Question

Ball A of mass $ 1\, \text{kg} $ moving along a straight line with a velocity of $ 4\, \text{m/s} $ hits another ball B of mass $ 3\, \text{kg} $ which is at rest. After collision, they stick together and move with the same velocity along the same straight line. If the time of impact of the collision is $ 0.1\, \text{s} $, then the force exerted on B is:

• A. 30 N
• B. 24 N
• C. 36 N
• D. 27 N

Hint:

Use impulse-momentum theorem: \( F = \frac{\Delta p}{\Delta t} \)

Answer 1

The Correct Option is A

Initial momentum: \[ P_i = 1 \cdot 4 + 3 \cdot 0 = 4\, \text{kg·m/s} \] Total mass after sticking = \( 1 + 3 = 4\, \text{kg} \) Final velocity: \[ v_f = \frac{4}{4} = 1\, \text{m/s} \] Change in momentum of B: \[ \Delta p_B = 3 \cdot 1 - 0 = 3\, \text{kg·m/s} \] Force exerted on B: \[ F = \frac{\Delta p}{\Delta t} = \frac{3}{0.1} = 30\, \text{N} \]