Laws of Motion — Physics STD 11 Science — Question
Maharashtra BoardEnglish MediumSTD 11 SciencePhysicsLaws of Motion3 Marks
Question
Define coefficient of restitution and obtain its value for an elastic collision and a perfectly inelastic collision.
✓
Answer
i. For two colliding bodies, the negative of ratio of relative velocity of separation to relative velocity of approach is called the coefficient of restitution.ii. Consider an head-on collision of two bodies of masses $m_1$ and $m_2$ with respective initial velocities $u_1$ and $u_2$. As the collision is head on, the colliding masses are along the same line before and after the collision. Relative velocity of approach is given as,
$u_a = u_2 – u_1$
Let $v_1$ and $v_2$ be their respective velocities after the collision. The relative velocity of recede (or separation) is then $v_s = v_2 – v_1$
$\therefore \quad e =-\frac{ v _{ s }}{ u _{ a }}=-\frac{ v _2- v _1}{ u _2- u _1}=\frac{ v _1- v _2}{ u _2- u _1}$
iii. For a head on elastic collision, According to the principle of conservation of linear momentum,
Total initial momentum = Total final momentum
$ \therefore \quad m_1 u _1+ m _2 u _2= m _1 v _1+ m _2 v _2$
$\therefore \quad m _1\left( u _1- v _1\right)= m _2\left( v _2- u _2\right) $
As the collision is elastic, total kinetic energy of the system is also conserved.
$ \therefore \frac{1}{2} m _1 u _1^2+\frac{1}{2} m _2 u _2^2=\frac{1}{2} m _1 v _1^2+\frac{1}{2} m _2 v _2^2 \ldots(4)$
$\therefore m _1\left( u _1^2- v _1^2\right)= m _2\left( v _2^2- u _2^2\right)$
$\therefore m _1\left( u _1+ v _1\right)\left( u _1- v _1\right)= m _2\left( v _2+ u _2\right)\left( v _2- u _2\right) $
$ u _1+ v _1= u _2+ v _2$
$u _2- u _1= v _1- v _2$
Substituting this in equation (1),
$e =\frac{ v _1- v _2}{ u _2- u _1}=1$
iv. For a perfectly inelastic collision, the colliding bodies move jointly after the collision, i.e.,
$ v _1= v _2$
$\therefore v _1- v _2=0 $
Substituting this in equation $(1), e =0$.
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