Question

A plank AB of mass $M$ and length $l$ is placed on a frictionless horizontal floor with its end B at some distance away from a fixed stop. A small block of mass $m$ is placed on the end A of plank and the block as shown in the figure. Coefficient of friction between the plank and the block is $\mu$. The plank is given a velocity towards the fixed stop. If the collision of the plank with the fixed stop is perfectly elastic, for what maximum value of $v_0$ the block will not slide off the plank?

• A. $\sqrt{\frac{\mu g l (M+m)}{2M}}$

Answer 1

Answer: (A)

$\sqrt{\frac{\mu g l (M+m)}{2M}}$

Answer 2

The problem asks for the maximum initial velocity $v_0$ of the plank such that the block does not slide off the plank. We need to consider the sequence of events.

Assumptions:

1. Initially, the plank and the block move together with velocity $v_0$. This is the most common interpretation when "the plank is given a velocity" and a block is "placed" on it, implying they start as a single unit, and the critical event for relative motion is a sudden change in motion like a collision. If they move together, there is no relative motion and thus no friction acting before the collision.
2. The collision of the plank with the fixed stop is perfectly elastic.

Phases of Motion:

Phase 1: Motion before collision

• The plank (mass $M$) and the block (mass $m$) move together with velocity $v_0$ towards the fixed stop.
• Since there is no relative motion, the friction force between them is static, and it is sufficient to prevent sliding.

Phase 2: Collision of the plank with the fixed stop

• The plank hits the fixed stop with velocity $v_0$.
• Since the collision is perfectly elastic, the plank's velocity instantaneously reverses direction and magnitude, becoming $-v_0$ (towards the left).
• At the instant of collision, the block does not directly interact with the stop. Its velocity remains $v_0$ (towards the right).

Phase 3: Motion after collision (relative motion between block and plank)

• Immediately after the collision:
  • Plank's velocity: $V_P = -v_0$ (left)
  • Block's velocity: $V_B = v_0$ (right)
• The block is at end A of the plank.
• The relative velocity of the block with respect to the plank is $V_{B/P} = V_B - V_P = v_0 - (-v_0) = 2v_0$. This relative velocity is directed towards the right (from A to B).
• Since the block is moving to the right relative to the plank, the kinetic friction force on the block acts to the left: $f_b = -\mu mg$.
• The acceleration of the block is $a_B = \frac{f_b}{m} = -\mu g$.
• By Newton's third law, the kinetic friction force on the plank acts to the right: $f_p = \mu mg$.
• The acceleration of the plank is $a_P = \frac{f_p}{M} = \frac{\mu mg}{M}$.
• The relative acceleration of the block with respect to the plank is $a_{B/P} = a_B - a_P = -\mu g - \frac{\mu mg}{M} = -\mu g \left(1 + \frac{m}{M}\right) = -\mu g \frac{M+m}{M}$.
• Since the relative velocity is positive (right) and the relative acceleration is negative (left), the block will slow down relative to the plank and eventually stop sliding.

Condition for the block not to slide off the plank:

• The block starts at end A and moves towards end B relative to the plank.
• For the block not to slide off, its relative displacement must be less than or equal to the length of the plank, $l$.
• The maximum value of $v_0$ occurs when the block just stops sliding relative to the plank exactly at end B. This means the relative displacement is $\Delta x_{rel} = l$, and the final relative velocity is $V_{rel,f} = 0$.
• Using the kinematic equation for relative motion: $V_{rel,f}^2 = V_{rel,i}^2 + 2 a_{rel} \Delta x_{rel}$ $0^2 = (2v_0)^2 + 2 \left(-\mu g \frac{M+m}{M}\right) l$ $0 = 4v_0^2 - 2\mu g l \frac{M+m}{M}$ $4v_0^2 = 2\mu g l \frac{M+m}{M}$ $v_0^2 = \frac{2\mu g l (M+m)}{4M}$ $v_0^2 = \frac{\mu g l (M+m)}{2M}$ $v_0 = \sqrt{\frac{\mu g l (M+m)}{2M}}$

This is the maximum value of $v_0$ for which the block will not slide off the plank.

The final answer is $\boxed{\sqrt{\frac{\mu g l (M+m)}{2M}}}$.

Explanation of the solution:

1. Initial State: The plank and block move together with velocity $v_0$ towards the fixed stop.
2. Collision: The plank undergoes a perfectly elastic collision with the fixed stop, reversing its velocity to $-v_0$. The block's velocity remains $v_0$.
3. Relative Motion: After the collision, the block moves right with $v_0$ and the plank moves left with $-v_0$. This creates a relative velocity of $2v_0$ (block relative to plank, to the right).
4. Friction and Acceleration: Kinetic friction acts to slow down the block relative to the plank and accelerate the plank. The relative acceleration is calculated.
5. No Sliding Condition: For the block not to slide off, its relative displacement (starting from end A, moving towards end B) must not exceed the plank's length $l$. The maximum $v_0$ occurs when the block just reaches end B and its relative velocity becomes zero.
6. Kinematics: Using the kinematic equation relating initial relative velocity, final relative velocity, relative acceleration, and relative displacement, the expression for $v_0$ is derived.