Question

The track shown is figure is frictionless. The block B of mass 2m is lying at rest and the block A or mass m is pushed along the track with some speed. The collision between A and B is perfectly elastic. With what velocity should the block A be started to get the sleeping man awakened?
Figure

Answer 1

Given:
Mass of the block, A = m
Mass of the block, B = 2m

Let the initial velocity of block A be u₁ and the final velocity of block A,when it reaches the block B be v₁.

Using the work-energy theorem for block A, we can write:
Gain in kinetic energy = Loss in potential energy

$$\begin{gathered} \therefore \left(\frac{1}{2}\right)m{v}_1^2-\left(\frac{1}{2}\right)m{u}_1^2=mgh \\ \Rightarrow v_1^2-u_1^2=2\mathrm{gh} \\ \Rightarrow v_1=\sqrt{2gh+u_1^2}...\left(1\right) \end{gathered}$$

Let the block B just manages to reach the man's head.
i.e. the velocity of block B is zero at that point.

Again, applying the work-energy theorem for block B, we get:
$$\begin{gathered} \left(\frac{1}{2}\right)\times 2m\times (0{)}^2-\left(\frac{1}{2}\right)\times 2m\times v^2=mgh \\ \Rightarrow \mathrm{v}=\sqrt{2\mathrm{gh}} \\ \mathrm{Therefore},\mathrm{before}\mathrm{the}\mathrm{collision}: \\ \mathrm{Velocity}\mathrm{of}\mathrm{A},u_{\mathrm{A}}=v_1 \\ \mathrm{Velocity}\mathrm{of}\mathrm{B},\mathit{}{u}_{\mathrm{B}}=0 \\ \mathrm{After}\mathrm{the}\mathrm{collision}: \\ \mathrm{Velocity}\mathrm{of}\mathrm{A},v_{\mathrm{A}}=v(\mathrm{say}) \\ \mathrm{Velocity}\mathrm{of}\mathrm{B},v_{\mathrm{B}}=\sqrt{2\mathrm{gh}} \end{gathered}$$

As the collision is elastic, K.E. and momentum are conserved.

$$\begin{gathered} m{v}_1+2m\times 0=mv+2m\sqrt{2gh} \\ \Rightarrow v_1-v=2\sqrt{2gh}...\left(2\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(\frac{1}{2}\right)m{v}_1^2+\left(\frac{1}{2}\right)2m\times (0{)}^2=\left(\frac{1}{2}\right)m{v}^2+\left(\frac{1}{2}\right)2m{\left(\sqrt{2gh}\right)}^2 \\ \Rightarrow v_1^2-v^2=2\times \sqrt{2gh}\times \sqrt{2gh}...\left(3\right) \\ \mathrm{Dividing}\mathrm{equation}(3)\mathrm{by}\mathrm{equation}(2),\mathrm{we}\mathrm{get}: \\ {v}_1+v=\sqrt{2gh}...\left(4\right) \\ \mathrm{Adding}\mathrm{the}\mathrm{equations}\left(4\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ 2v_1=3\sqrt{2gh} \\ \mathrm{Now}\mathrm{using}\mathrm{equation}\left(1\right)\mathrm{to}\mathrm{substitue}\mathrm{the}\mathrm{value}\mathrm{of}{v}_{\mathit{1}},\mathrm{we}\mathrm{get}: \\ \sqrt{2gh+u^2}=\left(\frac{3}{2}\right)\sqrt{2gh} \\ \Rightarrow 2gh+u^2=\left(\frac{9}{4}\right)(2gh) \\ \Rightarrow u=\sqrt{2.5gh} \end{gathered}$$

Block a should be started with a minimum velocity of $\sqrt{2.5gh}$ to get the sleeping man awakened.