Question

A bullet of mass m moving with a velocity v strikes a wooden block of mass M and gets embedded in the block which was initially at rest. The final speed of the system is ?

Answer 1

we can apply momentum conservation rule here which is

sum of initial momentum= sum of final momentum

now momentum of the bullet = $$\begin{gathered} p_1=m\times v \\ momentumoftheblockinitially{p}_2=0asitwasinrest \\ massofthetotalsystemafterbulletstrikeis(M+m) \\ letusconsiderthespeedofthecombinedsystemin{v}_1 \\ Sothemomentumwillbe(M+m)\times v_1 \\ now \\ m\times v+0=(M+m)\times v_1 \\ {v}_1=\frac{m\times v}{M+m} \\ Sothefinalvelocitywillbe\frac{m\times v}{M+m} \end{gathered}$$