(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] 2 Momentum Questions

Question #1

A dog of mass 10 kg is standing on a raft so that he is 20 m from shore. He walks 8 m along the raft towards shore and then halts. The raft has a mass of 40 kg, and we can assume that there is no friction between the raft and the water. How far is the dog from shore when he stops? [Answer: 13.6m]

Attempt:

i honestly have no idea. i'm thinking as the dog walks, the momentum is in one direction which means the momentum of the raft is in the other direction.

Question #2

A 1000 kg plane is trying to make a forced landing on the deck of a 2000 kg barge at rest on the surface of a calm sea. The only frictional force to consider is between the plane's wheels and the deck, and this braking force is constant and equal to one-quarter of the plane's weight. What must the minimum length of the barge be, in order that the plane can stop safely on deck, if the plane touches down just at the rear end of the deck with a velocity of 50m/s towards the front of the barge? [Answer: 340.7m]

Attempt: (please excuse me, i don't know how to input subscripts, etc)

m[tex]_{p}[/tex]=1000kg (plane)

v[tex]_{p}[/tex]=50 m/s [fwd]

m[tex]_{b}[/tex]=2000kg (barge)

v[tex]_{pb}[/tex]'=speed of plane and barge after inelastic collision?

m[tex]_{p}[/tex]v[tex]_{p}[/tex]=(m[tex]_{p}[/tex]+m[tex]_{b}[/tex])v[tex]_{pb}[/tex]'

(1000kg)(50 m/s [fwd])=(3000kg)v[tex]_{pb}[/tex]'

v[tex]_{pb}[/tex]'=16.7 m/s [fwd]

F[tex]_{f}[/tex]=force of friction= 0.25m[tex]_{p}[/tex]g=0.25(1000kg)(9.8N/kg)=2450N

Since F[tex]_{f}[/tex] is only force, F[tex]_{f}[/tex]=[tex]\Sigma[/tex]a

therefore, 2450=(1000kg)a

a=-2.45 m/s [fwd]

Use V[tex]_{2}[/tex][tex]^{2}[/tex]=V[tex]_{1}[/tex][tex]^{1}[/tex]+2a[tex]\Delta[/tex]d

Plane starts to accelerate negatively when it lands on barge, so

V[tex]_{1}[/tex]=16.7 m/s [fwd]

V[tex]_{2}[/tex]=0 m/s

but [tex]\Delta[/tex]d works out to be 56.7m

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# Homework Help: 2 Momentum Questions

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