# Laws of Motion - Class 11 - Physics

Renews every month. Cancel anytime

### Your personal doubt-solving assistant

Chatterbot AI gives you 100% accurate answers to your questions in an instant.

## Extra 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) $m g \frac{(\sin \lambda-\mu \cos \lambda)}{(\cos \alpha-\mu \sin \alpha)}$

### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now### Improve your grades!

Join English Chatterbox to access detailed and curated answers, and score higher than you ever have in your exams.

Sign up now