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
- A \(30 \mathrm{~N}\)
- B \(24 \mathrm{~N}\)
- C \(36 \mathrm{~N}\)
- D \(27 \mathrm{~N}\)
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,…
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