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AP EAMCET · PHYSICS · Center of Mass Momentum and Collision

Ball \(A\) of mass \(1 \mathrm{~kg}\) moving along a straight line with a velocity of \(4 \mathrm{~ms}^{-1}\) hits another ball \(B\) of mass \(3 \mathrm{~kg}\) 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 \(0.1 \mathrm{~s}\) then the force exerted on \(B\) is

  1. A \(30 \mathrm{~N}\)
  2. B \(24 \mathrm{~N}\)
  3. C \(36 \mathrm{~N}\)
  4. D \(27 \mathrm{~N}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(30 \mathrm{~N}\)

Step-by-step Solution

Detailed explanation

For ball \(A\), \(\begin{aligned} & m_A=1 \mathrm{~kg}, v_A=4 \mathrm{~ms}^{-1} \\ & \text { for ball } B, m_B=3 \mathrm{~kg}, v_B=0\end{aligned}\) \(\therefore\) Total momentum before collision,…