# Homework Help: Particles of mass 600g and 400g are connected by a light

1. Nov 18, 2009

### MrPickle

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

Particles of mass 600g and 400g are connected by a light, inextensible string passing over a smooth fixed pulley. Initially both masses hang vertically, 30cm above the ground. If the system is released from rest find the greatest height reached above ground by the 400g mass.

2. Relevant equations

F = ma and the suvat equations

3. The attempt at a solution

Using common sense I figured the highest it could go would be 60cm, because after the 600g mass has come down 30cm, it will be on the ground and cannot go down any more but the answer in the back of the book says 66cm. I can't see how the 600g mass can go down 36cm without going into the ground?

I'm not sure how to do it otherwise, my guess is to work out the acceleration using F = ma then use the suvat equations somehow.

2. Nov 18, 2009

### Staff: Mentor

Re: Pulleys

It's certainly true that the 600 g mass can only fall 30 cm. But when the 600 g mass hits the ground, does the other mass stop dead? What changes at that point? Hint: How fast are the masses moving when one hits the ground?

3. Nov 18, 2009

### MrPickle

Re: Pulleys

So you're saying it's going to roll or slide, that makes sense.

I could find the acceleration using F = ma, then I would have to use the suvat equations to find the speed at which it hits the ground, then use another suvat equation to find the distance travelled until it comes to a rest. I'm not sure what my acceleration would be in the last suvat equation, the same as what it fell as or g?

4. Nov 18, 2009

### Staff: Mentor

Re: Pulleys

I never said anything about rolling or sliding.

That's one way.
You tell me. Once the 600 g mass hits the ground, what forces act on the other mass?

5. Nov 18, 2009

### MrPickle

Re: Pulleys

Well how else is it going to move?

Weight, Normal contact force and the forward force? That's if there's no friction.

6. Nov 18, 2009

### Staff: Mentor

Re: Pulleys

From your description, I'm imagining a string draped over a pulley with a mass at each end hanging freely. No contact with anything but the string. No rolling or sliding. Am I wrong? (If so, please post a diagram.)

7. Nov 18, 2009

### MrPickle

Re: Pulleys

I was including the ground in my diagram. There was no diagram with the question

8. Nov 18, 2009

### Staff: Mentor

Re: Pulleys

You don't need to worry about the ground. What matters is what happens to the 400 g mass, the one that doesn't hit the ground.

9. Nov 18, 2009

### MrPickle

Re: Pulleys

But then the 600g mass is going into the ground?

Oh wait, is it like 'denting' the ground then? And the acceleration remains the same for the final suvat?

Thank you (:

10. Nov 18, 2009

### Staff: Mentor

Re: Pulleys

We don't care what happens to the 600g mass after it hits the ground. But once it hits, what's different about the motion of the 400g mass? Once the 600g mass hits the ground, what forces act on the 400g mass and what is its acceleration?

11. Nov 18, 2009

### MrPickle

Re: Pulleys

Weight and the Normal Contact force or is it the tension?

I'm not sure what the acceleration is going to be

Or is it Weight = mass x acceleration? So it'll be 9.8?

12. Nov 18, 2009

### Staff: Mentor

Re: Pulleys

While the 600g mass is falling, two forces act on the 400g mass: Gravity and the tension force. Once the 600g mass hits the ground, the string goes slack--no more tension. At that point, the only force on the 400g mass is gravity.

13. Nov 18, 2009

### MrPickle

Re: Pulleys

Okay, I've worked it out but I've gone wrong somewhere. I ended up getting a distance of -0.006 from my final suvat.

Here's my working out
Acceleration:
600g - T = 600a
T - 400g = 400a
T = 400a + 400g

600g - 400a - 400g = 600a
200g - 400a = 600a
200g = 1000a
$a = \frac{1}{5}ms^{-1}$

Speed at which it hits the ground:
$u = 0ms^{-1}, v = ?ms^{-1}, s = 0.3m, a = \frac{1}{5}ms^{-1}$
$v^2 = u^2 + 2as$
$v^2 = \frac{2}{5} \times 0.3 = 0.12$
$v^2 = \frac {\sqrt{3}}{5}ms^{-1}$

Distance it travels:
$u = \frac {\sqrt{3}}{5}ms^{-1}, v = 0ms^{-1}, a = 9.8ms^{-2}, s = ?m$
$v^2 = u^2 + 2as$
$0 = (\frac {\sqrt{3}}{5})^2 + 2\times9.8s$
$0 = 0.12 + 19.6s$
$19.6s = -0.12$
$s = -\frac{0.12}{19.6} = -\frac{3}{490}$

14. Nov 18, 2009

### Staff: Mentor

Re: Pulleys

Careful. The acceleration is (1/5)*g, not 1/5.

Right method, but incorrect acceleration from before.

Careful with signs: Here the acceleration is -9.8 m/s^2, since it acts down.

15. Nov 18, 2009

### MrPickle

Re: Pulleys

Gah, silly mistake.

Thanks a bunch for all the help.