A 450 gram cart and a 500 gram cart push off from each other on a 120 cm long track. If they reach the opposite ends of the track simultaneously, how far does the 450 cart move?

Starting from rest, two ice skaters push off against each other on smooth level ice (friction is negligible). One is a woman (m﻿1﻿ =51 kg), and one is a man (m﻿2﻿ =76 kg). The woman moves away with a velocity of 2.5 m/s. What is the recoil velocity of the man?

Relevance

1)

Since total initial momentum is 0, we have final momentum given by

m₁v₁+m₂v₂=0

0.45v₁+0.50v₂=0

0.45v₁ = -0.50v₂

v₁/v₂ = -0.50/0.45

v₁/v₂ = -10/9 ...(1)

Lets consider the 450g cart travels x cm distance

then the 500g cart will travel (120-x)cm

As their reach the same time we get

x/v₁ = (120-x)/v₂

x/(120-x) = v₁/v₂

from (1) we get

x/(120-x) = -10/9

9x = -1200 + 10x

19x = 1200

x = 1200/19 ≈ 63 cm <----That's how long the 450g cart will move.

2)

Since total initial momentum is 0, we have final momentum given by

m₁v₁+m₂v₂=0

51*2.4 + 76v₂ = 0

122.4 + 76v₂ = 0

76v₂ = -122.4

v₂ = -122.4/76 = -1.6 m/s <----- recoil velocity of the man

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• When two isolated initially at rest masses m1 and m2 push off of each other, they will have velocities in the ratio v1 / v2 = m2 / m1.

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