Two spheres A and B of masses $m_1$ and $m_2$ respectively collide. A is at rest initially and B is moving wit

Question

Two spheres A and B of masses $m_1$ and $m_2$ respectively collide. A is at rest initially and B is moving with velocity $v$ along x-axis. After collision B has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction. The mass A moves after collision in the direction:

Accepted Answer

1. **Identify the System:** This is a two-dimensional collision problem. Sphere A (mass $m_1$) is initially stationary ($u_A = 0$). Sphere B (mass $m_2$) moves along the x-axis with velocity $v$ ($u_B = v\hat{i}$). After collision, B moves perpendicular to the original direction (along y-axis) with velocity $v/2$ ($v_B = \frac{v}{2}\hat{j}$).
2. **Apply Conservation of Linear Momentum:** Total initial momentum equals total final momentum.
   $$ \vec{P}_{initial} = m_1 u_A + m_2 u_B = 0 + m_2 v \hat{i} = m_2 v \hat{i} $$
   $$ \vec{P}_{final} = m_1 \vec{v}_A + m_2 \vec{v}_B = m_1 \vec{v}_A + m_2 \frac{v}{2} \hat{j} $$
   Equating momenta: $m_2 v \hat{i} = m_1 \vec{v}_A + m_2 \frac{v}{2} \hat{j}$.
3. **Solve for $\vec{v}_A$:**
   $$ m_1 \vec{v}_A = m_2 v \hat{i} - m_2 \frac{v}{2} \hat{j} $$
   $$ \vec{v}_A = \frac{m_2 v}{m_1} \hat{i} - \frac{m_2 v}{2m_1} \hat{j} $$
4. **Determine Direction ($	heta$):** The angle $	heta$ with the x-axis is given by $	an 	heta = \frac{v_{Ay}}{v_{Ax}}$.
   $$ 	an 	heta = \frac{- \frac{m_2 v}{2m_1}}{\frac{m_2 v}{m_1}} = -\frac{1}{2} $$
   $$ 	heta = 	an^{-1}\left(-\frac{1}{2}ight) $$
   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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