Question

A block of mass $0.50\, kg$ is moving with a speed of $2.00\, m s ^{-1}$ on a smooth surface. It strikes another mass of $1.00\, kg$ which is at rest and then they move together as a single body. The energy which is lost during the collision is

• A. 0.16 J
• B. 1.00 J
• C. 0.67 J
• D. 0.34 J

Answer 1

Answer: (C)

Here, $m_{1}=0.50\, kg , u_{1}=2.00\, m\, s ^{-1}$
$m_{2}=1.00 \,kg , u_{2}=0$
If $v$ is the velocity of the combination after collision, then according to the principle of conservation of linear momentum,
$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v $
or $ v=\frac{m_{1} u_{1}}{m_{1}+m_{2}}=\frac{0.50 \times 2.00}{0.50+1.00}=\frac{1.00}{1.50}=\frac{2}{3} ms ^{-1}\left(\because u_{2}=0\right) $
Energy loss = Initial energy - Final energy
$ =\frac{1}{2} m_{1} u_{1}^{2}-\frac{1}{2}\left(m_{1}+m_{2}\right) v^{2}$
$=\frac{1}{2} \times 0.50 \times(2.00)^{2}-\frac{1}{2} \times(0.50+1.00) \times\left(\frac{2}{3}\right)^{2}$
$=1.00-\frac{1.50}{2} \times \frac{4}{9}=0.67\, J$