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Two spheres A and B of masses m1m_1 and m2m_2 respectively collide. A is at rest initially and B is moving with velocity vv along x-axis. After collision B has a velocity v2\frac{v}{2} in a direction perpendicular to the original direction. The mass A moves after collision in the direction:

A

same as that of B

B

opposite to that of B

C

θ=tan⁡−1(12)\theta = \tan^{-1}(\frac{1}{2}) to the x-axis

D

θ=tan⁡−1(−12)\theta = \tan^{-1}(-\frac{1}{2}) to the x-axis

Step-by-Step Solution

  1. Identify the System: This is a two-dimensional collision problem. Sphere A (mass m1m_1) is initially stationary (uA=0u_A = 0). Sphere B (mass m2m_2) moves along the x-axis with velocity vv (uB=vi^u_B = v\hat{i}). After collision, B moves perpendicular to the original direction (along y-axis) with velocity v/2v/2 (vB=v2j^v_B = \frac{v}{2}\hat{j}).
  2. Apply Conservation of Linear Momentum: Total initial momentum equals total final momentum. P⃗initial=m1uA+m2uB=0+m2vi^=m2vi^\vec{P}_{initial} = m_1 u_A + m_2 u_B = 0 + m_2 v \hat{i} = m_2 v \hat{i} P⃗final=m1v⃗A+m2v⃗B=m1v⃗A+m2v2j^\vec{P}_{final} = m_1 \vec{v}_A + m_2 \vec{v}_B = m_1 \vec{v}_A + m_2 \frac{v}{2} \hat{j} Equating momenta: m2vi^=m1v⃗A+m2v2j^m_2 v \hat{i} = m_1 \vec{v}_A + m_2 \frac{v}{2} \hat{j}.
  3. Solve for v⃗A\vec{v}_A: m1v⃗A=m2vi^−m2v2j^m_1 \vec{v}_A = m_2 v \hat{i} - m_2 \frac{v}{2} \hat{j} v⃗A=m2vm1i^−m2v2m1j^\vec{v}_A = \frac{m_2 v}{m_1} \hat{i} - \frac{m_2 v}{2m_1} \hat{j}
  4. Determine Direction (θ\theta): The angle θ\theta with the x-axis is given by tan⁡θ=vAyvAx\tan \theta = \frac{v_{Ay}}{v_{Ax}}. tan⁡θ=−m2v2m1m2vm1=−12\tan \theta = \frac{- \frac{m_2 v}{2m_1}}{\frac{m_2 v}{m_1}} = -\frac{1}{2} θ=tan⁡−1(−12)\theta = \tan^{-1}\left(-\frac{1}{2}\right) This indicates the direction is at an angle below the x-axis [Class 11 Physics, Ch 6, Sec 5.11.3].
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