Question

A body of mass $4\,kg$ moving with velocity $12\,m/s$ collides with another body of mass $6 \,kg$ at rest. If two, bodies stick together after collision, then the loss of kinetic energy of system is

• A. zero
• B. $288\,J$
• C. $172.8\,J$
• D. $144\,J$

Answer 1

Answer: (C)

$172.8\,J$

Answer 2

In an inelastic collision, kinetic energy is not conserved but the total energy and momentum remains conserved.
Momentum before collision = Momentum after collision
$m_{1} u_{1}+m_{1} u_{2}=m_{1} v_{1}+m_{2} v_{2}$
$\Rightarrow 4 \times 12=(4+6) v$
$\Rightarrow v=4.8 \,m s^{-1}$
Kinetic energy before collision $=\frac{1}{2} m_{1} u_{1}^{2}$
$=\frac{1}{2} \times 4 \times(12)^{2}$
$=288\, J$
Kinetic energy after collision $=\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}$
$=\frac{1}{2}(10)(4.8)^{2}$
$=115.2\, J$
$\therefore$ Loss in kinetic energy $=288 \,J -115.2 \,J$
$=172.8\, J$