Question

A 300. kg motorboat is coasting toward the dock at 0.50 m/s. Isaac, whose mass is 62.0 kg, jumps off the front of the boat with a speed of 3.0 m/s relative to the boat. What is the velocity of the boat after Isaac jumps?

Answer 1

0.087 m/s

-0.12 m/s

Step-by-step explanation:

The momentum is restored

(M₁ + M₂)V = M₁V₁ + M₂V₂...........Eqn 1

M₁ = Mass of the boat = 300 kg

M₂ = Mass of Isaac = 62 kg

V = collective velocity before Isaac jumped = 0.5 m/s

V₁ = Velocity of the boat after Isaac jumped = ?

V₂ = Velocity of Isaac after he jumped =

1st Scenaro; If Isaac jumps forward, V₂ = 2.5 m/s

2nd Scenaro; If Isaac jumps backward V₂ = 3.5 m/s

Using equation for Eqn one first scenario = (300 + 62)0.5 = 300V₁ + 62(2.5) = 0.086 m/s

Using equation for Eqn one second scenario = (300 + 62)0.5 = 300V₁ + 62(3.5) = - 0.12 m/s

Note: The negative sign indicates the jump was made backwards