Batman (91kg) fakks straight down from a bridge onto a boat?

Question

..., hitting the boat with a speed of 18m/s. If the boat and the dirtball Batman is after have a combined mass of 510 kg, and the boat is initially moving at a speed of 11 m/s, what is the speed of the boatafter batman lands on it?

Answer 1

Assuming that the boat is not moving under constant power, and neglecting the force of water resistence, the boat's new velocity would be 9.33 m/s.

We know this because of conservation of momentum. The momentum after the collision must be the same as the momentum before the collision. Since momentum is equal to mass * velocity, the boat's momentum before the collision is:

510kg * 11m/s = 5610kg*m/s.

Since the momentum is the same after the collision, we just divide the total momentum by the new combined weight to get the new velocity:

V = 5610kg*m/s / (510kg + 91kg) = 9.33m/s

The speed with which Batman hit the boat does not factor into the equation, unless we're interested in the boat's angular velocity. I'm assuming that you only want to know the boat's horizontal velocity over the water. Since Batman had no horizontal momentum (he was falling straight down), his movement would not affect the horizontal motion of the boat.

Answer 2

Note that the mass of the boat (without batman) is 510 - 91 = 419 kg

Initial horizontal momentum of batman = 0
Initial horizontal momentum of the boat = 419 kg * 11 m/s
= 4609 m-kg/sec

Final horizontal momentum of boat+dirtball batman = 510kg * speed

Momentum is conserved so the inital momentum must equal the final.

510s = 4609
Final speed of boat = 4609/510 = 9.04 meters/sec

Answer 3

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