Batman (mass = 96.1 kg) jumps straight down from a bridge into a boat (mass = 458 kg) in which a criminal is fleeing. The velocity of the boat is initially +11.3 m/s. What is the velocity of the boat after Batman lands in it?

The velocity of the boat after the batman lands in it is +9.26 m/s

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Note: The collision between the Batman and the boat is an inelastic collision.

m'u'+mu = V(m+m') Equation 1

Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.

make V the subject of equation 1

V = (m'u'+mu)/(m+m') Equation 2

Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.

Substitute these values into equation 2

V = [(96.1×0)+(458×11.2)]/(96.1+458)

V = 5129.6/554.1

V = +9.26 m/s