Question

A $2 \,kg$ ball moving at $24\, m\, s ^{-1}$ undergoes head on collision with a $4 \,kg$ ball moving in the opposite direction at $48 \,m \,s ^{-1} .$ If the coefficient of restitution is $\frac{2}{3},$ their velocities in $m\, s ^{-1}$ after collision are

• A. -56, -8
• B. -28, -4
• C. -14, -2
• D. -7, -1

Answer 1

Answer: (A)

Here, $m_{1}=2 kg , m_{2}=4 kg =2 m _{1}$
$u_{1}=24\, m\, s ^{-1}, u_{2}=-48 m s ^{-1}=-2 u_{1}$
where $u_{1}$ and $u_{2}$ be the velocities of masses $m_{1}$ and $m_{2}$ before collision.
Let $v_{1}$ and $v_{2}$ be the velocities of masses $m_{1}$ and $m_{2}$ after collision.
According to the law of conservation of linear momentum, we get
$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} ; m_{1} u_{1}+2 m_{1}\left(-2 u_{1}\right)$
$=m_{1} v_{1}+2\, m_{1} v_{2} $
$-3 u_{1}=v_{1}+2 v_{2} \dots$(i)
Coefficient of restitution, $e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}$
$v_{2}-v_{1}=e\left(u_{1}-u_{2}\right)=e\left(u_{1}-\left(-2 u_{1}\right)\right)$
$v_{2}-v_{1}=3 u_{1} e \dots$(ii)
Adding (i) and (ii), we get
$3 v_{2}=3 u_{1} e-3 u_{1}$ or $v_{2}=u_{1}(e-1) \dots$(iiI)
Substituting this value of $v_{2}$ in (i), we get
$v_{1}=-u_{1}(1+2 e) \dots$(iv)
Substituting the given values in (iii) and (iv), we get
$v_{2}=24\left(\frac{2}{3}-1\right)=-8 m s ^{-1}$
and $v_{1}=-24\left(1+2 \times \frac{2}{3}\right)=-56\, m \,s ^{-1}$