt' $$ $$ s' = -590 \cdot 70 = -41300 \text{ m} $$ (south)
Total position: $$ x = -8700 + (-41300) = -50000 \text{ m} $$ (south)
Summary
Position at $t = -5 \text{ s}$: $-50 \text{ m}$ north
Position at $t = 25 \text{ s}$: $-6000 \text{ m}$ south
Position at $t = 100 \text{ s}$: $-50000 \text{ m}$ south
A truck starts from rest and accelerates uniformly at $2.0 \mathrm{~m} \mathrm{~s}^{2}$. At $t=10 \mathrm{~s}, \mathrm{a}$ stone is dropped by a person standing on the top of the truck ( $6 \mathrm{~m}$ high from the ground). What are the (a) velocity, and (b) acceleration of the stone at $t=$ 11 s? (Neglect air resistance.)
Let's break the problem down into smaller but manageable parts:
Velocity of the truck at $t = 10$ seconds$$ v = u + a t $$ Here,
Initial velocity, $u = 0 , \mathrm{m/s}$ (since the truck starts from rest).
Acceleration, $a = 2.0 , \mathrm{m/s^2}$.
Time, $t = 10 , \mathrm{s}$.
Plugging these values into the equation gives: $$ v = 0 + 2.0 , \mathrm{m/s^2} \times 10 , \mathrm{s} = 20 , \mathrm{m/s} $$ Hence, the velocity of the truck at $t = 10$ seconds is $20 , \mathrm{m/s}$.
Velocity of the stone at $t = 11$ secondsAt $t = 10 , \mathrm{s}$, the stone has the same horizontal velocity as the truck, which is $20 , \mathrm{m/s}$. When the stone is dropped, it continues with this horizontal velocity while it accelerates downwards under gravity.
The time taken for the stone to fall from a height of $6 , \mathrm{m}$ can be calculated using the free fall equation: $$ h = \frac{1}{2} g t^2 $$ Solving for $t$ when $h = 6 , \mathrm{m}$ and $g = 9.8 , \mathrm{m/s^2}$ gives: $$ 6 = \frac{1}{2} \times 9.8 \times t^2 \implies t^2 = \frac{12}{9.8} \implies t \approx 1.1 , \mathrm{s} $$
The vertical velocity of the stone after $1.1 , \mathrm{s}$ can be found using: $$ v = g t = 9.8 , \mathrm{m/s^2} \times 1.1 , \mathrm{s} = 10.78 , \mathrm{m/s} $$
The total velocity of the stone at $t = 11$ seconds (considering both horizontal and vertical components) can be found using Pythagorean theorem: $$ v_{\text{total}} = \sqrt{(20)^2 + (10.78)^2} , \mathrm{m/s} $$
Acceleration of the stone after its releaseOnce released, the stone only has vertical acceleration due to gravity, which is $9.8 , \mathrm{m/s^2}$ downward.
Hence the final answers are: (a) Velocity: $$ v_{\text{total}} = \sqrt{(20)^2 + (10.78)^2} , \mathrm{m/s} \approx 22.8 , \mathrm{m/s} $$
(b) Acceleration: $$ 9.8 , \mathrm{m/s^2} \text{ downward} $$
A bob of mass $0.1 \mathrm{~kg}$ hung from the ceiling of a room by a string $2 \mathrm{~m}$ long is set into oscillation. The speed of the bob at its mean position is $1 \mathrm{~m} \mathrm{~s}^{-1}$. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
(a) When the String is Cut at One of Its Extreme Positions
At the extreme position, the bob is momentarily at rest because it's changing direction. If the string is cut at this position, the only force acting on the bob is gravity. Therefore, the bob will simply fall vertically downward from that position.
Summary of Bob's Motion:
Initial Velocity: $v = 0 , \text{m/s}$
Force Acting: Gravity (Weight) (mg)
Trajectory: Vertical fall under gravity.
(b) When the String is Cut at Its Mean Position
At the mean position, the bob has its maximum speed (1 , \text{m/s}). If the string is cut at this position, the bob will continue to move in the direction of its velocity at the moment the string is cut, due to inertia. This means it will move horizontally with the initial speed of (1 , \text{m/s}) while simultaneously accelerating downward due to gravity.
To calculate the trajectory, we need to consider the horizontal and vertical motions separately.
