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(A) For a one-dimensional elastic collision,the final velocities $v_1$ and $v_2$ are given by the formulas:$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \

Vedclass · Accepted Answer

$6\, m/s, 12\, m/s$

Vedclass · Answer

(A) For a one-dimensional elastic collision,the final velocities $v_1$ and $v_2$ are given by the formulas:$v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} ight) u_1 + \left( \frac{2m_2}{m_1 + m_2} ight) u_2$$v_2 = \left( \frac{2m_1}{m_1 + m_2} ight) u_1 + \left( \frac{m_2 - m_1}{m_1 + m_2} ight) u_2$Given: $m_1 = 10\, kg, u_1 = 10\, m/s, m_2 = 5\, kg, u_2 = 4\, m/s$.Substituting the values:$v_1 = \left( \frac{10 - 5}{10 + 5} ight) 10 + \left( \frac{2 	imes 5}{10 + 5} ight) 4 = \left( \frac{5}{15} ight) 10 + \left( \frac{10}{15} ight) 4 = \frac{50}{15} + \frac{40}{15} = \frac{90}{15} = 6\, m/s$.$v_2 = \left( \frac{2 	imes 10}{10 + 5} ight) 10 + \left( \frac{5 - 10}{10 + 5} ight) 4 = \left( \frac{20}{15} ight) 10 + \left( \frac{-5}{15} ight) 4 = \frac{200}{15} - \frac{20}{15} = \frac{180}{15} = 12\, m/s$.Thus,the velocities are $6\, m/s$ and $12\, m/s$.

$A$ ball of mass $10\, kg$ is moving with a velocity of $10\, m/s$. It strikes another ball of mass $5\, kg$ which is moving in the same direction with a velocity of $4\, m/s$. If the collision is elastic,their velocities after the collision will be,respectively

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