Center of Mass, Impulse, Momentum, and Collisions

Question

You are playing airsoft with your friends. You are carrying a 3.0 kg airsoft that fires a 0.20 g pellet with a velocity of 95 m/s. (a) What is the recoil velocity of the airsoft gun? (b) If you hold the airsoft gun firmly against your shoulder, what will be your recoil velocity? Assume that your mass is 50.0 kg.

Use GUESS METHOD

Answer 1

Given, 

Mass of airsoft, m₁ = 3.0 kg

Mass of Pallet, m₂ = 0.20g = 0.2/1000 kg = 2 x 10^-4 kg

Velocity of pallet, v₂ = 95 ms^-1

(a) Let v₁ be the recoil velocity of airsoft

By conservation of momentum

Momentum before firing=momentum after firing

 

m₁v₁ + m₂v₂ = 0     

Since, before firing, airsoft and pallet are at rest, and hence momentum before firing is zero.

v_{1 =} - m₂v₂ / m₁ = - 2 x 10^-4 kg x 95 ms^-1 / 3.0 kg

= - 6.3 x 10^-3 ms^-1

Here negative sign indicates that the recoil velocity is opposite to the velocity of the pallet.

b) (b) Let V be the recoil velocity when the airsoft gun is hold firmly against your shoulder. 

In this case, both airsoft and the person holding the airsoft will have same recoil velcoity. 

Mass of person, M = 50 kg

By conservation of momentum,

Momentum before firing = momentum after firing

Here, Momentum before firing = 0

Momentum after firing = Momentum of airsoft + Momentum of person + Momentum of Pallet

Therefore,

m₁V + MV + m₂v₂V = 0     

= - m₂v₂ / (m₁ + M)

= - 2 x 10^-4 x 95 / (3 + 50)

= - 3.58 x 10^-4 ms^-1

Here, negative sign indicate that recoil velocity is opposite to the velocity of Pallet.