Answer:
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