CollisionsMedium

Cannon recoil

A cannon of mass M = 600 kg, at rest, fires a projectile m = 3 kg at v_p = 120 m/s inclined at α = 25°. Find the recoil speed and the impulse of the ground's constraint reaction.

Given data

M = 600 kgm = 3 kgv_p = 120 m/sα = 25°

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Before firing everything is at rest: total horizontal momentum is zero. Firing is an internal process (the explosive forces are internal to the system) and no external horizontal force acts, so horizontal momentum is conserved and must stay zero. Therefore M·V (cannon, backward) = m·v_p·cosα (horizontal component of the projectile).
Write your working — the numerical result must appear in the text m/s
Vertically, momentum is NOT conserved, because the ground exerts an upward constraint reaction (it keeps the cannon from sinking). By the impulse–momentum theorem, the impulse of this reaction equals the vertical momentum gained by the projectile: J = m·v_p·sinα.

Full worked solution

Before firing everything is at rest: total horizontal momentum is zero. Firing is an internal process (the explosive forces are internal to the system) and no external horizontal force acts, so horizontal momentum is conserved and must stay zero. Therefore M·V (cannon, backward) = m·v_p·cosα (horizontal component of the projectile).
$V = \frac{m\,v_p\cos\alpha}{M} = \frac{3\cdot120\cdot\cos25°}{600}$
$V = \frac{326.3}{600} \approx \mathbf{0.54\,m/s}$ .
Vertically, momentum is NOT conserved, because the ground exerts an upward constraint reaction (it keeps the cannon from sinking). By the impulse–momentum theorem, the impulse of this reaction equals the vertical momentum gained by the projectile: J = m·v_p·sinα.
$J = m\,v_p\sin\alpha = 3\cdot120\cdot\sin25°$
$J = \mathbf{152\,kg\cdot m/s}$ .

Result:

V \approx 0.54

m/s,

J \approx 152

kg·m/s

Cannon recoil

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