Collisions in a harmonic oscillator

[Leo Consoli]

Homework Statement

The problem is from the Monbukagakusho exam.[/B]
An object of mass M is hanging by a light spring of force constant k from the ceiling. A small ball of mass m which moves vertically upward collides with the object. After the collision, the object and the small ball stick together and oscillate in simple harmonic motion. Before the collision, the object is at rest. The speed of the small ball just before the collision is denoted as v. The acceleration of gravity is denoted as g. Answer the following questions.
1) Find the amount of stretch of the spring from its natural length before the collision.
2) Find the speed of the object and the small ball just after the small ball collides with the object and they stick together.
3) Find the amount of decrease of the sum of kinetic energies of the small ball and the object, before and after the small ball collides with the object.
4) Find the period of the simple harmonic oscillation after the small ball and the object stick together.
5) During the simple harmonic oscillation of the small ball and the object which are stuck together, the spring is at its natural legth when the object is at its highest position. Find the kinect energy of the ball just before it collides with the object.

Homework Equations

Hooke's law.
Period of a harmonic oscilator.
Conservation of momentum.
Cinetic energy.

The Attempt at a Solution

I was able to solve from 1 to 4, but I left if here because I think it can help solving 5, here is what I did from 1 to 4:
1)

2)

3)

4)

Now as for the 5 question I can't really find an answer because according to the formula, the kinectic energy would be mv^2/2, but the alternatives are very different from this, the collision is inelastic so mechanic energy is not conserved so I can't find a relation between the mechanic energy of the ball, and the mechanic energy of the oscilating system with the ball and the object together, here are the alternatives. Thanks for any help.

 

[SammyS]

Homework Statement

The problem is from the Monbukagakusho exam.[/B]
An object of mass M is hanging by a light spring of force constant k from the ceiling. A small ball of mass m which moves vertically upward collides with the object. After the collision, the object and the small ball stick together and oscillate in simple harmonic motion. Before the collision, the object is at rest. The speed of the small ball just before the collision is denoted as v. The acceleration of gravity is denoted as g. Answer the following questions.
1) Find the amount of stretch of the spring from its natural length before the collision.
2) Find the speed of the object and the small ball just after the small ball collides with the object and they stick together.
3) Find the amount of decrease of the sum of kinetic energies of the small ball and the object, before and after the small ball collides with the object.
4) Find the period of the simple harmonic oscillation after the small ball and the object stick together.
5) During the simple harmonic oscillation of the small ball and the object which are stuck together, the spring is at its natural length when the object is at its highest position. Find the kinetic energy of the ball just before it collides with the object.

Homework Equations

Hooke's law.
Period of a harmonic oscillator.
Conservation of momentum.
Kinetic energy.

The Attempt at a Solution

I was able to solve from 1 to 4, but I left if here because I think it can help solving 5, here is what I did from 1 to 4:
1) View attachment 237157
2) View attachment 237158
3) View attachment 237159
4) View attachment 237160

Now as for question 5, I can't really find an answer because according to the formula, the kinetic energy would be mv^2/2, but the alternatives are very different from this, the collision is inelastic so mechanic energy is not conserved so I can't find a relation between the mechanic energy of the ball, and the mechanic energy of the oscillating system with the ball and the object together, here are the alternatives. Thanks for any help.

Hello @Leo Consoli , $\ \ \$

Use conservation of energy to get an expression for the K.E. of the ball and object immediately after the collision in terms of the change in potential energy going from that position to the highest position. From this K.E., you should be able to get the K.E. of the ball immediately before the collision.

 

[Leo Consoli]

Thank you, the change in potential energy would be from the deformation going from x to 0 in the highest position, so the expression for the conservation of energy would be something like : KE + PE= maxKE ?

 

[SammyS]

Thank you, the change in potential energy would be from the deformation going from x to 0 in the highest position, so the expression for the conservation of energy would be something like : KE + PE= maxKE ?

Actually I would start by looking at the total mechanical energy at two positions, breaking the mechanical energy down into three components:

• Gravitational potential energy
• Potential energy stored in the spring
• Kinetic energy of the combined object and ball.

A) At the highest position -

B) At the position of the collision, immediately after the collision -

Equate mechanical energy for A) and B), then solve for KE (at B), which is $\ \frac{1}{2} (M+m) (v_f)^2 \$.