Horizontal Motion:
Initial Horizontal Velocity: $v_{x} = 1 , \text{m/s}$
Acceleration: $0 , \text{m/s}^2$ (since there is no horizontal force after the string is cut)
Position along x-axis: $$ x = v_{x} t $$
Vertical Motion:
Initial Vertical Velocity: $v_{y0} = 0 , \text{m/s}$
Acceleration: $g = 9.8 , \text{m/s}^2$
Position along y-axis: $$ y = \frac{1}{2} g t^2 $$
Combining these: $$ y = \frac{1}{2} g \left(\frac{x}{v_{x}}\right)^2 = \frac{1}{2} (9.8) \left(\frac{x}{1}\right)^2 = 4.9 x^2 $$
Thus, the equation describing the trajectory is: $$ y = 4.9 x^2 $$
Summary of Bob's Motion:
Initial Velocity: $v = 1 , \text{m/s}$ horizontally
Force Acting: Gravity (Weight) (mg)
Trajectory: Parabolic motion described by $y = 4.9 x^2$
Conclusion
At One of Its Extreme Positions: The bob falls vertically downward.
At Its Mean Position: The bob follows a parabolic trajectory with the equation $ y = 4.9 x^2 $.
A man of mass $70 \mathrm{~kg}$ stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of $10 \mathrm{~m} \mathrm{~s}^{-1}$,
(b) downwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$,
(c) upwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Part (a): Lift moving upwards with a uniform speed of $10 \mathrm{~m} \mathrm{~s}^{-1}$
The reading on the scale is: $$ R = 70 \mathrm{~kg} \times 9.8 \mathrm{~m} \mathrm{~s}^{-2} = 686 \mathrm{~N} $$
Part (b): Lift moving downwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$
The reading on the scale is: $$ R = 70 \mathrm{~kg} \times (9.8 - 5) \mathrm{~m} \mathrm{~s}^{-2} = 336 \mathrm{~N} $$
Part (c): Lift moving upwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$
The reading on the scale is: $$ R = 70 \mathrm{~kg} \times (9.8 + 5) \mathrm{~m} \mathrm{~s}^{-2} = 1036 \mathrm{~N} $$
Part (d): Lift mechanism fails and it hurtles down freely under gravity
The reading on the scale is: $$ R = 70 \mathrm{~kg} \times (9.8 - 9.8) \mathrm{~m} \mathrm{~s}^{-2} = 0 \mathrm{~N} $$
In summary, the readings on the scale in each case are:
(a) $686 \mathrm{~N}$
(b) $336 \mathrm{~N}$
(c) $1036 \mathrm{~N}$
(d) $0 \mathrm{~N}$
Figure 4.16 shows the position-time graph of a particle of mass $4 \mathrm{~kg}$. What is the (a) force on the particle for $t<0, t>4 \mathrm{~s}, 0<t<4 \mathrm{~s}$ ? (b) impulse at $t=0$ and $t=4 \mathrm{~s}$ ? (Consider one-dimensional motion only).
Fig. 4.16
(a) Force on the particle
For $ t < 0 $:
The position-time graph indicates that the particle is at rest at $ x = 0 $. Therefore,
Velocity $ v = 0 $,
Acceleration $ a = 0 $.
Using Newton's Second Law, $$ F = m \cdot a = 4 \cdot 0 = 0 , \text{N} $$
For $ t > 4 , \text{s} $:
The graph shows that the particle's position is constant at $ x = 3 , \text{m} $. Therefore,
Velocity $ v = 0 $,
Acceleration $ a = 0 $.
Using Newton's Second Law, $$ F = m \cdot a = 4 \cdot 0 = 0 , \text{N} $$
For $ 0 < t < 4 , \text{s} $:
From the graph, the particle moves from $ x = 0 $ to $ x = 3 $ over $ 4 $ seconds. Therefore,
Velocity $ v = \frac{\Delta x}{\Delta t} = \frac{3 , \text{m}}{4 , \text{s}} = 0.75 , \text{m/s}$.
Since the velocity is constant,
Acceleration ( a = 0 ).
Using Newton's Second Law, $$ F = m \cdot a = 4 \cdot 0 = 0 , \text{N} $$
(b) Impulse at $ t = 0 $ and $t = 4 , \text{s} $
Impulse ( J ) can also be found from the change in momentum. $$ J = \Delta p = m \cdot \Delta v $$
At ( t = 0 ):
Initial velocity $ v_i = 0 $
Final velocity $ v_f = 0.75 , \text{m/s} $
$$ J = 4 \cdot (0.75 - 0) = 4 \cdot 0.75 = 3 , \text{Ns} $$
At $ t = 4 , \text{s}$:
Initial velocity $ v_i = 0.75 , \text{m/s} $
Final velocity $ v_f = 0 $
$$ J = 4 \cdot (0 - 0.75) = 4 \cdot (-0.75) = -3 , \text{Ns} $$
Summary
Force (F) on the particle:
For $ t < 0 $, $ F = 0 , \text{N}$,
For $ t > 4 , \text{s} $, $F = 0 , \text{N}$,
For $0 < t < 4 , \text{s} $, $ F = 0 , \text{N} $.
Impulse (J):
At $ t = 0 $, $ J = 3 , \text{Ns} $,
At $ t = 4 , \text{s} $, $ J = -3 , \text{Ns} $.
