# Homework Help: Speed before impact

1. Nov 21, 2009

### Dark Visitor

Mitch throws a 100 g lump of clay at a 500 g target, which is at rest on a horizontal surface. After impact, the target, including the attached clay, slides 2.1 m before stopping. If the coefficient of kinetic friction is $$\mu$$k = 0.35, find the speed of the clay before impact.

I don't know where to start in this problem. I know that the equation for kinetic friction is

fk = $$\mu$$kn

but I don't know what the normal force is. My only thought would be that n = mg, which means it would be .6 kg (5g + 1g = 6g). Is that correct? Any help would be appreciated. Please be detailed with each step, because I need to understand it and be able to show my work.

2. Nov 21, 2009

### Delphi51

n = mg = (0.1 kg + 0.5 kg)*g

It is a collision. You must use conservation of momentum.
Momentum before = momentum after

3. Nov 22, 2009

### Dark Visitor

Okay, well I have the kinetic friction equation complete. But I don't know where to go now.

4. Nov 22, 2009

### Dark Visitor

Anyone want to help? I need this preferably by tonight, but I have until tomorrow night.

5. Nov 22, 2009

### Delphi51

It is a collision. You must use conservation of momentum.
Momentum before = momentum after

6. Nov 22, 2009

### Dark Visitor

But how can I find either momentum? All I have is kinetic friction so far.

7. Nov 22, 2009

### Delphi51

Oh, I thought you had found the velocity after the collision using the friction approach. If you did not find the velocity, what did you find? If you show your work it avoids a lot of misunderstanding. If you found the friction force, use F = ma to find the acceleration and then use a motion formula to get the velocity.

8. Nov 22, 2009

### Dark Visitor

Well, I went back to our older posts, and realized I made a mistake. My friction formula had an error in it, which I have now fixed. Now I have:

n = mg = (.5 kg + .1 kg)(9.8 m/s2) = 5.88 N

But before I go plugging it all into "F = ma", which free body do I go by? The before they collide, or the after they collide?

9. Nov 22, 2009

### Delphi51

After they collide, the friction brings them to a stop in 2.1 meters.
Having found the normal force, use the friction formula to find the friction force.
Then F = ma to find acceleration.
Then find a motion formula with distance but not time in it so you can find the initial velocity of the combined lumps of clay after their collision.
Then momentum before = momentum after.

10. Nov 23, 2009

### Dark Visitor

I understand what I have to do now, but my question is how many forces are there in the "F = ma" equation? I see fk, n, and mctg. Is that right?

(fk is friction, mctg is weight of the clay and target combined)

I also realized I am missing the final velocity. I thought it would be zero, but I am unsure. Can you tell me whether zero is the final velocity or not?

Last edited: Nov 23, 2009
11. Nov 23, 2009

### Dark Visitor

Let me put what I have done on my own. I know something somewhere is wrong, I just can't figure out where, so please tell me where my mistake(s) are.

F = ma
fk + n - mctg = ma
(2.058 N) + (5.88 N) - (5.88 N) = (.6 kg) a
a = 3.43 m/s2

Then I found the equation with distance, but no time:

$$\omega$$f2 = $$\omega$$o2 + 2$$\alpha$$$$\Delta$$$$\theta$$

$$\omega$$o2 = wf2 - 2$$\alpha$$$$\Delta$$$$\theta$$

$$\omega$$o2 = 0 - 2 (3.43 m/s2)(2.1 m)

$$\omega$$o = $$\sqrt{}-14.406$$

12. Nov 23, 2009

### Delphi51

Only the horizontal forces cause the horizontal acceleration. So just
fk = ma

Yes, the final velocity is zero. But there is no rotation in this problem; you must use a linear formula, not a rotational formula. It is Vf² = Vi² + 2ad

13. Nov 23, 2009

### Dark Visitor

But that's still what I did in my last post, and something has to be wrong somewhere. How can I take a square root of a negative number? (if you do the velocity equation, since the final velocity is zero, it turns negative and you can't square root that...)

14. Nov 23, 2009

### Delphi51

What value of acceleration did you get when using only the horizontal forces?
The value of the acceleration is, of course, negative because it decelerates the object.

15. Nov 23, 2009

### Dark Visitor

I used this:

*(.35)(5.88 N) = 2.058 N*

fk = ma

2.058 N = (.6 kg) a

a = 3.43 m/s2

16. Nov 23, 2009

### Delphi51

Agree. Perhaps easier to understand that if you write
Fk = ma
μmg = ma
a = μg = .35*9.81 = 3.43 m/s²
Use a = -3.43 m/s² in the motion formula because the clay is decelerating.

17. Nov 23, 2009

### Dark Visitor

Okay, but I still get 3.8 m/s if I take the square root of 14.406, which 3.8 is nowhere near any of my answers. (remember it is multiple choice)

I realized I didn't post the choice options, so here they are:

* 45 m/s
* 36 m/s
* 27 m/s
* 23 m/s

Last edited: Nov 23, 2009
18. Nov 23, 2009

### Delphi51

I agree with 3.8! This is not the end; it is not even the beginning of the end. But it is, perhaps, the end of the beginning. (Winston Churchill after the Battle of Britain)

Now momentum before = momentum after

19. Nov 23, 2009

### Dark Visitor

Okay, but when I calculate the final momentum, I get zero because the final velocity was zero.

p = mv
p = (.6 kg)(0 m/s)

20. Nov 23, 2009

### Delphi51

The final final velocity is zero. But the velocity immediately after the collision is 3.8 m/s. Friction slows it to zero later.
Momentum before = momentum after
mv = M*3.8