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 $0.15v$. 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 $v=0.15v$
$mv=m(0.15v)+m{v}_2$
$v-0.15v=v_2$
$\Rightarrow 0.85v=v_{2}m/s\dots (i)$
Here, initial kinetic energy is
$\frac{1}{2}m{u}_1^2=\frac{1}{2}m{v}^2\dots (ii)$
Final kinetic energy is
$\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=\frac{1}{2}m(0.15v{)}^2+\frac{1}{2}m(0.85v{)}^2\dots (iii)$
The decrement in kinetic energy is
$\Delta KE=(KE{)}_i-(KE{)}_f$
$\Rightarrow \Delta KE=\frac{1}{2}m{v}^2-\left[\frac{1}{2}m(0\cdot 15v{)}^2+\frac{1}{2}m(0.85)\right]$
$\Delta KE=\frac{1}{2}m{v}^2\left[I-(0.15{)}^2+0.85^2\right]]\dots (iv)$
$\%$ 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 $KE=25\cdot 5\%\approx 25\%$