1.

A bullet of mass M = 1 kg, moving with a horizontal velocity of 8 m/s, hits and sticks to a block B of mass 4M= 4 kg which is initially at rest.Find the speed of the 4M block after collision.

Answer»

Answer:

1.6 m/s

Explanation:

Given:

Mass of the bullet = 1 kg

Velocity of the bullet = 8 m/s

Mass of the BLOCK = 4 kg

To FIND:

Speed of the COMBINED block and bullet after collision

Method to find:

First, we need to find out the total momentum of the carriages before the collision.

p = m * v  (Momentum = Mass * Velocity)

Momentum of the bullet:

p = 1 * 8

⇒ p = 8 kg m/s

Momentum of Block B:

p = 4 * 0 (Since it was at rest)

p = 0 kg m/s

Total momentum of the bullet and the block:

8 + 0 = 8 kg m/s

By applying the LAW of conservation of momentum, we can state that the sum of momentum of the 2 bodies before collision will be equal to the sum of momentum of the 2 bodies after the collision.

Now, we need to find the total mass of the bodies after the collision:

4 + 1 = 5 kg

Since they are 2 non-elastic bodies, their final velocity will be the same.

To find the final velocity,

p = m * v

⇒ 8 = 5 * v

⇒ v = 8/5 = 1.6 m/s

Hence, the speed of the 4M block after collision will be 1.6 m/s.


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