L_{min}=56.7mAnswer: 340.7mSolve 2 Momentum Questions: Quick Answers!

[subwaybusker]

[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$$_{p}$$=1000kg (plane)
v$$_{p}$$=50 m/s [fwd]
m$$_{b}$$=2000kg (barge)
v$$_{pb}$$'=speed of plane and barge after inelastic collision?

m$$_{p}$$v$$_{p}$$=(m$$_{p}$$+m$$_{b}$$)v$$_{pb}$$'
(1000kg)(50 m/s [fwd])=(3000kg)v$$_{pb}$$'
v$$_{pb}$$'=16.7 m/s [fwd]

F$$_{f}$$=force of friction= 0.25m$$_{p}$$g=0.25(1000kg)(9.8N/kg)=2450N
Since F$$_{f}$$ is only force, F$$_{f}$$=$$\Sigma$$a
therefore, 2450=(1000kg)a
a=-2.45 m/s [fwd]

Use V$$_{2}$$$$^{2}$$=V$$_{1}$$$$^{1}$$+2a$$\Delta$$d

Plane starts to accelerate negatively when it lands on barge, so
V$$_{1}$$=16.7 m/s [fwd]
V$$_{2}$$=0 m/s

but $$\Delta$$d works out to be 56.7m

 

[jae05]

Both the dog and the plane end up in the water. Who builds an 8m long raft and since when do planes land on barges. Hah.

But seriously, for the first problem, think of viewing the problem from two frames of reference, the river frame and the raft frame. From the river frame, we see that momentum of the raft-dog system is conserved (1st eqn). Then it's a simple task of using Gallilean velocity addition to get your 2nd equation to solve for the velocities of the dog and the raft. Can you imagine how much more irritating the question could be if the dog was traveling at 0.5c. :D

For the 2nd question, 16.7m/s is the speed of the final system. The friction is what brings the combined system to that speed. So using $$v_{plane}^2=u_{plane}^2+2as$$, your initial is 50m/s, and your final is 16.7m/s. And don't forget that by N3L, the boat also is acted upon by the frictional force, but in the opposite direction, so you need to subtract the distance moved by the boat from the distance moved by the plane to get your final answer.

 

[subwaybusker]

hey, sorry, i must be really dumb. i don't understand what you are saying for the first question. and for the second question, if the friction is between the deck and the plane's wheels then doesn't friction act after the plane has hit the barge? and what is N3L? sorry, I'm just really confused here

 

[jae05]

ok 2nd question first. when the plane hits the barge, not all the plane's momentum is transferred to the barge straight away. it's not an instantaneous effect. the physics you have effectively done with the momentum conservation, is the plane crashing into the deck. taking the initial speed of the plane to be 16.7m/s in your kinematic equation, means that instantaneously when the plane hits the deck, it slows down by 33.3m/s. i think that kind of decceleration is tantamount to crashing and burning.

so what happens, is that the plane hits the deck at 50m/s, and the frictional force slows the plane down, and as you have done, by momentum conservation, we see that the final ship + plane system travels at 16.7m/s. when the plane hits 16.7m/s, from the frame of reference of the ship, it has come to a stop. from the external reference frame, the plane and ship are still moving, but from the point of view of someone on the barge, or the pilot, the plane is essentially stationary.

then by N3L = Newton's third law, as the ship slows the plane down, the plane must speed the ship up. action-reaction. if you draw a force diagram, the friction acts backwards on the plane, and there must be another force equal and opposite to the frictional force that acts on the ship. so you need to work out the distance moved by the ship in the time taken for the plane to reach 16.7m/s, to subtract off from the total distance moved by the plane, to find the length of the barge.

for the first question, try to visualise that as the dog moves forward, it will push the raft backwards, because the momentum of the entire dog-raft system has to be zero (conservation). so even though the dog has walked 8m along the raft (say to the right), the raft has shifted in position in the external reference frame (to the left say $$d_{raft}$$m). so in the external frame, it would appear the dog would have moved only $$8-d_{raft}$$m. try to work it from there. :)

hope it helps!