Question

A 1200 kg cannon mounted on wheels shoots a 10kg projectile a with a muzzle speed of 725m\s at an angle of 36 degrees above the horizontal. What is the horizontal recoil speed of the cannon.

Answer 1

Solution:

Mass of the cannon, M = 1200 kg

Mass of the projectile, m = 10 kg

Initial speed of the projectile v = 725 m/s

Angle made by the projectile’s speed above the ground, θ = 36^o

The horizontal component of the projectile’s speed

v_x = v*cosθ

v_x = (725 m/s)*cos36^o

v_x = 586.54 m/s

Conservation of the linear momentum in the horizontal direction can be written as,

Initial momentum of the cannon-projectile just before shooting = final momentum of the cannon-projectile just after the shooting

M*V_i + m*v_xi = M*V + m*v_x          --------------------------------------------(1)

where V_i = initial speed of cannon and v_xi = initial horizontal speed of projectile

Since initially both the cannon and projectile were at rest, thus V_i = v_xi = 0 m/s

V = Recoil speed of the cannon just after the shooting

Thus equation (1) becomes

1200kg*0m/s + 10kg*0m/s = 1200kg*V + 10kg*586.54m/s

0 = 1200kg*V + 10kg*586.54m/s

V = -(10kg*586.54m/s) / 1200kg

V = - 4.89 m/s

Thus the cannon's horizontal recoil speed is - 4.89 m/s.

Negative sign indicates that the projectile’s and cannon’s speed are in opposite direction.

Hence the answer is 4.89 m/s (magnitude of the cannon’s recoil speed)