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

Question

Toppr Admin · Accepted Answer

Let the velocity of the bullet after the collision be $v_{b}$ and the velocity of the wooden block after the collision be $v_{w}$ .

Using conservation of energy for the wooden block,

$\frac{1}{2} m v_{w}^{2} = m g h$

$v_{w} = \sqrt{2 g h}$

$v_{w} = \sqrt{2 \times 9.8 \times 0.1} = 1.4 m / s$

Applying conservation of momentum before and after the collision,

$0.01 \times 400 = m_{w} v_{w} + m_{b} v_{b}$

$4 = 2 \times 1.4 + 0.01 v_{b}$

$v_{b} = 120 m / s$