Question

A block of mass m moving towards right with a velocity V strikes (head on) another block of mass M which is at rest connected to a spring. The coefficient of restitution for collision between the blocks is e = 0.5. Find the value of M for which the subsequent compression in the spring is maximum. There is no friction.

• A. m
• B. 2m
• C. $\frac{m}{2}$
• D. $\frac{m}{4}$

Answer 1

Answer: (A)

m

Answer 2

1. Collision Phase:

Let the mass of the first block be $m$ and its initial velocity be $V$. Let the mass of the second block be $M$ and its initial velocity be $0$. Let the velocities after collision be $v_m$ and $v_M$ for $m$ and $M$ respectively.

• Conservation of Momentum: $m V + M(0) = m v_m + M v_M$ $m V = m v_m + M v_M$ (Eq. 1)

• Coefficient of Restitution (e = 0.5): $e = (v_M - v_m) / (V - 0)$ $0.5 V = v_M - v_m$ $v_m = v_M - 0.5 V$ (Eq. 2)

Substitute (Eq. 2) into (Eq. 1): $m V = m (v_M - 0.5 V) + M v_M$ $m V = m v_M - 0.5 m V + M v_M$ $1.5 m V = (m + M) v_M$ $v_M = (1.5 m V) / (m + M)$

2. Spring Compression Phase:

The spring compression is maximum when the kinetic energy of block $M$ (which is connected to the spring) is fully converted into elastic potential energy of the spring. This happens when the velocity of $M$ momentarily becomes zero. We need to maximize $x_{max}$, where $(1/2) k x_{max}^2 = (1/2) M v_M^2$. Therefore, we need to maximize $M v_M^2$. Substitute the expression for $v_M$: $M v_M^2 = M * [(1.5 m V) / (m + M)]^2$ $M v_M^2 = (2.25 m^2 V^2) * [M / (m + M)^2]$ To maximize $M v_M^2$, we need to maximize the function $f(M) = M / (m + M)^2$.

3. Maximizing $f(M)$:

To find the maximum, we differentiate $f(M)$ with respect to $M$ and set the derivative to zero: $f'(M) = d/dM [M / (m + M)^2]$ Using the quotient rule: $f'(M) = [1 * (m + M)^2 - M * 2(m + M)] / [(m + M)^2]^2$ $f'(M) = [(m + M) - 2M] / (m + M)^3$ $f'(M) = (m - M) / (m + M)^3$ Set $f'(M) = 0$: $(m - M) / (m + M)^3 = 0$ This implies $m - M = 0$, so $M = m$. (We can verify this is a maximum by checking the second derivative, which will be negative at $M=m$).

4. Check for separation:

If $M = m$, then $v_m = V * (m - 0.5m) / (m + m) = V * (0.5m) / (2m) = 0.25 V$. And $v_M = (1.5 m V) / (m + m) = 0.75 V$. Since $v_M > v_m$, block $m$ separates from $M$ after the collision. Thus, the assumption that only $M$'s kinetic energy is converted to spring potential energy is valid.

The value of $M$ for which the subsequent compression in the spring is maximum is $m$.