Question

A bullet of mass 25 g and travelling horizontally at a speed of 200 m/s imbeds itself in a wooden block of mass 5 kg suspended by cords 3 m long. How far will the block swing from its position of rest before beginning to return?

Answer 1

Solution

Given data

Mass of bullet (m) =25g

Mass of block(M) =5Kg

Speed of bullet (v₁) =200 m/s

Length of cord=3m

diagram of this question can be drawn as

By momentum conservation

$mv_1=(m+M) v_2$

$v_2=\frac{mv_1}{m+M}$

$v_2=\frac{0.025\times200}{(0.025+5)}=0.99m/s$

By energy conservation

$\frac{(m+M) v_2^2}{2}=(m+M) gh$

$h=\frac{v_2^2}{2g}=\frac{(0.99) ^2}{2\times9.8}=0.05m$

So

block will move 0.05 m in y direction from initial position

displacement in x direction from initial position

$s_x=\sqrt{3^2-2.95^2}\\s_x=0.5m$

so total displacement from initial position is

$s=\sqrt{s_x^2+h^2}$

$S=\sqrt{0.05^2+0.5^2}\\s=0.5 m$

so block will move 0.5 m from its initial position.