Question

A bullet of mass 10 g travelling horizontally with
a velocity of 150 m/s strikes stationary wooden
block of mass 5 kg. Then they both move off
together in the same striaght line. Calculate the total
momentum just before the impact and just after the
impact. Also calculate the velocity of the combined
object.​

Answer 1

Given :

• Mass of bullet (m1) = 10 g

=> 10/1000

=> 0.01 kg

• Mass of block (m2) = 5 kg

• Initial velocity of bullet (u1) = 150 m/s

• Initial vlocity of block (u2) = 0 m/s

(Initial velocity of block is zero because block is stationary)

Now,

Momentum = mv

According to question,

Initial Momentum = m1u1 + m2u2

=> 0.01 × 150 + 5 × 0

=> 1.5 + 0

=> 1.5 kg m/s

According to law of conservation of momentum.

The total momentum of the system cannot be changed.

So,

Initial momentum = Final momentum

So, Final momentum also is 1.5 kg m/s.

Also, we have to calculate the combined velocity of the object.

Let combined velocity of object = v

So,

Final Momentum = m1v + m2v

=> 1.5 = (m1 + m2) v

=> 1.5 = (0.01 + 5)v

=> 1.5 = 5.01v

=> v = 0.299

=> v = 0.3 m/s (approx.)

∴ Total momentum just before the impact and just after the impact is 10.5 kg m/s.

Combined velocity of the object is 0.3 m/s.