A block of mass $0.50 \mathrm{~kg}$ is moving with a speed of $2.00 \mathrm{~m} / \mathrm{s}$ on a smooth…

A block of mass $0.50 \mathrm{~kg}$ is moving with a speed of $2.00 \mathrm{~m} / \mathrm{s}$ on a smooth surface. It strikes another mass of $1.00 \mathrm{~kg}$ and then they move together as a single body. The energy loss during the collision is
  1. $0.16 \mathrm{~J}$
  2. $1.00 \mathrm{~J}$
  3. $0.67 \mathrm{~J}$
  4. $0.34 \mathrm{~J}$

Solution

$ \begin{aligned} & m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\ & v=2 / 3 \mathrm{~m} / \mathrm{s} \\ & \text { Energy loss }=\frac{1}{2}(0.5) \times(2)^2-\frac{1}{2}(1.5) \times\left(\frac{2}{3}\right)^2=0.67 \mathrm{~J} \end{aligned} $

Asked in: JEE Main 2008

Practice more Center of Mass Momentum and Collision questions on Aicharya