Question
Two bodies undergo one$-$dimensional, inelastic, head$-$on collision. Obtain an expression for the magnitude of impulse.

Answer

$i.$ When two bodies undergo collision, the linear momentum delivered by the first body to the second body must be equal to the change in momentum or impulse of the second body and vice versa.
$\therefore$ Impulse,
$|J|=\left|\Delta p_1\right|=\left|\Delta p_2\right|$
$=\left|m_1 v_1-m_1 u_1\right|=\left|m_2 v_2-m_2 u_2\right| .....(1)$
$ii.$ Substituting $v_1=\frac{m_1-e_2}{m_1+m_2} u_1+\frac{(1+e) m_2}{m_1+m_2} u_2$,
$v_2=\frac{m_2-e_1}{m_1+m_2} u_2+\frac{(1+e) m_1}{m_1+m_2} u_1$
In equation $(1)$ and solving, we get,
$|J|=\left(\frac{m_1 m_2}{m_1+m_2}\right)(1+e)\left(\left|u_1-u_2\right|\right)=\mu(1+e) u_{\text {relative }}$
$u _{\text {relative }}=\left| u _1- u _2\right|=$ velocity of approach.

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