Question

A stone of mass 2 kg moving with a velocity of 4 m/s collides with another stone of mass 4 kg, which is at rest. If, after the collision, they move with the same velocity, then what would be the combined kinetic energy of the two stones after the collision?

• A. 5.33 J
• B. 13.67 J
• C. 16. 25 J
• D. 17.37 J

Answer 1

The correct option is A 5.33 J

Mass and velocity of the stones before and after collision are as shown in the diagram.


Using law of conservation of momentum:
Momentum before collision = Momentum after collision
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\Rightarrow 2times4+4\times 0=(2+4)v(\because v_1=v_2=v)\Rightarrow v=\frac{8}{6}m/s\text{K.E. after collision}=\frac{1}{2}(m_1+m_2)v^2\Rightarrow \text{K.E.}=\frac{1}{2}(2+4){\left(\frac{8}{6}\right)}^2=5.33J$
Hence, the correct answer is option (a).