Two bodies of masses $10 \mathrm{~kg}$ and $20 \mathrm{~kg}$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $\mathrm{F}=600 \mathrm{~N}$ is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
Step-by-step solution:
Calculate the acceleration:
Using Newton's second law: $$ a = \frac{F}{m} = \frac{600 \mathrm{~N}}{30 \mathrm{~kg}} = 20 \mathrm{~m/s^2} $$
Case (i): Force applied to $ \mathrm{A} ) $10 kg mass:
When the force is applied to mass $ \mathrm{m}_A $ , the tension $ T$ in the string can be determined by considering the mass $ \mathrm{m}_B $: $$T = \mathrm{m}_B \cdot a $$Substituting the known values: $$ T = 20 \mathrm{~kg} \cdot 20 \mathrm{~m/s^2} = 400 \mathrm{~N} $$
Case (ii): Force applied to $ \mathrm{B} $$20 kg mass$:
When the force is applied to mass $ \mathrm{m}_B $, the tension ( T ) in the string can be determined by considering the mass $ \mathrm{m}_A $: $$ T = \mathrm{m}_A \cdot a $$ Substituting the known values: $$T = 10 \mathrm{~kg} \cdot 20 \mathrm{~m/s^2} = 200 \mathrm{~N} $$
Summary
Case (i): Force applied to ( A ) (10 kg mass): The tension in the string is 400 N.
Case (ii): Force applied to ( B ) (20 kg mass): The tension in the string is 200 N.
Two masses $8 \mathrm{~kg}$ and $12 \mathrm{~kg}$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Results
Acceleration of the masses: $$ a = 1.96 , \text{m/s}^2 $$
Tension in the string: $$ T = 11.76 , \text{N} $$
Summary
The acceleration of the masses is 1.96 m/s².
The tension in the string is 11.76 N.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
To show that if a nucleus at rest disintegrates into two smaller nuclei, the products must move in opposite directions, we can use the law of conservation of momentum.
Conservation of Momentum
Before disintegration, the total momentum of the system is zero because the nucleus is at rest. After disintegration, the total momentum of the system must still be zero, as there are no external forces acting on the system. Let's denote the two smaller nuclei by $ A $ and $ B $, with masses $ m_A $ and $ m_B $, and velocities $ \mathbf{v}_A $ and $ \mathbf{v}_B $, respectively.
According to the conservation of momentum: $$ \text{Initial momentum} = \text{Final momentum} $$ Since the initial momentum is zero, we have: $$ 0 = m_A \mathbf{v}_A + m_B \mathbf{v}_B $$
Analyzing the Equation
The equation $ m_A \mathbf{v}_A + m_B \mathbf{v}_B = 0 $ can be rewritten as: $$ m_A \mathbf{v}_A = -m_B \mathbf{v}_B $$
This indicates that the product $ m_A \mathbf{v}_A $ has the same magnitude but opposite direction compared to $ m_B \mathbf{v}_B $. Therefore, the velocities $ \mathbf{v}_A $ and $ \mathbf{v}_B $ must be in opposite directions to satisfy the momentum conservation condition when their magnitudes are weighted by their respective masses.
Conclusion
Hence, we conclude that if a nucleus disintegrates into two smaller nuclei, the products must move in opposite directions to conserve momentum.
Two billiard balls each of mass $0.05 \mathrm{~kg}$ moving in opposite directions with speed $6 \mathrm{~m} \mathrm{~s}^{-1}$ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
To find the impulse imparted to each billiard ball due to the collision, we can use the concept of impulse, which is defined as the change in momentum.
Given:
Mass of each billiard ball, $m = 0.05 , \text{kg}$
Initial speed of each ball, $v = 6 , \text{m/s}$ (in opposite directions)
Let's assume the ball moving to the right has a positive velocity and the ball moving to the left has a negative velocity initially. After collision, both balls rebound with the same speed but in opposite directions.
Initial momenta:
Ball 1: $p_{1,\text{initial}} = m \cdot v = 0.05 , \text{kg} \cdot 6 , \text{m/s} = 0.3 , \text{kg} \cdot \text{m/s}$
Ball 2: $p_{2,\text{initial}} = m \cdot (-v) = 0.05 , \text{kg} \cdot (-6) , \text{m/s} = -0.3 , \text{kg} \cdot \text{m/s}$
Final momenta:
Ball 1 (rebounding): $p_{1,\text{final}} = m \cdot (-v) = 0.05 , \text{kg} \cdot (-6) , \text{m/s} = -0.3 , \text{kg} \cdot \text{m/s}$
Ball 2 (rebounding): $p_{2,\text{final}} = m \cdot v = 0.05 , \text{kg} \cdot 6 , \text{m/s} = 0.3 , \text{kg} \cdot \text{m/s}$
Change in momentum (Impulse):
For Ball 1: $$ \text{Impulse} = p_{1,\text{final}} - p_{1,\text{initial}} = -0.3 , \text{kg} \cdot \text{m/s} - 0.3 , \text{kg} \cdot \text{m/s} = -0.6 , \text{kg} \cdot \text{m/s} $$
For Ball 2: $$ \text{Impulse} = p_{2,\text{final}} - p_{2,\text{initial}} = 0.3 , \text{kg} \cdot \text{m/s} - (-0.3) , \text{kg} \cdot \text{m/s} = 0.6 , \text{kg} \cdot \text{m/s} $$
The impulse imparted to each ball due to the collision is $0.6 , \text{kg} \cdot \text{m/s}$. Since impulse involves direction, the impulse is $-0.6 , \text{N} \cdot \text{s}$ for Ball 1 and $0.6 , \text{N} \cdot \text{s}$ for Ball 2. Impulse magnitude for both balls is the same: 0.6 $\text{N} \cdot \text{s}$.
