Question

A bullet of mass 0.1 kg and travelling at a speed of 500 ms${}^{-1}$ strikes a block of mass 2 kg which is suspended by a string of length 5 m. The centre of gravity of the block is found to raise a vertical distance of 2.0 m. What is the speed of the bullet after is emerges from the block?

• A. $15m{s}^{-1}$
• B. $373.6m{s}^{-1}$
• C. $100m{s}^{-1}$
• D. $50m{s}^{-1}$

Answer 1

Answer: (B)

$373.6m{s}^{-1}$

Given : $m_1=0.1kg$ $m_2=2kg$ $u_1=500$ $m{s}^{-1}$ $u_2=0$

Let the velocities of the bullet and the block just after the collision be $v_1$ and $v_2$ respectively.

The height upto which the block is raised $h=2$ m

$\therefore$ From work-energy theorem for the block, $\frac{1}{2}{m}_2{v}_2^2=m_{2}gh$

$⟹v_2=\sqrt{2gh}$

$v=\sqrt{2\times 10\times 2}=2\sqrt{10}$ $m{s}^{-1}$

Applying conservation of momentum just before and just after the collision :

$m_1{u}_1+0=m_1{v}_1+m_2{v}_2$

$0.1\left(500\right)=0.1v_1+\left(2\right)\times 2\sqrt{10}$

$⟹v_1=373.51$ $m{s}^{-1}$