Question

A ball moving with a velocity $v$, collides head on with a stationary second ball of same mass. After the collision, the velocity of the first ball is reduced to $0.15\, v$. The kinetic energy of the system is decreased nearly by

• A. $20 \%$
• B. $25 \%$
• C. $30 \%$
• D. $40 \%$

Answer 1

Answer: (B)

Given,

$\Rightarrow $ Momentum is conserved,
$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{1} v_{2}$
Given $m_{1}=m_{2}=m, u_{1}=v$ and $u_{2}=0$,
also $ v=0.15 \,v$
$m v =m(0.15 v)+m v_{2} $
$v-0.15 v =v_{2} $
$\Rightarrow \, 0.85v =v_{2}\, m / s\,\,\,\,\,\dots(i)$
Here, initial kinetic energy is
$\frac{1}{2} m u_{1}^{2}=\frac{1}{2} m v^{2}\,\,\,\,\,\dots(ii)$
Final kinetic energy is
$\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2}=\frac{1}{2} m(0.15 v)^{2}+\frac{1}{2} m(0.85 v)^{2} \,\,\,\,\,\dots(iii)$
The decrement in kinetic energy is
$\Delta\, KE =( KE )_{i}-( KE )_{f}$
$\Rightarrow \, \Delta \,KE =\frac{1}{2} m v^{2}-\left[\frac{1}{2} m(0 \cdot 15 v)^{2}+\frac{1}{2} m(0.85)\right]$
$\left.\Delta \,KE =\frac{1}{2} m v^{2}\left[ I -(0.15)^{2}+0.85^{2}\right]\right]\,\,\,\,\,\dots(iv)$
$\%$ change in kinetic energy $=\frac{\Delta KE }{( KE )_{i}} \times 100$
$=\frac{\frac{1}{2} m v^{2}\left[1-(\cdot 15)^{2}-(\cdot 85)^{2}\right]}{\frac{1}{2} m v^{2}}$
$\%$ decrease in $KE =25 \cdot 5 \% \approx 25 \%$