A shell of mass $0.020 \mathrm{~kg}$ is fired by a gun of mass $100 \mathrm{~kg}$. If the muzzle speed of the shell is $80 \mathrm{~m} \mathrm{~s}^{-1}$, what is the recoil speed of the gun?
The recoil speed of the gun is:
$$ v_{\text{gun}} = -0.016 , \text{m/s} $$
The negative sign indicates that the direction of the gun's recoil is opposite to the direction of the shell's movement.
Thus, the recoil speed of the gun is 0.016 m/s in the direction opposite to that of the shell.
A batsman deflects a ball by an angle of $45^{\circ}$ without changing its initial speed which is equal to $54 \mathrm{~km} / \mathrm{h}$. What is the impulse imparted to the ball ? (Mass of the ball is $0.15 \mathrm{~kg}$.)
To find the impulse imparted to the ball, we need to calculate the change in momentum. The angle of deflection is $45^\circ$, mass of the ball is $0.15 \text{ kg}$, and speed is $15 \text{ m/s}$.
Impulse Calculation
Initial and Final Velocity Components:
Since the angle of deflection is $45^\circ$, the initial velocity components are:
$ v_{i_x} = 15 \text{ m/s} $
$ v_{i_y} = 0 \text{ m/s} $
The final velocity components after deflection:
$ v_{f_x} = 15 \cos(45^\circ) = 15 \times \frac{ \sqrt{2}}{2} = 15 \times 0.7071 = 10.61 \text{ m/s} $
$ v_{f_y} = 15 \sin(45^\circ) = 15 \times \frac{ \sqrt{2}}{2} = 15 \times 0.7071 = 10.61 \text{ m/s} $
Change in Momentum:
$$ \Delta \mathbf{p} = m \Delta \mathbf{v} $$
The initial momentum vector, (\mathbf{p_i}): $$ \mathbf{p_i} = m \mathbf{v_i} = 0.15 \times (15 \hat{i} + 0 \hat{j}) = 2.25 \hat{i} \text{ kg} \cdot \text{m/s} $$
The final momentum vector, (\mathbf{p_f}): $$ \mathbf{p_f} = m \mathbf{v_f} = 0.15 \times (10.61 \hat{i} + 10.61 \hat{j}) = 1.5915 \hat{i} + 1.5915 \hat{j} \text{ kg} \cdot \text{m/s} $$
The change in momentum: $$ \Delta \mathbf{p} = \mathbf{p_f} - \mathbf{p_i} = (1.5915 - 2.25) \hat{i} + 1.5915 \hat{j} = -0.6585 \hat{i} + 1.5915 \hat{j} \text{ kg} \cdot \text{m/s} $$
Magnitude of Impulse:
$$ |\Delta \mathbf{p}| = \sqrt{(-0.6585)^2 + (1.5915)^2} \approx \sqrt{0.4337 + 2.5339} \approx \sqrt{2.9676} \approx 1.723 \text{ kg} \cdot \text{m/s} $$
Hence, the impulse imparted to the ball is approximately 1.723 N·s.
A stone of mass $0.25 \mathrm{~kg}$ tied to the end of a string is whirled round in a circle of radius $1.5 \mathrm{~m}$ with a speed of $40 \mathrm{rev} . / \mathrm{min}$ in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200 \mathrm{~N}$ ?
The maximum speed $ v_{\max} $ with which the stone can be whirled around if the string can withstand a maximum tension of$ 200 , \text{N} $ is approximately: $$ 34.6 , \text{m/s} $$
Summary
Tension in the String (at $40 , \text{rev/min}$): $6.6 , \text{N}$
Maximum Speed Under Maximum Tension: $34.6 , \text{m/s}$
If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
(a) the stone moves radially outwards,
(b) the stone flies off tangentially from the instant the string breaks,
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
When the string breaks suddenly, the stone is no longer constrained by the centripetal force that was keeping it in circular motion.
