Question

A plank AB of mass $M$ and length $l$ is placed on a frictionless horizontal floor with its end B in contact with a fixed stop. A small block of mass $m$ is placed on the end A of plank as shown in the figure. Coefficient of friction between the plank and the block is $\mu$. The block is given a velocity towards the fixed stop. If after a perfectly elastic impact with the fixed stop the block returns to the end A of the plank and stops sliding, deduce suitable expression for the velocity $v_0$ initially given to the block.

Answer 1

v_0 = \sqrt{2\mu gl \left( \frac{2M+m}{M} \right)}

Answer 2

The problem can be divided into three phases:

1. Block slides on the plank from A to B (towards the fixed stop).
2. Block undergoes a perfectly elastic impact with the fixed stop.
3. Block slides back on the plank from B to A until it stops sliding relative to the plank.

Phase 1: Block slides from A to B

Initially, the block of mass $m$ is given a velocity $v_0$ towards the fixed stop. The plank of mass $M$ is at rest with its end B in contact with the fixed stop. Since the floor is frictionless and the plank is against a fixed stop, the plank remains stationary during this phase.

The block experiences a kinetic friction force $f_k = \mu mg$ acting opposite to its motion (i.e., towards the left). The acceleration of the block is $a_b = -\mu g$.

Let $v_1$ be the velocity of the block just before it hits the fixed stop (at end B). The block travels a distance $l$. Using the kinematic equation $v^2 = u^2 + 2as$:

$v_1^2 = v_0^2 + 2(-\mu g)l$ $v_1^2 = v_0^2 - 2\mu gl$ (Equation 1)

Phase 2: Perfectly elastic impact of the block with the fixed stop

The block hits the fixed stop at end B with velocity $v_1$ (towards the right). Since the impact is perfectly elastic, the block rebounds with the same speed but in the opposite direction.

So, immediately after impact, the velocity of the block becomes $-v_1$ (towards the left). The plank remains at rest as it was not directly involved in the impact and was not moving.

Phase 3: Block slides from B back to A, plank starts moving

After impact, the block moves left with initial velocity $u_b = -v_1$. The plank is now free to move (as the block is moving away from the stop) and its initial velocity is $u_p = 0$.

Friction acts between the block and the plank. Since the block is moving left relative to the plank, the kinetic friction force on the block is $f_k = \mu mg$ (acting towards the right). By Newton's third law, the friction force on the plank is $f_k = \mu mg$ (acting towards the left).

Conservation of Momentum:

Consider the system of the block and the plank. Since the floor is frictionless and the friction between the block and plank is an internal force, the total horizontal momentum of the block-plank system is conserved during this phase.

Initial momentum of the system (just after impact): $P_i = m(-v_1) + M(0) = -mv_1$. The block stops sliding when its velocity becomes equal to the velocity of the plank. Let this common final velocity be $v_f$. Final momentum of the system: $P_f = (m+M)v_f$.

By conservation of momentum: $-mv_1 = (m+M)v_f$ $v_f = -\frac{mv_1}{M+m}$ (The negative sign indicates the common velocity is towards the left).

Energy Dissipation due to Friction:

The block returns to end A of the plank, meaning its relative displacement with respect to the plank is $l$.

The energy dissipated due to friction is given by the product of the friction force and the relative distance moved: $E_{diss} = f_k \times (\text{relative distance}) = \mu mg l$.

Work-Energy Theorem for the system:

The change in the total kinetic energy of the system is equal to the negative of the energy dissipated by friction. $KE_{final} - KE_{initial} = -E_{diss}$ $KE_{initial} = \frac{1}{2} m u_b^2 + \frac{1}{2} M u_p^2 = \frac{1}{2} m (-v_1)^2 + \frac{1}{2} M (0)^2 = \frac{1}{2} m v_1^2$. $KE_{final} = \frac{1}{2} (m+M) v_f^2 = \frac{1}{2} (m+M) \left( -\frac{mv_1}{M+m} \right)^2$ $KE_{final} = \frac{1}{2} (m+M) \frac{m^2 v_1^2}{(M+m)^2} = \frac{1}{2} \frac{m^2 v_1^2}{M+m}$.

Now, substitute these into the work-energy equation: $\frac{1}{2} \frac{m^2 v_1^2}{M+m} - \frac{1}{2} m v_1^2 = -\mu mg l$ Multiply by 2 and divide by $m$: $v_1^2 \left( \frac{m}{M+m} - 1 \right) = -2\mu g l$ $v_1^2 \left( \frac{m - (M+m)}{M+m} \right) = -2\mu g l$ $v_1^2 \left( \frac{-M}{M+m} \right) = -2\mu g l$ $v_1^2 \frac{M}{M+m} = 2\mu g l$ $v_1^2 = \frac{2\mu g l (M+m)}{M}$ (Equation 2)

Combining results from Phase 1 and Phase 3:

Substitute $v_1^2$ from Equation 2 into Equation 1: $\frac{2\mu g l (M+m)}{M} = v_0^2 - 2\mu gl$ $v_0^2 = 2\mu gl + \frac{2\mu g l (M+m)}{M}$ Factor out $2\mu gl$: $v_0^2 = 2\mu gl \left( 1 + \frac{M+m}{M} \right)$ $v_0^2 = 2\mu gl \left( \frac{M + M+m}{M} \right)$ $v_0^2 = 2\mu gl \left( \frac{2M+m}{M} \right)$

Finally, solve for $v_0$: $v_0 = \sqrt{2\mu gl \left( \frac{2M+m}{M} \right)}$

The final answer is $\boxed{\sqrt{2\mu gl \left( \frac{2M+m}{M} \right)}}$.

Explanation of the solution:

1. Initial motion (Block A to B): The block moves towards the fixed stop. The plank is stationary. Friction opposes block's motion, causing deceleration. The velocity $v_1$ just before impact is found using kinematics.
2. Elastic Impact: The block hits the fixed stop and rebounds with velocity $-v_1$. The plank remains unaffected.
3. Return motion (Block B to A): The block moves left, and the plank, now free, also starts moving left due to friction from the block.
   • Momentum Conservation: For the block-plank system, horizontal momentum is conserved as there are no external horizontal forces. This gives the final common velocity $v_f$ when sliding stops.
   • Energy Conservation with Friction: The initial kinetic energy of the system (block only) minus the final kinetic energy (block+plank moving together) equals the energy dissipated by friction. The energy dissipated is $\mu mg$ times the relative distance $l$.
4. Combining: Equate the expression for $v_1^2$ from the initial motion to the expression for $v_1^2$ derived from the return motion and energy conservation to solve for $v_0$.