Question

02**. A bullet of mass 10 g and travelling at a speed of 500 m/s strikes a block of mass 2 kg which is suspended by a string of length 5 m. The bullet goes through the block in a very short time and the centre of gravity of the block is found to rise a vertical distance of 10 cm. What is the speed of the bullet just after it emerges from the block? ​

Answer 1

Answer: The speed of the bullet just after it emerges from the block is 5 m/s.

Step-by-step explanation:

To find the speed of the bullet just after it emerges from the block, we can use the principle of conservation of momentum and the principle of conservation of energy.

Let's denote the speed of the bullet just after it emerges from the block as v' (in m/s).

Conservation of momentum:

Before the collision, the bullet's momentum is given by:

Momentum before = Mass of the bullet * Initial velocity of the bullet

= (10 g) * (500 m/s) = 0.01 kg * 500 m/s = 5 kg m/s

After the collision, the bullet emerges from the block with a speed of v', so its momentum is:

Momentum after = Mass of the bullet * Velocity of the bullet after emerging = (10 g) * v'

The block gains momentum and rises, but since the string is attached and the block moves in a circular arc, the total momentum change is zero. Therefore, we can set up the momentum equation as follows:

Momentum before = Momentum after

5 kg m/s = (10 g) * v'

v' = 5 kg m/s / (10 g)

v' = 5 m/s

So, the speed of the bullet just after it emerges from the block is 5 m/s.