A woman of mass 44 kg jumps off the bow of a 69 kg canoe that is intially at rest.
If her velocity is 2.5 m/s to the right, what
is the velocity of the canoe after she jumps?
Answer in units of m/si.

v₃ = 1.59 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of linear momentum.

That is, the momentum is conserved before and after the jump.

Before the jump, we have the mass of the woman and the canoe without speed (at rest). After the jump you have the woman moving to the right and the canoe moving to the left.

where:

m₁ = mass of the woman = 44 [kg]

m₂ = mass of the canoe = 69 [kg]

v₁ = velocity of the canoe at rest = 0

v₂ = velocity of the woman after jumping = 2.5 [m/s]

v₃ = velocity of the canoe after jumping = [m/s]

answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.

real, smaller and upside-down.   so the answer should be b.