Question

Two spheres \( S_1 \) and \( S_2 \) of masses \( m_1 \) and \( m_2 \) collide. Initially \( S_1 \) is at rest and \( S_2 \) moves with velocity \( v \) along x-axis. After collision \( S_2 \) has velocity \( \frac{v}{2} \) in a direction perpendicular to the original direction. The motion of \( S_1 \) after collision is

• A. velocity magnitude \( \frac{m_2 v}{m_1}\frac{\sqrt{5}}{2} \)
• B. direction \( \theta = \tan^{-1}\left(-\frac{1}{3}\right) \)
• C. direction makes angle \( \theta \) such that \( \theta = \tan^{-1}\left(\frac12\right) \) or \( \tan^{-1}\left(-\frac12\right) \)
• D. velocity magnitude \( \frac{m_1}{2m_2} v\sqrt{5} \)

Hint:

2D collisions: \begin{itemize} \item Apply momentum conservation in x and y separately. \end{itemize}

Answer 1

The Correct Option is A

Concept: Conservation of momentum (vector) Initial momentum: \[ \vec{P}_i = m_2 v \hat{i} \] Step 1: {\color{red}Final momentum of \( S_2 \).} After collision, velocity is perpendicular to x-axis (say along y-axis): \[ \vec{v}_2 = \frac{v}{2}\hat{j} \] Momentum: \[ \vec{P}_2 = m_2 \frac{v}{2}\hat{j} \] Step 2: {\color{red}Momentum of \( S_1 \).} Let velocity of \( S_1 \) be \( \vec{u} = u_x \hat{i} + u_y \hat{j} \). Using conservation: \[ m_1 u_x = m_2 v \] \[ m_1 u_y = -\frac{m_2 v}{2} \] Step 3: {\color{red}Velocity magnitude.} \[ u = \sqrt{u_x^2 + u_y^2} = \frac{m_2 v}{m_1}\sqrt{1 + \frac14} = \frac{m_2 v}{m_1}\frac{\sqrt{5}}{2} \] So (A) correct. Step 4: {\color{red}Direction.} \[ \tan\theta = \frac{u_y}{u_x} = -\frac{1}{2} \] Angle could be in 4th quadrant: \[ \theta = \tan^{-1}\left(\pm\frac12\right) \] So (C) correct.