Question

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 1

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.

Answer 2

Given:

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.

To find:

Speed of 4M after collision ?

Calculation:

First of all, the speed of 4M block is equivalent to the speed of the 4M and M block together as they stick together (completely inelastic collision).

Let speed of the system after collision be v :

Applying Conservation of Momentum:

$\rm\therefore \: Mu + (4M)0 = (M + 4M)v$

$\rm\implies \: Mu +0 = (5M)v$

$\rm\implies \: Mu = (5M)v$

$\rm\implies \: u = 5v$

$\rm\implies \: v = \dfrac{u}{5}$

$\rm\implies \: v = \dfrac{8}{5}$

$\rm\implies \: v = 1.6 \: m {s}^{ - 1}$

So, velocity of the block combination will be 1.6 m/s.