Question

A target of mass 400g moving with a speed of 10m/s horizontally it's hit by a bullet of mass .01 kg moving in opposite direction.If both the bullet and target comes to rest.calculate the velocity of the bullet at the time of striking the target

Answer 1

Mass of target =400g
=0.4kg
Velocity = 10m/s

Mass of the bullet=0.01kg
Let it's velocity be v

As law of conservation of momentum says
0.4 × 10 = 0.01 ×v
100 =v
So, the velocity of the bullet at the time of striking the target was 100m/s.

Answer 2

Answer:

Explanation:

Solution :-

Given :-

Mass of the bullet = 0.01 kg

Mass of the target = 400 g

= 400 × 10⁻³ kg

Initial speed of the target = 10 m/s

Final speed of the bullet = 0 m/s

Final speed of the target = 0 m/s

To Calculate :-

Initial speed of the bullet = ??

Formula to be used :-

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

= (0.01 × 0) + (400 × 10⁻³ × 0)

or (0.01 u₁) + 4 = 0

or 0.01 u₁ = - 4

or u₁ = - 4/0.01

u₁ = - 400 m/s.

Hence, the velocity of the bullet at the time of striking the target is  - 400 m/s.