# Homework Help: Coefficient of restitution help

1. May 10, 2008

### rock.freak667

[SOLVED] Coefficient of restitution help

1. The problem statement, all variables and given/known data
A light elastic string AB, of natural length 0.5m and modulus of elasticity 12N, has a particle of mass 0.3 kg attached to B. The end A is attached to a fixed point at a distance of 0.7m above a horizontal table. The coefficient of restitution between the particle and the table is e. The particle is released from rest at A, hits the table and rebounds to a height of 0.22m above the table. Find the value of e,
assuming that there is no air resistance.

2. Relevant equations

$e= \frac{\Delta v}{\Delta u}$
KE +PE= constant
3. The attempt at a solution

So using the law of conservation of mechanical energy:
When the mass is moving down the total energy is

$$E=\frac{1}{2} \frac{\lambda}{a} x^2 +mgh+ \frac{1}{2}m(u-0)^2$$

$$E=\frac{1}{2}\frac{12}{0.5}(0.2)^2 +(0.3)(9.81)(0.7)+ \frac{1}{2}(0.3)u^2$$

When the mass hits the table and comes back up

$$E=\frac{1}{2}\frac{12}{0.5}(0.2)^2 + (0.3)(9.81)(0.22) + \frac{1}{2}(0.3)(eu)^2$$

and those two are equal and they should give me e, but I have an unknown,u, as well. What did I forget to do?

2. May 11, 2008

### alphysicist

Hi rock.freak667,

Can you describe where your two energy equations are being applied at? For example, in your first equation, the first term has a length of 0.2 which would indicate that the equation is being written for when the mass is at the table, but the height is 0.7 which indicates the mass is at its starting point.

If you do write the energy equation for the beginning and ending points, then you can find the total energy on each side of the bounce, since you know the velocity there. If you write the energy equation just before and just after the bounce, you can also find the velocity right before and after the bounce.

Remember that the energy is conserved along the falling motion, and it is separately conserved along the rising motion, but the energy before the bounce and after the bounce are not the same.

3. May 11, 2008

### rock.freak667

So then for the first equation.
$mgh=\frac{1}{2}kx^2 +\frac{1}{2} mv^2$

4. May 12, 2008

### alphysicist

That looks right to me; the second equation will have the same form. What does that gives?

5. May 12, 2008

### rock.freak667

But after impact. It still has the elastic pe which goes into gpe and ke

so that

$\frac{1}{2}kx^2=mgh+\frac{1}{2}mv^2$

and v=eu.

I did that and after solving I got e=0.5 but the answer is $\frac{1}{3}$

6. May 12, 2008

### alphysicist

No, I think you'll have two equations of exactly the same form. For the case after the bounce, the initial energy (right after the bounce) is elastic potential energy and kinetic energy, and the final energy (at the highest point after the bounce) is gravitational potential. So the equation is the same for the other side:

$$\frac{1}{2}kx^2 + \frac{1}{2}mv^2 = mgh$$

just with different numbers for v and h.

7. May 12, 2008

### rock.freak667

Well that makes some more sense..thanks, I'll try it in a while.

8. May 12, 2008

### rock.freak667

ok well with your equation I get 0.33 to 2dp...but the answer has 1/3 exact. Would my answer be the same since I took g=9.81 and round off errors?

EDIT: wait...I was told g=10...well I got the answer thankssss

But I still am a bit unsure of the elastic and kinetic energy after impact. at impact, it has elastic, I get that. But how does it it have kinetic as well?

Last edited: May 12, 2008
9. May 12, 2008

### alphysicist

The coefficient of restitution relates the speeds just before and just after contact with the table (just before the collision and just after the collision). So the v in the second energy equation is after the force from the table has caused the ball to turn around and head back upwards.

10. May 12, 2008

### rock.freak667

oh thanks then!