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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For a one-dimensional elastic collision,the final velocities $v_1'$ and $v_2'$ are given by:$v_1' = \left( \frac{m_1 - m_2}{m_1 + m_2} ight) v_1 + \left( \frac{2m_2}{m_1 + m_2} ight) v_2$$v_2' = \left( \frac{2m_1}{m_1 + m_2} ight) v_1 + \left( \frac{m_2 - m_1}{m_1 + m_2} ight) v_2$Given: $m_1 = 10 \ kg, v_1 = 10 \ m/s, m_2 = 5 \ kg, v_2 = 4 \ m/s$.Calculating $v_1'$:$v_1' = \left( \frac{10 - 5}{10 + 5} ight) 10 + \left( \frac{2 	imes 5}{10 + 5} ight) 4 = \left( \frac{5}{15} ight) 10 + \left( \frac{10}{15} ight) 4 = \frac{10}{3} + \frac{8}{3} = \frac{18}{3} = 6 \ m/s$.Calculating $v_2'$:$v_2' = \left( \frac{2 	imes 10}{10 + 5} ight) 10 + \left( \frac{5 - 10}{10 + 5} ight) 4 = \left( \frac{20}{15} ight) 10 + \left( \frac{-5}{15} ight) 4 = \frac{40}{3} - \frac{4}{3} = \frac{36}{3} = 12 \ m/s$.Thus,the velocities are $6 \ m/s$ and $12 \ m/s$.

$A$ sphere of mass $10 \ kg$ moving with a velocity of $10 \ m/s$ undergoes an elastic collision with a body of mass $5 \ kg$ moving in the same direction with a velocity of $4 \ m/s$. What are their velocities after the collision?

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