Question

An 825-kg cannon rests on a frictionless surface and fires a 1.12-kg projectile at 124 meters per second to the right. What is the velocity of the cannon immediately after the projectile is fired?(a) 0.124 m/s to the left. (b) 0.152 m/s to the left.(c) 0.168 m/s to the left.(d) 0.174 m/s to the left.

Answer 1

Given,

Mass of the cannon, m₁=825 kg

Mass of the projectile, m₂=1.12 kg

The velocity of the projectile, v₂=124 m/s

According to the law of conservation of momentum, the momentum of the projectile fired should be equal to the momentum of the cannon.

i.e.,

`m_1v_1=m_2v_2`

Therefore the velocity of the cannon is given by,

`v_1=(m_2v_2)/(m_1)`

On substituting the known values in the above equation,

`v_1=(1.12*124)/(825)=0.168\text{ m/s}`

Therefore the velocity of the cannon immediately after the projectile is shot is 0.168 m/s