WBJEE · Physics · Center of Mass Momentum and Collision
A spherical ball \(A\) of mass 4 kg, moving along a straight line strikes another spherical ball \(B\) of mass \(1 \mathrm{kg}\) at rest. After the collision, \(A\) and \(B\) move with velocities \(y_{1} \mathrm{ms}^{-1}\) and \(v_{2} \mathrm{ms}^{-1}\) respectively making angles of \(30^{\circ}\) and \(60^{\circ}\) with
respect to the original direction of motion of \(A .\) The ratio \(\frac{y_{1}}{v_{2}}\) will be
- A \(\frac{\sqrt{3}}{4}\)
- B \(\frac{4}{\sqrt{3}}\)
- C \(\frac{1}{\sqrt{3}}\)
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}}{4}\)
Step-by-step Solution
Detailed explanation
Along \(y\) -axis momentum remains zero. Here \[ \begin{aligned} 4 v_{1} \sin 30^{\circ} &=v_{2} \sin 60^{\circ} \\ \frac{y_{1}}{v_{2}} &=\frac{\sqrt{3}}{4} \end{aligned} \]
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