 

[Abhishek11235]

Now let's go in reverse. We know that if mass M is hanged then it stretches spring by x= Mg/k. Hence,if the particle hits the block then there is momentum conservation(Why? There is net external force of gravity on this system. Think) Now,from momentum conservation determine the velocity of combine system. We now invoke work-energy principal taking system as Block+mass+Earth+spring . Hence we have:

∆W = ∆K + ∆U +∆H.
Since in this system no external force acts we have ∆W=0 ,∆K= -1/2mv^2, ∆U= mgx+1/2kx^2. Now here x=mg/k.

Hence you can find v. Use momentum conservation to find v' (The speed of mass)

 

[Delta2]

Now let's go in reverse. We know that if mass M is hanged then it stretches spring by x= Mg/k. Hence,if the particle hits the block then there is momentum conservation(Why? There is net external force of gravity on this system. Think) Now,from momentum conservation determine the velocity of combine system. We now invoke work-energy principal taking system as Block+mass+Earth+spring . Hence we have:

∆W = ∆K + ∆U +∆H.
Since in this system no external force acts we have ∆W=0 ,∆K= -1/2mv^2, ∆U= mgx+1/2kx^2. Now here x=mg/k.

Hence you can find v. Use momentum conservation to find v' (The speed of mass)

What is $\Delta H$?? Other than that I mostly agree, except that where you say $m$ we should put $m+M$ while $x=Mg/k$. But now we have almost reveal the solution to the OP. At least I think so.

 

[Leo Consoli]

Thanks for the help, so basically I have to use conservation of energy to get an equation for Vf, then plug vf into the conservatio of momentum equation to find v of the small ball, then through Ke=mv^2/2 find the kinectic energy of the ball, what I didnt get is why is x=Mg/k and not (M+m)g/k.

 

[SammyS]

Thanks for the help, so basically I have to use conservation of energy to get an equation for Vf, then plug vf into the conservation of momentum equation to find v of the small ball, then through Ke=mv^2/2 find the kinectic energy of the ball, what I didnt get is why is x=Mg/k and not (M+m)g/k.

The collision occurs a distance x from the equilibrium position of the spring itself - the same value of x from your answer to Question #1.

Most of what @Abhishek11235 suggests that you do, you have already done to answer Questions 1 and 2.

 

[Leo Consoli]

I got to this point, it looks like the alternatives but its still not there.

 

[SammyS]

I got to this point, it looks like the alternatives but its still not there.

That's not quite right.

Total energy at (immediately after) the collision:

Kinetic: You should know this. General form is (½)(mass) (velocity)² .
Potential Energy (two contributions: One of them could be zero.)

PE for the stretched spring. (Don't forget the factor of ½ .)
PE as gravitational PE.​

Total energy at the highest point. (Position at which spring is neither stretched nor compressed):

Kinetic: You should know this. General form is (½)(mass) (velocity)²
Potential Energy (two contributions: One of them is zero)

PE for the spring.
PE as gravitational PE.​

Equate these.

You seem to be able to substitute for $X$ just fine. For the square of $v_f\,,\,$ use the square of your answer to question 2 exactly as it is given there.

 

[Leo Consoli]

How can I find the velocity at the highest point? From the theory of the SHM it would be the maximum velocity since X=0, but how can I find it?
Immediately after the collision the velocity would be vf, PE for the streched spring would be (KX^2)/2, and the gravitational PE could be zero if the reference would be the object, at the highest point I can't seem to find the velocity, the PE for the spring is 0 since X=0 and the gravitational PE would be (M+m)gx.

 

[Delta2]

The velocity at the highest point is simply zero, that's why it is the highest point, it can't go higher cause the velocity is zero there and then the velocity starts increasing with the mass going downwards.

 

[Leo Consoli]

Then if my idea in the 11th comment is right the result after equating the mechanic energies would be:

 

[Leo Consoli]

KX^2 times 1/2

 

[SammyS]

Then if my idea in the 11th comment is right the result after equating the mechanic energies would be:
View attachment 237221

Check your algebra.

(Answer to question re PE of spring follows.)

KX^2 times 1/2

 

[Leo Consoli]

Thanks, it was a typing mistake:

 

[SammyS]

Thanks, it was a typing mistake:
View attachment 237222

Very good.

Now substitute.
For $v_f\,:$

For $X\,:$

.
Simplify as well as solving for $v$.

 

[Leo Consoli]

Thank you and everyone here for the patience, I was finally able to solve it, the answer was letter f, here it is, thank you again:

 

[Abhishek11235]

What is $\Delta H$?? Other than that I mostly agree, except that where you say $m$ we should put $m+M$ while $x=Mg/k$. But now we have almost reveal the solution to the OP. At least I think so.

∆H is entropical loss here. Sorry for revealving the solution. Actually this is my first answer to post on PF.