Question

A bullet of mass 20 g, travelling at a speed of 350 ms−1, strikes a steel plate at an angle of 30º with the plane of the plate. It ricochets off at the same angle, at a speed of 320 ms−1. What is the magnitude of the impulse that the steel plate gives to the projectile? If the collision with the plate takes place over a time ∆t = 10−3 s, what is the average force exerted by the plate on the bullet?

Answer 1

m = 20 grams = 0.02 kg

Impulse = J = mΔv

J =m * (Vf - Vi)
J = 0.02*( 320@30 - 350@330)
J = 0.02*( 320@30 + 350@150)
OR:
J = 6.4@30 + 7@150

∑Px
6.4* Cos( 30)+ 7* Cos(150) = -0.5196 kg*m/s

∑Py
6.4* Sin( 30)+ 7* Sin(150) = 6.7kg*m/s

Bullet Impulse magnitude sqr(Px^2+Py^2) = 6.720 kg*m/s

Answer 2

See diagram.

mass of bullet = 0.020 kg
The component of velocity of the bullet in the direction parallel to the steel plate changes in magnitude from:  350 cos30° to 320 Cos 30°.
=> change in momentum = 0.020 *[ 320 - 350 ] * √3/2 = - 0.30 √3 kg-m/s

The component in the perpendicular direction changes from  - 350 Sin 30°  m/s to +320 Sin30°  m/s.
=> change in momentum = 0.020 * [320 - (-350) ] * 1/2 = 6.70 kg-m/s

So Impulse = change of momentum during the impact  (vector addition..)
                   = √[6.70² + 0.27] = 
                    = 6.7201 kg-m/s

Force = impulse /Δt = 6.7201/0.001 = 6,720.1 N