Question

13. A block of mass 200Kg approaches a man with a velocity of 10ms He fires bullets
each of mass 10gm with a velocity of 200 ms. The number of bullets he must fired​

Answer 1

Answer:

20g is the right answer

Answer 2

Given :

Mass of block = 200kg

Velocity of block = 10m/s

Mass of the bullet = 10g = 0.01kg

Velocity of bullet = 200m/s

To Find :

The number of bullets the man must fire to stop the block.

Solution :

❖ Momentum is defined as the product of mass and velocity.

• It is a vector quantity having both magnitude as well as direction.
• SI unit : N s

♦ Initial momentum of bullet :

➙ p₁ = n × m₁u₁

➙ p₁ = n × 0.01 × 200

➙ p₁ = 2n

Here n denotes number of bullets.

♦ Initial momentum of block :

➙ p₂ = m₂u₂

➙ p₂ = 200 × (-10)

➙ p₂ = -2000 N s

Here negative sign shows opposite direction.

Final momentum of the system will be zero as both come to rest after collision.

No net force acts on the system, linear momentum is conserved$.$

➙ 2n - 2000 = 0

➙ 2n = 2000

➙ n = 1000 bullets