a bullet of mass0.01 kg and travelling with a speed of 500m/s strikes a block of mass 2 kg which is suspended by a string of length 5m. The center of gravity of the block is found to raise vertical distance of 0.1m. The speed of bullet after emerging from block is:

100 220m/s

110

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

According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is given by the product of its mass and velocity:

Momentum (p) = mass (m) * velocity (v)

Let's denote the initial velocity of the bullet as v1, and the final velocity of the bullet after emerging from the block as v2. The mass of the bullet is given as 0.01 kg.

Before the collision:
The momentum of the bullet before the collision = m * v1 = 0.01 kg * 500 m/s

After the collision:
The momentum of the bullet after the collision = m * v2 = 0.01 kg * v2

Since the total momentum before and after the collision should be equal, we have:

0.01 kg * 500 m/s = 0.01 kg * v2

Simplifying the equation, we get:

500 m/s = v2

Therefore, the speed of the bullet after emerging from the block is 500 m/s.