An empty canoe, mass mc= 25kg, is floating away from a dock i calm water with a

Question

An empty canoe, mass mc= 25kg, is floating away from a dock i calm water with a speed of vc=0.80m/s in the +x direction. When the canoe's owner (mass mp= 75kg) realizes this, he gets a running start of vp= 2.40m/s in the +x direction and jumps into the canoe.

A) What is the velocity of the caoe-person system just after he lands in the canoe? Indicate the direction with a sign and your anwser ignore frictional forces.

B) What is the change in momentum of the canoe only during this "collision"? Be sure to include the correct sign

C) What is the change in momentum of the person only during this "collision"? Be sure to include the correct sign

Explanation / Answer

a)
use conservation of momentum,
m1*v1i + m2*v2i = (m1+m2)*vf
25*0.80 + 75*2.40 = (25+75)*vf
vf = 2 m/s
Answer: 2m/s

b)
vi = 0.8 m/s
vf = 2 m/s
change in momentum = m* (vf-vi)
= 25 * (2-0.8)
= 30 Kgm/s

c)
vi = 2.4 m/s
vf = 2 m/s
change in momentum = m* (vf-vi)
= 75 * (2-2.4)
= - 30 Kgm/s

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