According to Newton's First Law of Motion, an object in motion will continue in its state of motion unless acted upon by an external force. At the instant the string breaks, there is no longer a centripetal force acting on the stone.
The stone will continue to move in the direction it was moving at the instant the string broke, which is tangential to the circular path. Therefore, the correct description of the stone's trajectory after the string breaks is:
(b) the stone flies off tangentially from the instant the string breaks.
This is due to the fact that the stone's velocity vector at the moment of the break is tangent to the circle.
Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawn mower than to push it,
(d) a cricketer moves his hands backwards while holding a catch.
Let's examine each of the scenarios using the laws of motion explained in the chapter.
(a) A horse cannot pull a cart and run in empty space
A horse can pull a cart on the ground because of the frictional force provided by the ground. When the horse exerts a force on the ground backward, the ground exerts an equal and opposite force on the horse, pushing it forward, as per Newton's Third Law of Motion. In empty space, there is no ground to push against, so no forward force can be generated. Hence, the horse cannot pull the cart.
(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly
When a bus stops suddenly, the lower part of the passengers' bodies, which is in contact with the seat, experiences a deceleration due to the friction between the seat and the passengers. However, due to inertia (Newton's First Law of Motion), the upper part of their bodies continues to move forward with the same speed as the bus was moving before. This makes the passengers feel as if they are being thrown forward.
(c) It is easier to pull a lawn mower than to push it
When pulling a lawn mower, the force applied is at an angle to the horizontal. This means the vertical component of the pulling force reduces the normal force and consequently the frictional force between the mower and the ground. When pushing, on the other hand, the vertical component of the force acts to increase the normal force, thereby increasing the frictional force. Less friction makes it easier to pull the mower than to push it.
(d) A cricketer moves his hands backwards while holding a catch
When a cricketer catches a ball, moving the hands backward increases the time duration over which the ball comes to a stop. By Newton’s Second Law of Motion and the principle of impulse, the change in momentum is the same, but spreading this change over a longer time period reduces the force experienced by the hands. This minimizes the impact and reduces the chances of injury.
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Ask Chatterbot AIExtra Questions - Laws of Motion | NCERT | Physics | Class 11
A block is released from the state of rest from a wedge as shown. The block and the wedge are of the same mass. If all surfaces are smooth, the total horizontal displacement of the block with respect to earth before it falls on the earth is given by $xR$. Then $x =$
Take $\sqrt{2} = 1.41$.
The problem involves determining the total horizontal displacement of a block sliding off a wedge when both are released from rest and all surfaces are smooth. Key steps involve using physics principles like momentum conservation and energy conservation.
Initial Setup:
Let the displacement of the wedge when the block reaches the bottom be $x$.
Final Position Equation:
Assuming final positions and contribution to motion, we use the formula for final position $x_f$ of the system: $$ x_f = \frac{a + a_1 + R - 2x}{2} $$ Here, $a$ and $a_1$ would generally denote horizontal displacements of the block and wedge; however, in this scenario, we equate initial and final conditions which simplifies to: $$ x_i = x_f $$ So the horizontal displacement $x$, just as the block leaves the wedge is $\frac{R}{2}$.
Parabolic Path and Time of Fall:
Once the block leaves the wedge, it follows a parabolic path. Using the equations of motion, the time $t$ for the block to hit the ground is determined by: $$ R = \frac{1}{2} g t^2 $$ leading to: $$ t = \sqrt{\frac{2R}{g}} $$
Displacement During Fall:
Let $x'$ be the horizontal displacement during fall, and $V_m$ the initial horizontal velocity of the block: $$ x' = V_m t = \sqrt{\frac{2R}{g}} \cdot V_m $$ From the conservation of momentum: $$ m V_m = m V_M \Rightarrow V_m = V_M (equal velocities for block and wedge) $$ Using energy conservation: $$ mgR = mV_m^2 \Rightarrow V_m = \sqrt{gR} $$ Therefore, $$ x' = \sqrt{\frac{2R}{g}} \times \sqrt{gR} = \sqrt{2} R $$
Total Horizontal Displacement:
The total displacement $x_{\text{total}}$ is the sum of the displacement while on the wedge and while falling: $$ x_{\text{total}} = x + x' = 0.5R + \sqrt{2}R = (0.5 + 1.41)R = 1.91R $$
Thus, in the expression given for total horizontal displacement as $xR$, here $x = 1.91$.
Find the minimum force $\mathrm{P}$ required to move a block of weight $\mathrm{W}$ up the incline plane.
A) $m g \frac{(\sin \lambda+\mu \cos \lambda)}{(\cos \alpha-\mu \sin \alpha)}$ B) $m g \frac{(\sin \alpha+\mu \cos \alpha)}{(\cos \alpha+\mu \sin \alpha)}$ C) $m g \frac{(\sin \lambda+\mu \cos \lambda)}{(\cos \alpha+\mu \sin \alpha)}$ D) $m g \frac{(\sin \lambda-\mu \cos \lambda)}{(\cos \alpha-\mu \sin \alpha)}$
The correct answer is Option C.
