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
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
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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.
Updated On: Jun 22, 2026
- \(30\,\text{N}\)
- \(24\,\text{N}\)
- \(36\,\text{N}\)
- \(27\,\text{N}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use conservation of linear momentum.
Before collision, ball \(A\) has mass \[ m_1=1\,\text{kg} \] and velocity \[ u_1=4\,\text{m s}^{-1} \] Ball \(B\) has mass \[ m_2=3\,\text{kg} \] and velocity \[ u_2=0 \] After collision, both balls stick together and move with common velocity \(v\).
Using conservation of momentum, \[ m_1u_1+m_2u_2=(m_1+m_2)v \] Substituting values, \[ 1\times 4+3\times 0=(1+3)v \] \[ 4=4v \] \[ v=1\,\text{m s}^{-1} \]
Step 2: Find change in momentum of ball \(B\).
Initial velocity of ball \(B\) is \[ 0 \] Final velocity of ball \(B\) is \[ 1\,\text{m s}^{-1} \] Change in momentum of ball \(B\) is \[ \Delta p=m_2(v-u_2) \] \[ \Delta p=3(1-0) \] \[ \Delta p=3\,\text{kg m s}^{-1} \]
Step 3: Use impulse-force relation.
Impulse is equal to change in momentum: \[ F\Delta t=\Delta p \] Given, \[ \Delta t=0.1\,\text{s} \] Therefore, \[ F=\frac{\Delta p}{\Delta t} \] \[ F=\frac{3}{0.1} \] \[ F=30\,\text{N} \]
Step 4: Final conclusion.
Therefore, the force exerted on \(B\) is \[ \boxed{30\,\text{N}} \]
Before collision, ball \(A\) has mass \[ m_1=1\,\text{kg} \] and velocity \[ u_1=4\,\text{m s}^{-1} \] Ball \(B\) has mass \[ m_2=3\,\text{kg} \] and velocity \[ u_2=0 \] After collision, both balls stick together and move with common velocity \(v\).
Using conservation of momentum, \[ m_1u_1+m_2u_2=(m_1+m_2)v \] Substituting values, \[ 1\times 4+3\times 0=(1+3)v \] \[ 4=4v \] \[ v=1\,\text{m s}^{-1} \]
Step 2: Find change in momentum of ball \(B\).
Initial velocity of ball \(B\) is \[ 0 \] Final velocity of ball \(B\) is \[ 1\,\text{m s}^{-1} \] Change in momentum of ball \(B\) is \[ \Delta p=m_2(v-u_2) \] \[ \Delta p=3(1-0) \] \[ \Delta p=3\,\text{kg m s}^{-1} \]
Step 3: Use impulse-force relation.
Impulse is equal to change in momentum: \[ F\Delta t=\Delta p \] Given, \[ \Delta t=0.1\,\text{s} \] Therefore, \[ F=\frac{\Delta p}{\Delta t} \] \[ F=\frac{3}{0.1} \] \[ F=30\,\text{N} \]
Step 4: Final conclusion.
Therefore, the force exerted on \(B\) is \[ \boxed{30\,\text{N}} \]
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