Question

A boy of mass 50 kg running at 5 m/s jumps on to a 20 kg trolley travelling in the same direction at 1.5 m/s. What is their common velocity?

Answer 1

Here the momentum of the system will remain conserved as no external forces are involved.

Total momentum before the boy jumps on the trolley = $M_{\mathrm{Boy}}{u}_{\mathrm{Boy}}+M_{\mathrm{Trolley}}{u}_{\mathrm{Trolley}}=50\mathrm{kg}\times 5\mathrm{m}/\mathrm{s}+20\mathrm{kg}\times 1.5\mathrm{m}/\mathrm{s}=280\mathrm{kg}\mathrm{m}/\mathrm{s}$

Let the common velocity of the boy and the trolley is 'v_Common'.

Total momentum after the boy jumps on the trolley = $\left(M_{\mathrm{Boy}}+M_{\mathrm{Trolley}}\right)v_{\mathrm{Common}}=\left(50\mathrm{kg}+20\mathrm{kg}\right)\times v_{\mathrm{Common}}=70\mathrm{kg}\times v_{\mathrm{Common}}$

Accordding to momentum conservation, Initial momentum = Final momentum

$$\begin{gathered} \therefore 280\mathrm{kg}\mathrm{m}/\mathrm{s}=70\mathrm{kg}\times {\mathrm{v}}_{\mathrm{Common}} \\ \Rightarrow {\mathrm{v}}_{\mathrm{Common}}=\frac{280\mathrm{kg}\mathrm{m}/\mathrm{s}}{70\mathrm{kg}}=4\mathrm{m}/\mathrm{s} \\ \Rightarrow {\mathrm{v}}_{\mathrm{Common}}=4\mathrm{m}/\mathrm{s} \end{gathered}$$