Question

A bullet of mass $M$ is fired with a velocity of $50{ms}^{-1}$ at an angle $\theta$ with the horizontal. At the highest point of trajectory it collides with a bob of mass $3M$ suspended vertically by a massless string of length $\frac{10}{3}m$ and gets embedded into it. After the collision the string moves through an angle ${120}^{\circ }$, what is the angle of throw $\theta$.

• A. ${\cos }^{-1}\frac{2}{5}$
• B. ${\cos }^{-1}\frac{3}{5}$
• C. ${\cos }^{-1}\frac{4}{5}$
• D. ${\cos }^{-1}\frac{1}{5}$

Answer 1

Answer: (C)

${\cos }^{-1}\frac{4}{5}$

Before collision, bullet momentum at $A=M\left(50\right)\cos \theta$
After collision, bob momentum at $A=(3M+M)v$

according to conservation of linear momentum $M\left(50\right)\cos \theta =(3M+M)vorv=\frac{50\cos \theta }{4}$

velocity $v$ is responsible for rotating the bob in vertical circle and raises it to height $=h=\frac{10}{3}+\frac{10}{3}\sin 30=5m$

Also kinetic energy at $A$ is getting converted into potential energy at $B,$ so according to law of conservation of energy,

$\frac{1}{2}\left(4M\right)v^2=\left(4M\right)ghor\frac{1}{2}{\left(\frac{50\cos \theta }{4}\right)}^2=10\times 5$

$or{\cos }^2\theta =\frac{16}{25}or\theta ={\cos }^{-1}\frac{4}{5}$