A bullet of mass $10g$ moving horizontally with a velocity of $400m/s$ strikes a wooden block of mass $2kg$ which is suspended by a light inextensible string of length $5m$. As a result, the centre of gravity of the block is found to rise a vertical distance of $10cm$. The speed of the bullet after it emerges out horizontally from the block will be

The correct option is A $120m/s$

Given,
Mass of the wooden block ($m_1$) is $2kg$
Mass of the bullet ($m_2$) is $10g$
Velocity $=400m/s$
Length of the string $=5m$

Let the velocity of the bullet and the block after the collision will be $v_2$ and $v_1$ respectively

The block rises $10cm=0.1m$

According to conservation of energy,

P.E at the maximum height = K.E at the bottom of the block,

$m_{1}gh=\frac{1}{2}{m}_1{v}_1^2$
$v_1=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.1}$$=1.4m/s$

Let $u_1$ be the initial velocity of the bullet. Then, according to conservation of momentum,

$m_2{u}_2=m_1{v}_1+m_2{v}_2$
$v_2=\frac{m_2{u}_2-m_1{v}_1}{m_2}$
$=\frac{0.01\times 400-2\times 1.4}{0.01}=120m/s$