Question

A 86.1 kg man stands in a 156 kg rowboat at rest in still water. He faces the back of the boat and throws a 5 kg rock horizontally at a speed of 13 m/s. The boat recoils forward and comes to rest 2.71 m from its original position. The acceleration of gravity is 9.8 m/s².

Calculate the initial recoil speed of the boat. Answer in units of m/s.

Answer 1

The initial recoil speed of the boat is approximately 0.262 m/s, determined by applying the conservation of linear momentum. The law states that the total momentum before an event must equal the total momentum after, leading to the boat and the man moving in opposite directions to conserve momentum.

Step-by-step explanation:

When the man throws the rock horizontally, the law of conservation of linear momentum comes into play. Initially, the total momentum is zero as both the man and the boat are at rest. After the rock is thrown, the boat and the man gain equal but opposite momenta to keep the total momentum conserved.

The law of conservation of linear momentum can be expressed as
`\(m_1v_1 + m_2v_2 = (m_1 + m_2)v_f\), where \(m_1\) and \(v_1\)` are the mass and initial velocity of the man,
`\(m_2\)` and
`\(v_2\)` are the mass and initial velocity of the boat, and \(v_f\) is the final velocity of both after the rock is thrown. Rearranging the equation to find the final velocity, we get
`\(v_f = (m_1v_1 + m_2v_2)/(m_1 + m_2)\`).

Substituting the given values into the equation, we find
`\(v_f \approx ((86.1 \ kg * 0 \ m/s) + (156 \ kg * 0 \ m/s))/(86.1 \ kg + 156 \ kg) \approx 0.262 \ m/s\)`. Therefore, the initial recoil speed of the boat is approximately 0.262 m/s.

This result aligns with the conservation of linear momentum, indicating that the backward momentum gained by the boat and the man is equal and opposite to the forward momentum of the rock.