# Homework Help: Particle Hanging by a String is Hit from Below by Another Particle

1. Apr 3, 2013

### jmlibunao

1. The problem statement, all variables and given/known data
A light elastic string has a natural length 1 m. One end of the string is attached to the fixed point O and a particle P of mass 4 kg is suspended from the other end of the string. When hanging in equilibrium, P is 1.2 m below O.

#1 Find the modulus of elasticity of the string.

When P is hanging in equilibrium, it is hit from below by a particle Q, of mass 2 kg, which is travelling vertically upwards. Immediately after the impact, P moves vertically upwards with a velocity u m/s. When the string is just taut, P is still moving vertically upwards with a velocity of √10 m/s.

#2 Find the value of u.

Given that Q is moving with a velocity of 4√3 m/s upwards before it hits P,

#3 show that it is momentarily at rest just after impact.

#4 Find the position of the lowest point, with respect to the equilibrium point, reached by P
in the subsequent motion.

2. Relevant equations
Young's modulus of elasticity (λ)
λ = (F/A)/(ΔL/Lo)
F - force
A - cross sectional area
ΔL - change in length
Lo - original length

g = 10 m/s^2

I'm also guessing you're gonna be using the formulas for Conservation of Energy, Ki + Pi = Kf + Pf, right?

I'm not so sure what other equations can be used

3. The attempt at a solution
So the cross sectional area, A, was not given in the problem. I just assigned a variable A for it so my λ = [(4 kg)(10 m/s^2)/A] / [(0.2)/(1)] = 200A

Sadly this is what I can do for now. Still thinking about the rest. Can anyone help?

Last edited: Apr 3, 2013
2. Apr 3, 2013

### Sunil Simha

To find the force per unit extension (k), consider the initial equilibrium of the ball. Write down the f.b.d. and balance the forces to get k. You don't need the young's modulus.

Once you've obtained that, use conservation of energy to find u and also remember to include the potential energy stored in the string in the equation.

3. Apr 4, 2013

### jmlibunao

I'm sorry, I don't quite follow this. How would I be able to get the force per unit extension by considering the initial equilibrium? The modulus of elasticity is being asked by the problem.

I can't recall a formula that gives the potential energy when hanging from a string. I can't use P = mgh since height, h, was not given.

4. Apr 4, 2013

### Sunil Simha

You see, the original length of the string was 1m right? When the ball was hung by it , the string was extended to 1.2 m . Assuming that k was the force constant (Elastic Force applied by string due to unit extension), the forces acting on the ball are kx (x is the extension) and mg. So now the ball is in equilibrium, the forces are balanced and thus k can be found.

When the elastic string is extended, it stores some energy and this energy is called potential energy of the string. Its expression can be found by following the definition of work done to extend the string.

Energy stored E = -∫F.dx where F = -kx. (the integral is a simple exercise )

Edit: You don't need h to find the gravitational potential energy because you only need the difference in gravitational potential energy in the initial and final state. The difference can be calculated because the change in altitude of the ball is given.

Last edited: Apr 4, 2013
5. Apr 4, 2013

### jmlibunao

Okay, thanks for clearing that up! So I can just use kx = mg and just solve for k. But what about the modulus of elasticity? You say I don't need it but the problem is asking for it.

Okay, now I remember. The potential energy stored in the string would be E = (1/2)kx^2, the same equation used for springs. The difference that you're talking about would be the Work, right? I also don't get the part where the problem says "When the string is just taut, P is still moving vertially upwards with a velocity of √10 m/s." Why would the string be taut? If P was hit from below then it would move upward, "un-stretching" the string, right? I think I'm having a hard time just understanding what the problem is giving me.

6. Apr 4, 2013

### Sunil Simha

Where?

The work done on the string is stored as its potential energy. The string would be just taut when there is just a bit of tension (kx) in it. That is, when it just has 0 extension.

7. Apr 4, 2013

### jmlibunao

It's the #1 question.

So I computed my k = 200. To compute for u can I use the following equation?

(1/2)(200)(0.2^2) + (1/2)(4 kg)(u^2) = (1/2)(4 kg)(10)

The velocity u is just after the impact so the string would still be stretched at that moment so I included (1/2)(200)(0.2^2) on the left side with u. On the right side it would only be the kinetic energy (1/2)(4 kg)(10) alone since (1/2)(k)(x^2) would be 0

8. Apr 4, 2013

### Sunil Simha

I think the equation should be $\frac{kx^2}{2} + \frac{mu^2}{2} = mgx + \frac{mv^2}{2}$. The ball is gaining gravitational potential energy but is losing kinetic energy the spring potential energy.