Question

A woman of mass 44 kg jumps off the bow of

Answer 1

Answer:

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.

$(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})$

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]

$(44+69)*0= (44*2.5)-(69*v_{3})\\110 = 69*v_{3}\\v_{3}=1.59[m/s]$