Question

A metal ball of mass 2 kg moving with a velocity of 36 km/h has a head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is :

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

Answer 1

Answer: (C)

Key Idea: The first key idea is that final velocity v is given by the law of conservation of momentum. Initial momentum = Final momentum $ \therefore $ $ {{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=({{m}_{1}}+{{m}_{2}})v $ Given: $ {{v}_{1}}=36km/h=36\times \frac{5}{18}=10m/s $ $ {{v}_{2}}=0 $ $ {{m}_{1}}=2kg, $ $ {{m}_{2}}=3kg $ $ \therefore $ $ v=\frac{{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}}{{{m}_{1}}+{{m}_{2}}} $ $ =\frac{2\times 10+3\times 0}{2+3} $ or $ v=\frac{20}{5}=5m/s $ Loss in kinetic energy $ =\frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2}-\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{v}^{2}} $ $ =\frac{1}{2}\times 2\times {{(10)}^{2}}+0-\frac{1}{2}(2+3)\times {{(4)}^{2}} $ $ =100-40=60J $