To determine the minimum force $\mathrm{P}$ needed to move the block up an inclined plane, we consider the equilibrium of forces along and perpendicular to the plane. The frictional force, which opposes the movement, acts downward along the plane, and is at its maximum (limiting) value since the block is just about to move.
Let $\alpha$ be the angle of the incline, $\mu$ the coefficient of friction, and $\mathrm{W} = m g$ the weight of the block. The normal reaction ($N$) and the forces along the inclined plane need to be considered:
The normal force is given by: $$ N = \mathrm{W} \cos \alpha - P \sin \alpha \quad \text{(equation I)} $$
The force balance along the incline gives: $$ P \cos \alpha = \mathrm{W} \sin \alpha + \mu N $$
Substituting for $N$ from equation I: $$ P \cos \alpha = \mathrm{W} \sin \alpha + \mu (\mathrm{W} \cos \alpha - P \sin \alpha) $$ Which rearranges to: $$ P (\cos \alpha + \mu \sin \alpha) = \mathrm{W} (\sin \alpha + \mu \cos \alpha) $$ Solving for $P$, we find: $$ P = \mathrm{W} \frac{\sin \alpha + \mu \cos \alpha}{\cos \alpha + \mu \sin \alpha} $$
This equation is present in Option C, confirming that it is the correct choice.
A curved road of $50 \mathrm{~m}$ radius is banked at the correct angle for a given maximum speed of $\mathrm{v} \mathrm{m/s}$. If the radius of curvature of the road is changed to $200 \mathrm{~m}$ while keeping the same banking angle, the maximum safe speed on the road is:
A) $2 \mathrm{v} \mathrm{m/s}$
B) $0.5 \mathrm{v} \mathrm{m/s}$
C) $3 \mathrm{v} \mathrm{m/s}$
D) $4 \mathrm{v} \mathrm{m/s}$
The correct answer is Option A: $2v$ m/s.
To understand the solution, let's start by analyzing the forces on a car on a banked road. The primary forces are the gravitational force $(mg)$ acting downwards and the normal force $(N)$ from the banked road surface. When a road is banked at angle $\theta$, the normal force $N$ can be resolved into two components:
a horizontal component $N \sin \theta$ providing the necessary centripetal force for circular motion,
a vertical component $N \cos \theta$ balancing the gravitational force.
Based on these considerations:
The equilibrium in the vertical direction gives $N \cos \theta = mg$.
The requirement for the circular motion provides $N \sin \theta = \frac{mv^2}{r}$.
By dividing these two equations: $$ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{v^2}{rg} $$ This formula gives the condition that $\tan \theta$ is directly proportional to $\frac{v^2}{r}$.
When the radius $r$ is changed from $50$ m to $200$ m, keeping the banking angle $\theta$ constant implies $\tan \theta$ remains unchanged. Therefore: $$ \frac{\left(v^{\prime}\right)^2}{200g} = \frac{v^2}{50g} $$
Simplify and solve for $v'$: $$ \left(v^{\prime}\right)^2 = 4v^2 \implies v' = 2v $$
Thus, the maximum safe speed on the road becomes $2v$ m/s when the radius of curvature is increased to 200 m, keeping the same banking angle.
When a person jumps off a boat, there is no force acting on the boat, and thus it will stay still.
A) True
B) False
The correct option is B) False.
According to Newton's third law of motion, "Every action has an equal and opposite reaction."
When a person jumps off a boat, they exert a force on the boat that pushes the boat backward. In response, the boat also exerts an equal and opposite force on the person. This force interaction causes the boat to move, contrary to the implication that the boat would stay still.
A bullet of $5$ g is fired from a pistol of $1.5$ kg. If the recoil velocity of the pistol is $1.5$ m/s, what is the final momentum of the pistol?
A) $10$ kg*m/s
B) $2.5$ kg*m/s
C) $1.5$ kg*m/s
D) $2.25$ kg*m/s
To find the final momentum of the pistol after it is fired, we utilize the principle that momentum is the product of mass and velocity. This can be mathematically expressed as:
$$ \text{Momentum} = \text{Mass} \times \text{Velocity} $$
Here, the mass of the pistol is given as $1.5$ kg and the recoil velocity is $1.5$ m/s. Plugging these values into the formula gives:
$$ \text{Momentum} = 1.5 , \text{kg} \times 1.5 , \text{m/s} $$
$$ \text{Momentum} = 2.25 , \text{kg} \cdot \text{m/s} $$
Thus, the final momentum of the pistol is $2.25$ kg·m/s.
The correct answer is D) $2.25$ kg·m/s.
Who designed locomotives which could pull really heavy loads?
