Question

An 80 kg fisherman jumps from a dock into a 100 kg rowboat which is not moving. If the velocity of the fisherman is 4 m/s when he jumps into the boat, what is the final velocity of the fisherman and the boat?

Answer 1

v = 1.78 m/s

Step-by-step explanation:

Momentum of the fisherman before = mass of the fisherman x velocity of the fisherman

= 80 kg x 4 m/s

= 320 kg·m/s

Momentum of the boat before = mass of the boat x velocity of the boat

= 100 kg x 0 m/s

= 0 kg·m/s

Total momentum before = Momentum of the fisherman before + Momentum of the boat before

= 320 kg·m/s + 0 kg·m/s

= 320 kg·m/s

Total mass after = mass of the fisherman + mass of the boat

= 80 kg + 100 kg

= 180 kg

Total momentum before = Total momentum after

320 kg·m/s = 180 kg x v

v = 320 kg·m/s / 180 kg

v = 1.78 m/s

Answer 2

1.78 m/s

Step-by-step explanation:

We again have to use the inelastic collision formula, which is m1*v1 + m2*v2 = (m1+m2)*vf. The question gives us m1 = 80kg, v1 = 4 m/s, and m2 = 100kg. Plugging this into the equation gets us: 80 * 4 = (80 + 100) * vf. Solving for vf, we get: vf = 320/180 = 1.78 m/s.