Question

A ball of mass 10 g moving perpendicular to the plane of the wall strikes it and rebounds in the same line with the same velocity. If the impulse experienced by the wall is 0.54 Ns, the velocity of the ball is

• A. 27 ms$^{-1}$
• B. 3.7 ms$^{-1}$
• C. 54 ms$^{-1}$
• D. 37 ms$^{-1}$
• E. 5.4 ms$^{-1}$

Hint:

When an object rebounds with the same speed, the change in momentum is always $2mv$. This simple shortcut helps avoid common errors with sign conventions 40].

Answer 1

The Correct Option is A

Concept: This problem is based on the impulse-momentum theorem, which states that the impulse applied to an object is equal to its change in linear momentum.
• Impulse ($J$): $J = \Delta p = p_f - p_i$ 171].
• Momentum ($p$): $p = mv$.
• Sign Convention: If the initial velocity is $+v$, then the rebound velocity in the opposite direction is $-v$.

Step 1: Calculate the change in momentum.
Given mass $m = 10 \text{ g} = 0.01 \text{ kg}$. Let the velocity be $v$. Initial momentum: $p_i = m(v) = 0.01v$. Final momentum: $p_f = m(-v) = -0.01v$. Change in momentum ($\Delta p$): $p_f - p_i = -0.01v - 0.01v = -0.02v$. Magnitude of change in momentum: $|\Delta p| = 0.02v$.

Step 2: Solve for velocity using the given impulse.
The impulse $J = 0.54 \text{ Ns}$. According to the theorem: \[ J = |\Delta p| \] \[ 0.54 = 0.02v \] \[ v = \frac{0.54}{0.02} = \frac{54}{2} = 27 \text{ ms}^{-1} \]