A bullet B of mass mB traveling with a speed v0 = 1400 m/s ricochets off a fixed steel plate A of mass mA. Let mA ≫ mB so that it can be assumed that mBmA→0 . If the incidence angle of the bullet is θ = 15° and the COR of the impact is e = 0.5, determine the rebound angle ϕ , as well as the bullet’s rebound speed.

Rebounce angle is 345°

Rebounce speed is 989.95m/s

Explanation:

Calculate the x  component of the velocity of the bullet before impact by using the following relation:

Vbx= Vb Cos thetha

Here,  is the initial velocity of the bullet, Vo = 1400m/s and is the incidence angle of the bullet.= theta = 15°

Substituting

Vbx = Cos15 ×1400 = 1352.30m/s

Calculate the y component using the relation:

Vby = Vo Sin theta

Vby = sin 15° × 1400

Vby = 362.35m/s

The rebounce angle = 360 - incidence angle

Rebounce angle =( 360 - 15)° = 345°

The rebound speed V' = Vby - Vbx

V' = (1352.30 - 362.35)m/s

V' = 989.95 m/s

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