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A Mass ‘m1’ Travelling With Speed Speed ‘u’ On A Horizontal Plane Hits Another Mass ‘m2’ Which Is At Rest. If The Coefficient Of Restitution Is ‘e’, Calculate The Loss Of Kinetic Energy.

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Ans.

On a horizontal plane, a mass m_1 hits another mass m_2 . u and zero are the velocities of the masses m_1 and m_2 respectively.

Let, v_1 and v_2 be their respective velocities after collision.

According to the conservation of linear momentum,

m_1u+m_1v_1+m_2v_2

If e be the coefficient of restitution, then

\displaystyle{e=\frac{v_2-v_1}{u}}

After collision, the loss of kinetic energy is

Total kinetic energy before collision – Total kinetic energy after collision

= \frac{1}{2}m_1u^2-\left(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\right)

= \frac{1}{2(m_1+m_2)}\left[(m_1+m_2)m_1v_1^2+(m_1+m_2)m_2v_2^2\right]

= \frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left(m_1^2v_1^2+m_2^2v_2^2+m_1m_2(v_1^2+v_2^2)\right)\right]

= \frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left({(m_1v_1+m_2v_2)}^2-2m_1m_2v_1v_2+m_1m_2(v_1^2+v_2^2)\right)\right]

= \frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left({(m_1v_1+m_2v_2)}^2+m_1m_2{(v_1-v_2)}^2\right)\right]

= \frac{1}{2(m_1+m_2)}\left[(m_1^2u^2+m_1m_2u^2)-\left(m_1^2u^2+m_1m_2{(v_1-v_2)}^2\right)\right]
[ using equation (1) ]

= \frac{1}{2(m_1+m_2)}\left[m_1m_2u^2-m_1m_2{(v_1-v_2)}^2\right]

= \frac{1}{2(m_1+m_2)}\left[m_1m_2u^2-m_1m_2e^2u^2\right]
[ using equation (2) ]

= \displaystyle{\frac{1}{2}\frac{m_1m_2}{(m_1+m_2)}(1-e^2)u^2}

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