# Distance a Block Moves up a Ramp

A mass m is moving at a speed of Vo and slides smoothly onto a sloped big block of mass M. The big block can also move on the table surface. Assume that everything moves without friction. After the small block reaches the height h on the slope It slides down. Find the height h.
Problem Number 15-16 From:

2. Ug = mgh and KE = .5mVo^2

3. The Attempt at a Solution :
.5(m+M)Vo ^ 2 = mgh

solve for h = [(m+M)v^2]/(2mg)

Which isn't the correct answer in the answer set, I am assuming I am factoring in how the big block moves into my equation wrong. How would I set this up to receive the answer given in the solution set?
Thanks.

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Doc Al
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3. The Attempt at a Solution :
.5 (m+M) v ^ 2 = mgh
I don't quite understand what you're doing here. What's v? The initial velocity of the small mass? Or the final velocity of both?

I assume you want to apply conservation of energy. So compare the initial and final energies. (That's part of what you need to do.)

I don't quite understand what you're doing here. What's v? The initial velocity of the small mass? Or the final velocity of both?

I assume you want to apply conservation of energy. So compare the initial and final energies. (That's part of what you need to do.)
Sorry, v is the initial velocity of the small block, there is a diagram in the PDF I linked to as well.

For the energy conversion, what is wrong with what I did? (I converted the KE of small block and the big sloped block, because the big sloped block will move when the small block hits it, and then converted it into gravitational potential)

Doc Al
Mentor
For the energy conversion, what is wrong with what I did? (I converted the KE of small block and the big sloped block, because the big sloped block will move when the small block hits it, and then converted it into gravitational potential)
(1) Do you think the speed of the wedge is the same as the initial speed of the small block?
(2) Do you think that the speed of both is zero when the small block reaches the highest point?

Again, to apply conservation of energy: compare the total mechanical energy of the system before the small block touches the wedge to what it is when it's at its highest point.

(1) Do you think the speed of the wedge is the same as the initial speed of the small block?
(2) Do you think that the speed of both is zero when the small block reaches the highest point?

Again, to apply conservation of energy: compare the total mechanical energy of the system before the small block touches the wedge to what it is when it's at its highest point.
1. Yes, I was assuming they would move at the same speed, which may be my problem.

2. The speed of the small block will be zero when the block reaches the highest point. Not sure what will happen to the big sloped block.

So for the energy conversion

1. Total Mechanical Energy of the System before Small Block Touches Wedge = Total Mechanical Energy of the System When Block is at it's highest point

2. .5(m)Vo^2 = ?

The left side of that formula has to be right as before it touches the big block will not be moving

3. .5(m)Vo^2 = mgh + ?

Then adding the gravitational potential energy of the small block when it reaches its height on the big block. At this point the small block will have no kinetic energy.

What else needs to be added to the right side? .5(M)Vo^2, because the the big sloped block never stopped moving?

Would the two blocks move slower when they collide? Is that what I'm missing?

Thanks.

Doc Al
Mentor
1. Yes, I was assuming they would move at the same speed, which may be my problem.
That would violate energy conservation. Initially, only the small block is moving.

2. The speed of the small block will be zero when the block reaches the highest point. Not sure what will happen to the big sloped block.
This is another incorrect assumption. Hint: What else is conserved besides energy?

So for the energy conversion

1. Total Mechanical Energy of the System before Small Block Touches Wedge = Total Mechanical Energy of the System When Block is at it's highest point

2. .5(m)Vo^2 = ?

The left side of that formula has to be right as before it touches the big block will not be moving
Good.

3. .5(m)Vo^2 = mgh + ?

Then adding the gravitational potential energy of the small block when it reaches its height on the big block. At this point the small block will have no kinetic energy.
Again, do not assume that.

What else needs to be added to the right side? .5(M)Vo^2, because the the big sloped block never stopped moving?
That would imply that the wedge is moving at the initial speed of the small mass--that would violate conservation of energy. (Assuming M > m.)

See my hint above.

Momentum is also conserved, so I believe I understand the situation. Initially the first block will be moving at Vo then it will collide with the wedge, in which the wedge will move at a slower speed than the smaller block. Because the smaller block moves faster, it will go up the ramp, in which its kinetic energy will be converted into gravitational potential energy.

I will try this in equation form and see if it works. I still do not understand why at the highest point, the small block will have kinetic energy. Wouldn't the kinetic energy be converted into gravitational potential at the highest point?

Doc Al
Mentor
Momentum is also conserved, so I believe I understand the situation. Initially the first block will be moving at Vo then it will collide with the wedge, in which the wedge will move at a slower speed than the smaller block. Because the smaller block moves faster, it will go up the ramp, in which its kinetic energy will be converted into gravitational potential energy.
The original KE becomes the KE + PE in the final configuration.

I will try this in equation form and see if it works. I still do not understand why at the highest point, the small block will have kinetic energy. Wouldn't the kinetic energy be converted into gravitational potential at the highest point?
Not totally. And if the speed of the small mass was zero at the top, it would still be moving with respect to the wedge (which you agree must be moving)--so it couldn't have reached the highest point yet. Hint: What must be true about the relative speed of the two masses when the small mass reaches the highest point?

The original KE becomes the KE + PE in the final configuration.

Not totally. And if the speed of the small mass was zero at the top, it would still be moving with respect to the wedge (which you agree must be moving)--so it couldn't have reached the highest point yet. Hint: What must be true about the relative speed of the two masses when the small mass reaches the highest point?
So there is still kinetic energy in the small block because it is on the wedge which is still moving.

So when the small block begins moving faster than the wedge when they collide. The small block will climb the wedge. Once it reaches the highest point, the wedge and the small block will be moving at the same speed?

Doc Al
Mentor
So there is still kinetic energy in the small block because it is on the wedge which is still moving.

So when the small block begins moving faster than the wedge when they collide. The small block will climb the wedge. Once it reaches the highest point, the wedge and the small block will be moving at the same speed?
Exactly!

Exactly!
Great! Will try that out. Thanks for the help!