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Answer 1

Question:- A gun of mass 10 kg fires horizontally a bullet of mass 500g with a speed of 100m/s find:

a) Recoil speed of gun

b) Impulse imparted to bullet

Answer:

V = -5 m/s

Δp = 50 kg-m-s⁻¹

Explanation:

Let the mass of gun and bullet be M and m and recoil velocity be V and v respectively.

Given that

M = 10 kg

m = 500g = 500/1000 kg = 0.5 kg

v = 100 m/s

The initial momentum of system will be zero as before firing system is in rest.

Initial Momentum = 0

Final Momentum = MV + mv

                            = 10V + (0.5×100)

                            = 10V + 50

By Law of conservation of momentum,

Final Momentum = Initial Momentum

10V + 50 =  0

10V = -50

V = -50/10

V = -5 m/s

Here negative sign shows that direction of velocity of gun is opposite of direction of velocity if bullet

Impulse imparted to bullet,

Δp = Final Momentum of bullet - Initial Momentum of bullet

Δp = mv - 0

Δp = 100×0.5 - 0

Δp = 50 kg-m-s⁻¹