Question

A 96.2 kg cannon at rest contains a 9.2 kg cannon ball. When fired, the cannon ball leaves the cannon with a speed of 85.8 m/s. What is the recoil speed of the cannon?

Answer 1

8.29 m/s

Step-by-step explanation

Given:

• The mass of the cannon, m₁ = 95.2 kg

,

• The mass of the cannonball, m₂ = 9.2 kg

,

• The initial speed of cannon-cannonball system, u₁ = u₂ = 0 m/s

,

• The final speed of the cannonball, v₂ = 85.8 m/s

Unknown:

• The recoil speed of the cannon, v₁

By the law of conservation of momentum, we have,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

The initial speeds are both 0 m/s, so the left side of the equation is equal to 0,

`0=m_1v_1+m_2v_2`

Solving for v₁,

`v_1=(-m_2v_2)/(m_1)`

Replace with the known values and solve,

`v_1=(-9.2kg\cdot85.8m/s)/(95.2kg)\approx-8.29m/s`

The minus sign indicates that the direction of the velocity is opposite to the direction of motion of the cannonball. Hence, the recoil speed of the cannon is 8.29 m/s