Question

A 10 kg ball and a 20 kg ball approach each other with velocities 20 m/s and 10 m/s respectively. What are their velocities after collision, if the collision is perfectly elastic

Answer 1

Mass of first ball, $\sf{m_1=10\ kg.}$

Mass of second ball, $\sf{m_2=20\ kg.}$

Initial velocity of first ball, $\sf{u_1=20\ m\,s^{-1}.}$

Initial velocity of second ball, $\sf{u_2=-10\ m\,s^{-1}.}$

Negative sign is because the two balls approach each other in opposite directions.

Since the collision is perfectly elastic, the velocity of the first ball after collision is,

$\sf{\longrightarrow v_1=\left(\dfrac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\dfrac{2m_2}{m_1+m_2}\right)u_2}$

$\sf{\longrightarrow v_1=\left(\dfrac{10-20}{10+20}\right)20-\left(\dfrac{2\times20}{10+20}\right)10}$

$\sf{\longrightarrow\underline{\underline{v_1=-20\ m\,s^{-1}}}}$

Negative sign indicates the first ball moves towards its initial position after collision.

And the velocity of the second ball after collision is,

$\sf{\longrightarrow v_2=\left(\dfrac{m_2-m_1}{m_2+m_1}\right)u_2+\left(\dfrac{2m_1}{m_2+m_1}\right)u_1}$

$\sf{\longrightarrow v_2=-\left(\dfrac{20-10}{20+10}\right)10+\left(\dfrac{2\times10}{20+10}\right)20}$

$\sf{\longrightarrow\underline{\underline{v_2=10\ m\,s^{-1}}}}$

So the second ball moves towards its initial position too, after collision.

Answer 2

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