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Laws of Motion — Physics STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion4 Marks
Question
Two bodies undergo one$-$dimensional, perfectly inelastic, head$-$on collision. Derive an expression for the loss in the kinetic energy.
✓

Answer

$i.$ Let two bodies $A$ and $B$ of masses $m_1$ and $m_2$ move with initial
velocity $\overrightarrow{ u }_1$ and $\overrightarrow{ u }_2$, respectively such that particle $A$ collides head on with particle $B$ i.e., $u _1> u _2$.
$ii.$ If the collision is perfectly inelastic, the particles stick together and move with a common velocity $\overrightarrow{ v }$ after the collision along the same straight line.
loss in kinetic energy = total initial kinetic energy $–$ total final kinetic energy.
$iii.$ By the law of conservation of momentum,
$m_1u_1 + m_2u_2 = (m_1 + m_2)v$
$\therefore V=\frac{ m _1 u _1+ m _2 u _2}{m_1+ m _2}$
$iv.$ Loss of Kinetic energy,
$\Delta K.E. =\left(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\right)-\frac{1}{2}\left(m_1+m_2\right) v^2$
$=\left(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\right)-\frac{1}{2}\left(m_1+m_2\right)\left[\frac{m_1 u_1+m_2 u_2}{m_1+m_2}\right]^2$.
$=\frac{ m _1^2 u _1^2+ m _1 m_2 u _2^2+ m _1 m_2 u _1^2}{2\left(m_1+ m _2\right)}+\frac{ m _2^2 u _2^2- m _1^2 u _1^2- m _2^2 u _2^2-2 m_1 m_2 u _1 u _2}{2\left(m_1+ m _2\right)}$
$=\frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(u_1-u_2\right)^2$
$iv.$ Both the masses and the term $(u_1 – u_2)^2$ are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as $e = 0,$ the loss is maximum.

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