# Homework Help: Block collision with a spring

1. Apr 25, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

A 1.0-kg block slides on a frictionless, horizontal air track with an initial velocity of v = 2.0 m/s. It collides with another 1.0-kg block, which is initially at rest. One end of an ideal massless spring of spring constant k = 200 N/m is attached to the second block, so that the moving block does not collide directly with the motionless block, but with the free end of the spring, causing it to compress. As the spring compresses, it causes the first block to slow down, and the second block to start moving.

What is the maximum compression of the spring during the interaction? (Hint: both blocks are never simultaneously at rest, so don't assume (1/2) kx^2max = (1/2) mv^2 )

2. Relevant equations
$m_a v_a + m_b v_b = m_a v_{af} + m_b v_{bf}$

$K_1 + U_1 = K_2 + U_2$

3. The attempt at a solution

This one has been giving me trouble. I want to split it up into parts. At first I was gonna have the first part be an energy problem where the block compresses the spring but I cant do that because as the problem says the other block will also be moving. Then I wanted to do the first part as the collision using momentum but i cant take their speed after the collision to be the same because the problem says the spring causes the first block to slow down as the second starts moving. I am really confused as to how to set up this problem. Please help point me in the right direction

2. Apr 26, 2014

### ehild

Here you need to follow the collision process, when the blocks interact through the spring. As the problem says, the first block slows down while the second one starts moving and speeds up. The spring is compressed in that process, till the velocities become equal. What happens afterwards?

The two blocks and the spring make a closed system. What do you know about the motion of the centre of mass? What is the common velocity of the blocks when they move together?

You can use conservation of momentum and conservation of energy during the whole process.

ehild

3. Apr 26, 2014

### toothpaste666

Ok so I can start out the problem by figuring out the common velocity of both blocks after the collision? the second block starts at rest and the masses are the same and cancel out so

Va= Vf+Vf = 2Vf and Va = 2 so

2 = 2Vf

so the velocity of the whole system moving together is 1. As they move the spring transfers the kinetic energy of the first block to potential energy in the spring which then transfers it back to kinetic energy in the second block. And this is happening as they all move together? So the first block slows and the second speeds up until the velocities are equal. After that the velocity of the second block is faster so they separate?

4. Apr 26, 2014

### ehild

Yes,after the moment reaching common velocity they start to separate.
What about the elastic energy of the spring?

ehild

5. Apr 26, 2014

### toothpaste666

it will all be transferred to kinetic energy at once and that would speed up the second block?

6. Apr 26, 2014

### ehild

It does not happen at once. See the pictures. What happens after the question mark?
When the blocks completely separate, the string becomes relaxed, and all its elastic energy transforms to kinetic energy of the blocks.
When is the elastic energy of the spring maximum?

ehild

#### Attached Files:

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7. Apr 26, 2014

### toothpaste666

When its fully compressed or extended. In this case i dont think it gets there because the second block starts moving

8. Apr 26, 2014

### ehild

Can the spring extend longer than its relaxed length? It is attached to one block only. So the spring energy is maximum at maximum compression. What is the kinetic energy of the blocks then?

ehild

9. Apr 26, 2014

### toothpaste666

The spring would be at its maximum compression right before they separate, so when the blocks have the same velocity. And after they separate the second block would be moving with the first blocks original velocity because there is no friction. So if Mf is the mass of the whole system when it is moving together, then

.5(Mf)(Vf)^2 + .5(k)(x)^2 = 0 + .5(Mb)(Va)^2 + 0

the left hand side of the equation is the energy of the system of the two blocks right after the collision when they are moving together with the speed Vf that we figured out before and with the spring as its potential energy and the right side of the equation represents the system after the blocks separate. The first 0 is because the first block is now at rest, the final zero is because now the spring is at its normal position with no energy and the middle part is because the second block is now moving with the first blocks original speed. Since x is the only unknown I can solve for x.

Would this give me the right answer?

10. Apr 26, 2014

### ehild

Yes, the spring is maximum compressed when the blocks have the same velocity. You can get the maximum compression from the equation you wrote, but instead of the state when the blocks already separated, you could have used the original situation. Both are correct.

ehild

11. Apr 27, 2014

Thanks!