Batman (mass = 95 kg) jumps straight down from a bridge into a boat (mass = 560 kg) in which a criminal is fleeing. The velocity of the boat is initially +10 m/s. What is the velocity of the boat after Batman lands in it?
Or save yourself the trouble and use the following formula.
I just did this problem.
(m1*v01)/(m1+m2)
mindy's is right
Assuming the criminal's weight is included in that of the boat (560 kg), calculate the initial kinetic energy using
KE0 = (1/2)Mv0²
Calculate the new kinetic energy
KE1 = (1/2)(M+m)v²
Equate KE0 and KE1 and solve for v.
don't try this ^^ there wrong
To solve this problem, we can apply the principle of conservation of momentum. The momentum before Batman jumps into the boat is equal to the momentum after he lands in it.
Let's assume the velocity of Batman before jumping into the boat is zero since he is at rest initially. So, before the jump, the total momentum of the system is given by:
Total initial momentum = (Mass of Batman) * (Velocity of Batman) + (Mass of Boat) * (Velocity of Boat)
Since Batman is at rest, his initial velocity is 0. Therefore, the initial momentum is:
Total initial momentum = (Mass of Batman) * 0 + (Mass of Boat) * (Velocity of Boat)
= 0 + (95 kg) * (10 m/s)
= 950 kg*m/s
After Batman jumps into the boat, the total momentum of the system is given by:
Total final momentum = (Mass of Batman + Mass of Boat) * (Velocity of Boat after landing)
Since Batman and the boat move together after he lands, their velocities will be the same. Let's assume the final velocity of the boat after Batman lands in it is v.
Total final momentum = (Mass of Batman + Mass of Boat) * (Velocity of Boat after landing)
= (95 kg + 560 kg) * v
= 655 kg * v
According to the principle of conservation of momentum, the total initial momentum equals the total final momentum. Therefore:
Total initial momentum = Total final momentum
950 kg*m/s = 655 kg * v
To find the velocity of the boat after Batman lands in it (v), we can rearrange the equation:
v = (950 kg*m/s) / (655 kg)
v ≈ 1.45 m/s
Therefore, the velocity of the boat after Batman lands in it is approximately 1.45 m/s.