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(A) The final velocity $v_1$ of the first particle after a one-dimensional elastic collision is given by the formula:$v_1 = \left( \frac{m_1 - m_2}{m_

Vedclass · Accepted Answer

$2 \, m/s$ in the original direction

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(A) The final velocity $v_1$ of the first particle after a one-dimensional elastic collision is given by the formula:$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2$Given $m_1 = m$ and $m_2 = M$. Since $m Substituting these into the formula:$v_1 = \left( \frac{0 - M}{0 + M} \right) u_1 + \left( \frac{2M}{0 + M} \right) u_2$$v_1 = -u_1 + 2u_2$Given $u_1 = 6 \, m/s$ and $u_2 = 4 \, m/s$ (both in the same direction):$v_1 = -6 + 2(4) = -6 + 8 = 2 \, m/s$The positive sign indicates that the lighter particle moves in the original direction with a speed of $2 \, m/s$.

$A$ particle of mass $m$ is moving with a horizontal speed of $6 \, m/s$ as shown in the figure. If $m << M$,then for a one-dimensional elastic collision,the speed of the lighter particle after the collision will be:

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