Question

A metal ball of mass 2 kg moving with speed of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after collision, both the balls move as a single mass, then the loss in K.E. due to collision is

• A. 100 J
• B. 140 J
• C. 40 J
• D. 60 J

Answer 1

Answer: (D)

60 J

Answer 2

Mass of metal ball = 2 kg;
Speed of metal ball (v$_1$) = 36 km/h = 10 m/s and
mass of stationary ball = 3 kg.
Applying law of conservation of momentum,
$m_1v_1+m_2v_2=(m_1+m_2)v$
or, $v=\frac{m_1v_1+m_2v_2}{m_1+m_2}=\frac{(2 \times 10)+(3 \times 0)}{2+3}=\frac{20}{5}$
= 4 m/s
Therefore loss of energy
$\, \, \, \, \, \, =\bigg[\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\bigg]-\frac{1}{2}+(m_1+m_2)v^2$
=$\bigg[\frac{1}{2} \times 2 \times (10)^2 +\frac{1}{2} \times 3 (0)^2\bigg] -\frac{1}{2} \times (2+3)\times(4)^2$
= 100 - 40 = 60 J.