Question

Ball \(A\) of mass \(1\,\text{kg}\) moving along a straight line with a velocity of \(4\,\text{m s}^{-1}\) 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. \(30\,\text{N}\)
• B. \(24\,\text{N}\)
• C. \(36\,\text{N}\)
• D. \(27\,\text{N}\)

Hint:

In collision problems, first use conservation of momentum to find the final velocity, then use impulse: \[ F\Delta t=\Delta p \] to calculate the average force.

Answer 1

The Correct Option is A

Step 1: Apply conservation of linear momentum.
Before collision: Ball A (mass $m_1 = 1$ kg, velocity $u_1 = 4$ m/s), Ball B (mass $m_2 = 3$ kg, velocity $u_2 = 0$). Since they stick together (perfectly inelastic collision): \[ m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \] \[ 1 \times 4 + 3 \times 0 = (1 + 3)v \] \[ 4 = 4v \implies v = 1 \text{ m/s} \]
Step 2: Find the change in momentum of ball B.
Ball B was initially at rest ($u_2 = 0$) and after collision it moves at $v = 1$ m/s. Change in momentum of B: \[ \Delta p_B = m_2(v - u_2) = 3(1 - 0) = 3 \text{ kg m/s} \]
Step 3: Apply the impulse-momentum theorem.
The impulse equals the change in momentum: \[ F \cdot \Delta t = \Delta p_B \] Given time of impact $\Delta t = 0.1$ s: \[ F \times 0.1 = 3 \] \[ F = \frac{3}{0.1} = 30 \text{ N} \]
Step 4: Understand what force this represents.
This 30 N is the average force exerted on ball B by ball A during the collision. By Newton's third law, ball B also exerts 30 N on ball A in the opposite direction.
Step 5: Verify the calculation.
Momentum transferred to B = $3 \times 1 = 3$ N\cdots. Time = 0.1 s. Force = $3/0.1 = 30$ N. Correct.
Step 6: State the final answer.
The force exerted on ball B during the collision is 30 N. \[ \boxed{30 \text{ N}} \]