Question

A bullet of mass $0.01\text{kg}$ and travelling at a speed of $500\text{m/s}$ strikes a block of $2\text{kg}$ which is suspended by a string of length 5 m. The centre of gravity of the block is found to rise a vertical distance of 0.1 m. What is the speed of the bullet in $\text{m/s}$ after it emerges from the block?

Answer 1

Suppose the velocity of the bullet of mass $m$ is $v$ and it strikes the block of mass M. After collision, the linear velocity of the block is V and that of the bullet is v'.
Applying law of conservation of linear momentum, we get
$mv=MV+m{v}^{\prime }$
or $500\times 0.01=2V+0.01v^{\prime }$
or 5 = 2V + 0.01 v' (i)
Now, the kinetic energy gained by the block $\frac{1}{2}M{V}^2$ raises it to a height h where it gains gravitational potential energy Mgh. By conservation of energy, we get
$\frac{1}{2}M{V}^2=\text{Mgh}$
or $V=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.1}=1.4\text{m/s}$
Putting the value of V in Eq. (i), we get
$5=2\times 1.4+0.01v^{\prime }$ or v' = $220\text{m/s}$