Solve Implus and Momentum for Block A Velocity at t=1

[PaintballerCA]

Question:
In the figure below, block B is moving with a speed of 3 ft/s at t=0. Block A has a weight of 10 lb and B has a weight of 3 lb. If the surface A is on is smooth, what is the velocity of A at t=1?

Attempt:
Given the pully set-up, V(A) should be twice that of V(B) since for any distance travled by B will require twice that distance in rope. So, 2V(B) = V(A). My other equation is the principle equation of impluse and momentum:
M(1)V(1) + integral(F*dt) = M(2)V(2)



For this problem, that gives:

[M(A)V(A1) + M(B)V(B1)] + integral(F*dt) = [M(A)V(A2) + M(B)V(B2)]

2V(B) = V(A)

For this problem, the force acting on the system is that of gravity on block B, so integral(F*dt) is a constant force integrated over the change in time (1 second), for an impluse of 3 lb*s.

Problem:

When I do this, I get a final velocity of 14.8 ft/s for block A, which is wrong (block A should have a velocity of 10.5 ft/s after 1 second) and I am not sure where I am making a mistake.

 

[Doc Al]

Given the pully set-up, V(A) should be twice that of V(B) since for any distance travled by B will require twice that distance in rope. So, 2V(B) = V(A).

Good.

My other equation is the principle equation of impluse and momentum:
M(1)V(1) + integral(F*dt) = M(2)V(2)

How interesting! This might be the first time I've seen someone trying to use this approach on this kind of problem.

For this problem, that gives:

[M(A)V(A1) + M(B)V(B1)] + integral(F*dt) = [M(A)V(A2) + M(B)V(B2)]

2V(B) = V(A)

For this problem, the force acting on the system is that of gravity on block B, so integral(F*dt) is a constant force integrated over the change in time (1 second), for an impluse of 3 lb*s.

The problem is that (1) momentum is a vector, and (2) there are multiple forces acting on the blocks, not just gravity. You could use this approach, if you applied it separately to each mass and accounted for all the forces acting.

I suggest that instead of using this approach, that you analyze the forces acting on each mass, apply Newton's 2nd law to each (separately), and solve for the acceleration of mass A.

 

[PaintballerCA]

I was thinking for doing that, since the acceleration of A is twice that of B.

The fact that momentum is a vector and not a scaler did enter my mind, but my argument for treating it as a scaler was the pully system; I kind of treated the rope itself as an axis. If the system was changed to a set up like this, I don't see how it would effect the mechanics of the system. In this case, both momentums would be in the same axis.

The only forces I can account for are gravity on block B, gravity and the normal force on block A (which is predendicular to the direction of travel and will therefore have no effect) and the tension in the rope. I think my really issue here is figuring out the tension in the rope. This has gotten down to a concept problem. Since there is no friction in this problem, the force pulling block A along will determine the rate at which A accelerates. My problem is figuring out that force. If block B was not moving, then the rope would have 1.5 lb of tension in it (2 forces equal to 1.5 lb going up to counter the 3lb going down). Since block B moves, there is a net force (and thereby a net acceleration) on it going down.

I know that if I can find the acceleration of one block, I can find it for the other. If I have the acceleration of a block, I can use the principle of impluse and momentum on each block (the force on the block being its acceleration times its mass).

 

[Doc Al]

If you knew the acceleration of the blocks, you'd be done (with a bit of kinematics). No need to use impulse and momentum if you already know the acceleration.

To find the acceleration, first analyze the forces acting on each block. What forces act on block A? On block B? Then apply Newton's 2nd law to each block. Combine the two equations to solve for the accelerations and the tension in the rope. Don't forget to include the constraint that the acceleration of A is twice that of B.

If you prefer to use impulse and momentum instead of Newton's 2nd law, to find the change in speed instead of the acceleration, no problem. Same idea.

 

[PaintballerCA]

Ah, I get it. The force acting on each block is different, so even if they were on the same axis, one would still need to apply two different impulse and momentum equations.

So, in the case of block A, the forces acting on the block are gravity, the normal force, and the force from the rope, which I'll call F(A). The force on block B will be F(B) which is its weight. That means that F(net)=F(B)-F(A) where F(net) is the net force acting on block B. If F(net) is divided by the mass of block B, then one would get the acceleration of block B, correct?

This means that F(net)=M(B)*(g)-M(A)*a(A)=M(B)*a(B). Since a(A)=2a(B), this gives:

F(net)=M(B)*(G)-M(A)*2a(B)=M(B)*a(B)

I can solve for a(B) and then for a(A) and finally apply the impulse and momentum equations for each.

This right or have I missed something?

 

[Doc Al]

So, in the case of block A, the forces acting on the block are gravity, the normal force, and the force from the rope, which I'll call F(A).

What are you calling F(A)? For simplicity, I'd call the tension in the rope T.

The force on block B will be F(B) which is its weight.

What about the tension in the ropes pulling on it?

That means that F(net)=F(B)-F(A) where F(net) is the net force acting on block B. If F(net) is divided by the mass of block B, then one would get the acceleration of block B, correct?

I'm confused as to why you are mixing up the forces acting on different bodies. (Resist the temptation to do it in your head!) Keep them separate, apply Newton's law, then combine the resulting equations.

What's the net force on block A? Apply Newton's 2nd law.
What's the net force on block B? Apply Newton's 2nd law.
Combine and solve for the acceleration.

Once you have the acceleration, use v_f = v_i + at.

 

[PaintballerCA]

Solved it.
One reason why I used the principle of impulse and momentum is because this is the section we are covering in dynamics. I would of figured that using the principle of impulse and momentum would have made this type of problem easier to solve.