A) James Hargreaves
B) James Watt
C) George Stephenson
D) John Kay
The correct answer is C) George Stephenson.
George Stephenson was instrumental in designing locomotives capable of hauling heavy loads efficiently. He developed a locomotive that could transport hefty loads over a 64-kilometer stretch between Liverpool and Manchester at a speed of 46 kilometers per hour. At that time, this was significantly faster than other available options for transporting heavy loads.
A bullet of mass $60$ g is fired with a velocity of $60$ m/s from a gun of mass $10$ kg. Calculate the recoil velocity of the gun.
To solve for the recoil velocity of the gun, we apply the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act upon it.
Let's define:
$m_1 = 0.06 \text{ kg}$ (mass of the bullet, converted from grams to kilograms),
$v_1 = 60 \text{ m/s}$ (velocity of the bullet),
$m_2 = 10 \text{ kg}$ (mass of the gun),
$v_2$ (recoil velocity of the gun that we need to find).
The initial momentum of the system is entirely due to the moving bullet (the gun is initially at rest), and by conservation of momentum, the momentum before and after the firing must be equal. Thus, the equation for the conservation of momentum is given by:
$$ m_1 v_1 = m_2 v_2 $$
Substituting the values:
$$ 0.06 \cdot 60 = 10 \cdot v_2 $$
Calculating $v_2$:
$$ v_2 = \frac{0.06 \times 60}{10} = 0.36 \text{ m/s} $$
Therefore, the recoil velocity of the gun is $0.36$ m/s.
If the mass of an object is $m$, velocity $v$, acceleration $a$, and applied force is $F$, then the momentum is given by:
A $\mathrm{p} = \frac{m}{v}$
B $\mathrm{p} = m \times a$
C $\mathrm{p} = m \times v$ (D) $p = \frac{a}{m}$
The correct answer is C $\mathrm{p} = m \times v$.
Momentum is mathematically defined as the product of an object's mass and velocity. Thus, the formula for momentum is: $$ p = m \cdot v $$ Here, $p$ represents momentum, $m$ is the mass of the object, and $v$ is the velocity at which the object is moving.
Adjoining figure shows two blocks $A$ and $B$ pushed against the wall with the force $F$. The wall is smooth but the surfaces in contact of $A$ and $B$ are rough. Which of the following is true for the system of blocks to be at rest against the wall:
A. $F$ should be more than the weight of $P \& Q$
B. $F$ should be equal to the weight of $P \& Q$
C. $F$ should be less than the weight of $P \& Q$
D. The system cannot be in equilibrium
To determine the conditions under which the system of blocks $A$ and $B$ will be at rest against the wall, let's analyze the forces acting on the blocks.
Force $F$: This force acts horizontally towards the right.
Rough Surfaces Between A and B: Since the surfaces in contact between $A$ and $B$ are rough, friction ($f_v$) will act to resist relative motion. The frictional force, $f_v$, can be expressed as: $$ f_v = \mu F $$ where $\mu$ is the coefficient of friction and $F$ is the normal force exerted by block $A$ on block $B$.
Normal Force: The normal force here is equal to the applied force $F$, because $F$ is the force pushing block $B$ against block $A$.
Equilibrium Condition of Block A: For block $A$ to be in equilibrium, the frictional force $f_v$ must balance the weight of block $A$: $$ f_v = mg_A $$ Substituting in $f_v$: $$ \mu F = mg_A $$
Weight of Block B: The weight of block $B$ ($mg_B$) acts downwards. Since the wall is smooth, there is no vertical force component from the wall to balance the weight of block $B$.
Since the surface in contact between block $B$ and the wall is smooth, there is no friction to counterbalance the downward force due to the weight of block $B$. This absence of a balancing vertical force means that the system cannot achieve equilibrium.
Thus, the correct answer is:
D. The system cannot be in equilibrium.
Adjoining figure shows two blocks $A$ and $B$ pushed against the wall with the force $F$. The wall is smooth but the surfaces in contact of $A$ and $B$ are rough. Which of the following is true for the system of bloods to be at rest against the wall:
A $F$ should be more than the weight of $P \& Q$
B $F$ should be equal to the weight of $P \& Q$
C $F$ should be less than the weight of $P \& Q$
D The system cannot be in equilibrium
To determine the correct condition for the system of blocks $A$ and $B$ to be at rest against the wall, let's analyze the forces at play:
Applied Force ($F$):
The force $F$ is pushing block $B$ towards the wall, which in turn pushes block $A$ against the wall.
Wall Interaction:
The wall is smooth, which means there is no friction force between block $A$ and the wall.
Friction Between Blocks $A$ and $B$:
The contact surface between block $A$ and block $B$ is rough, so there is a frictional force $f_v$ between them.
Normal Force between $A$ and $B$:
The normal force between blocks $A$ and $B$ is equal to the applied force $F$.
