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 $=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2-\frac{1}{2}(m_1+m_2)v^2$ $=\frac{1}{2}\times 2\times {(10)}^2+0-\frac{1}{2}(2+3)\times {(4)}^2$ $=100-40=60J$