Question

A woman of mass 49 kg jumps off the bow

of a 78 kg canoe that is intially at rest.
If her velocity is 4.5 m/s to the right, what
is the velocity of the canoe after she jumps?
Answer in units of m/s i.
x x

Answer 1

The woman-canoe system has zero initial momentum. When the woman jumps from the canoe, the total momenta of the woman and canoe is conserved, so that

0 = (49 kg) (4.5 m/s) + (78 kg) v

where v is the velocity of the canoe. Solving for v, we find

(78 kg) v = - (49 kg) (4.5 m/s)

v = - 49/78 (4.5 m/s)

v ≈ -2.8 m/s

which is to say, the boat is given a velocity of 2.8 m/s to the left.