Question

Q. A bullet of mass m strikes an obstruction and deviates off at 60$°$ to its original direction. If its speed is also changed from u to v, find the magnitude of the impulse acting on the bullet.

Answer 1

dear student as impulse = change in momentum
$$\begin{gathered} initialmomentum=m\overrightarrow{u} \\ finalmomentum=m\overrightarrow{v} \\ ImpulseI=m\overrightarrow{v}-m\overrightarrow{u}=m\overrightarrow{v}+(-m\overrightarrow{u}) \\ nowanglebetween\overrightarrow{v}andnegativeof\overrightarrow{u}is120degree \\ somagnitudeofchangeinmomentum=|∆\overrightarrow{p}|=\sqrt{(mu{)}^2+(mv{)}^2+2mumv\cos 120} \\ =m\sqrt{v^2+u^2+2uv*(-1/2)}=m\sqrt{v^2+u^2-uv}answer \end{gathered}$$