Frictional Force ($f_v$):
The frictional force $f_v$ is given by $f_v = \mu N$, where $\mu$ is the coefficient of friction and $N$ is the normal force. Here, $N = F$, so $f_v = \mu F$.
Considering the forces:
Block $A$ experiences the frictional force $f_v$ upwards, which balances its weight $m_A g$ (where $m_A$ is the mass of block $A$).
Therefore, for block $A$ to be in equilibrium, $f_v = \mu F = m_A g$.
However, for block $B$:
The weight of block $B$ ($m_B g$) is acting downward.
Since the wall is smooth, there is no vertical force to balance the weight of block $B$. The horizontal force $F$ only creates a normal force but does not contribute to balancing the vertical forces.
Conclusion:
Because the system does not have any mechanism to balance the vertical force due to the weight of block $B$, particularly since there is no friction between block $A$ and the wall, the system cannot be in equilibrium.
Therefore, the correct answer is:
D. The system cannot be in equilibrium.
A man of mass 62 kg is standing on a stationary boat of mass 238 kg. The man is carrying a sphere of mass 0.5 kg in his hands. If the man throws the sphere horizontally with a velocity of $12 ms^{-1}$, find the velocity with which the boat will move (in magnitude).
A. $0.02 ~ms^{−1}$
B. $0.5 ~ms^{−1}$
C. $0.04 ~ms^{−1}$
D. $0.06 ~ms^{−1}$
To determine the velocity of the boat when a man throws a sphere horizontally, we'll use the principle of the conservation of linear momentum. Initially, the entire system (man, boat, and sphere) is at rest, so the total initial momentum is zero.
Given Data:
Mass of man: $62 , \text{kg}$
Mass of boat: $238 , \text{kg}$
Mass of sphere: $0.5 , \text{kg}$
Velocity of sphere: $12 , \text{m/s}$ (when thrown horizontally)
Steps to Solve:
Initial Momentum:Since everything is at rest initially, the initial momentum of the system is: $$ p_i = 0 , \text{kg} \cdot \text{m/s} $$
Final Momentum of the Sphere:When the man throws the sphere, its final momentum is: $$ p_{\text{sphere}} = m_{\text{sphere}} \times v_{\text{sphere}} = 0.5 , \text{kg} \times 12 , \text{m/s} = 6 , \text{kg} \cdot \text{m/s} $$
Final Momentum of the Boat-Man System:Let $v$ be the velocity of the boat in the backward direction after the man throws the sphere. The combined mass of the boat and the man is: $$ m_{\text{total}} = m_{\text{man}} + m_{\text{boat}} = 62 , \text{kg} + 238 , \text{kg} = 300 , \text{kg} $$ The momentum of the boat-man system is: $$ p_{\text{boat-man}} = 300 , \text{kg} \times v $$
Applying Conservation of Momentum:According to the conservation of momentum: $$ \text{Total Initial Momentum} = \text{Total Final Momentum} $$ Therefore: $$ 0 = 6 , \text{kg} \cdot \text{m/s} + 300 , \text{kg} \times v $$ Solving for $v$: $$ 300v = -6 $$ $$ v = -\frac{6}{300} = -0.02 , \text{m/s} $$
Magnitude of the Boat's Velocity:The negative sign indicates that the direction of the boat's velocity is opposite to the direction in which the sphere is thrown. For the magnitude: $$ |v| = 0.02 , \text{m/s} $$
Final Answer:
Therefore, the velocity with which the boat will move in magnitude is 0.02 m/s.
Option:
Option A
Figure shows two blocks tied by a string. A variable force $F=5$ t is applied on block A. The coefficient of friction for blocks A and B are 0.5 and 0.6 respectively. Find the tension in the string at time $t=2$ sec (in newtons).
The Free Body Diagram (FBD) of the blocks is as shown in the problem statement.
Given:
The variable force applied on block A is $ F = 5t $.
At $ t = 2 ,\text{s} $: $$ F = 5t = 5 \times 2 = 10 , \text{N} $$
From the FBD, we have the normal forces as: $$ N_A = 20 , \text{N} \ N_B = 10 , \text{N} $$
The maximum value of the frictional force on block A is given by: $$ f_{\text{Amax}} = \mu N_A = 0.5 \times 20 = 10 , \text{N} $$
Thus, at $ t = 2 ,\text{s} $: $$ F = f_{\text{Amax}} $$
This implies that the system will be at rest because the applied force equals the maximum frictional force.
Therefore, the frictional force $ f_A $ will be equal to $ F $, i.e., $ 10 , \text{N}$.
Now, considering the tension in the string $ T $: $$ T + f_A = F \ T + 10 = 10 \ T = 0 $$
Hence, the tension in the string at time $ t = 2 ,\text{s} $ is: $$ \boxed{0 , \text{N}} $$
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