Batman (mass = m kS) jumps straight down from a bridge into a boat (mass - 510kS) in which a criminal is fleeing. The velocity of the boat is initially + 11 m/s. What is the velocity of the boat after Batman lands in it?
To solve this problem, we'll need to apply the law of conservation of momentum. According to this law, the total momentum before Batman jumps is equal to the total momentum after he lands in the boat.
The momentum of an object is given by the product of its mass and velocity. Therefore, the initial momentum of Batman is:
P1 = m * 0 (as Batman is initially at rest)
And the initial momentum of the boat is:
P2 = mboat * 11 m/s
After Batman lands in the boat, the combined system will have one velocity, which we'll call Vfinal. The final momentum of the system is:
Pfinal = (m + mboat) * Vfinal
According to the law of conservation of momentum:
P1 + P2 = Pfinal
m * 0 + mboat * 11 m/s = (m + mboat) * Vfinal
Simplifying the equation gives us:
mboat * 11 m/s = (m + mboat) * Vfinal
Now, we can solve for Vfinal:
Vfinal = (mboat * 11 m/s) / (m + mboat)
Given that the mass of Batman (m) is kS and the mass of the boat (mboat) is 510kS, we can substitute these values into the equation:
Vfinal = (510kS * 11 m/s) / (kS + 510kS)
Simplifying further, we get:
Vfinal = (5610kS*m/s) / 511kS
Now, we have to consider the units. We need to multiply the fractions to obtain m/s as the final unit:
Vfinal = (5610 * m/s) / 511
Therefore, the velocity of the boat after Batman lands in it is (5610 / 511) m/s, or approximately 10.98 m/s.