Question

A body of mass $0.05$ kg travelling with a velocity of $5$ m/s makes an angle of ${30}^o$ with vertical wall as shown in figure. The body rebounds with same speed making an angle of ${30}^o$ with the wall. Find the change in momentum of the body(in Kg-m/s).

Answer 1

Given mass of the object $m=0.05kg$ and velocity of the object before collision $V_1$=5 m/s=velocity of the object after collision $V_2$

We know that net momentum of the object before and after collision will be equal in order to satisfy law of conservation of momentum

$p_1=p_2$=mass$\times$velocity=0.05$\times$5 =0.25 kg m/s

Now from figure we get that total angle $\theta$=${30}^0$+${30}^0$=${60}^0$

Using the cosine law , the magnitude of change in linear momentum

$\Delta p=\sqrt{p_1+p_2-2p_1{p}_2\cos {60}^0}=\sqrt{(0.25{)}^2+{\left(0.25\right)}^2-2{(0.25}^2)\cos {60}^0}=0.25kgm/s$