Find Speed of Clay before Impact: Friction Coefficient \muk = 0.35

[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

f_k = \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.

 

[Delphi51]

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

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

 

[Dark Visitor]

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

 

[Dark Visitor]

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

 

[Delphi51]

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

 

[Dark Visitor]

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

 

[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.

 

[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?

 

[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.

 

[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 f_k, n, and m_ctg. Is that right?

(fk is friction, m_ctg 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?

 

[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
f_k + n - m_ctg = 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 = w_f² - 2\alpha\Delta\theta

\omega_o² = 0 - 2 (3.43 m/s²)(2.1 m)

\omega_o = \sqrt{}-14.406

Please help me figure out where my mistake(s) are...

 

[Delphi51]

F = ma
fk + n - mctg = ma

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

 

[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...)

 

[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.

 

[Dark Visitor]

I used this:

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

f_k = ma

2.058 N = (.6 kg) a

a = 3.43 m/s²

 

[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.

 

[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

 

[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

 

[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)

 

[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

 

[Dark Visitor]

I don't quite understand... I know momentum before = momentum after. But what does the "M" in your equation stand for? Momentum or mass? And what numbers go in for that because we don't know initial velocity and final velocity is zero.

(please don't get off yet. I will be back in like 20-30 minutes. I really need to finish this problem)

 

[Delphi51]

M is for mass. P is for momentum.
The usual procedure after writing momentum before = momentum after is to write and mv for each moving object. Then you put in the numbers you know. The v after the collision is 3.8, NOT ZERO. Leave the unknown v before so you can solve for it.

 

[Dark Visitor]

Okay, well doing it out on paper, I got this:

p = mv
p_f = (.6 kg)(3.8 m/s)
p_f = 2.28

m_fv_f = m_iv_i
2.28 = (.1 kg) v_i
v_i = 22.8 m/s

 

[Delphi51]

All right! Looks good to me.

 

[Dark Visitor]

Okay, thank you so very much. I really appreciate the help. Thanks also for waiting for me online. It seems like most people get off before they finish helping me and then I don't finish what needed to be done cause I didn't understand.

 

[Delphi51]

Most welcome. Yes, nice to get it done in one evening - so many of these conversations run over days or I never hear how it worked out at all.

 

[Dark Visitor]

Well, 2 evenings lol, but thanks a lot. When it comes to Physics, I need an extra push. SO to speak... Yeah that's how it is for me too. I had 2 other posts on here from this weekend, and they never got finished. But it's okay I guess. One is better than nothing.