Where Do Balls Collide When Dropped and Thrown Up Simultaneously?

[Fede Aguilera]

Homework Statement

A ball A is dropped from the top of a building of height h and simultaneously a ball B is thrown upwards and both balls collide. After the collision the ball A has the double the velocity of ball B
Determine the fraction of the building where the balls collide.

Homework Equations

mA.VA+mB.VB=mA.2V'+mB.V'
VA=-g.t
VB=Vo-g.t

The Attempt at a Solution

I tried with conservation of momentum, conservation of energy, but I can't get rid of the masses, they are always there and can't get them out of the equations, so I can't finish the problem.
Is there a special case when an object gets the double of the other's object velocity after a collision? I looked for it but I didn't find anything


PS: The answer is a fraction (obviously) but there are only numbers in it, so that's the reason I put it here and not in the Advanced Physics Homework.


Thanks in advance.

 

[haruspex]

I think you have to assume the masses are the same. You are already assuming work is conserved, which seems the greater assumption, but I think you need that too.

 

[Fede Aguilera]

I think you have to assume the masses are the same. You are already assuming work is conserved, which seems the greater assumption, but I think you need that too.

Are you sure about that? Then the problem will state that "An equal ball B is thrown simultaneously" but I'll try that out and see where it takes me. Thanks.

 

[haruspex]

Are you sure about that? Then the problem will state that "An equal ball B is thrown simultaneously" but I'll try that out and see where it takes me. Thanks.

I analysed the number of unknowns and number of equations, and it seemed to you needed to assume both conservation of work and equal masses. On that basis i obtained a solution, and it appears entirely consistent.

 

[Fede Aguilera]

Can you tell me which equations you used? I can't conclude the problem yet.

 

[nasu]

I analysed the number of unknowns and number of equations, and it seemed to you needed to assume both conservation of work and equal masses. On that basis i obtained a solution, and it appears entirely consistent.

How about initial speed of B? Will the result be independent of it, even assuming equal mass and elastic collision?

 

[haruspex]

How about initial speed of B? Will the result be independent of it, even assuming equal mass and elastic collision?

In this problem, no specific values emerge for anything - it's all a matter of ratios. There are specific ratios between all the speeds. If the speeds before collision are v_A down and v_B up, and the speeds after collision w_A up and w_B down, you are given w_A = 2*w_B. It follows from conservation laws and equal masses that 2*v_A = v_B. you will also find that the initial upward speed of B is 3*w_B.

 

[nasu]

Are you saying that you determined the fraction of the building's height with just the given data?
What is this fraction?

 

[Fede Aguilera]

I got it. Thank you haruspex!

@Nasu: If you care the answer is

Spoiler

1/3 from the top or 2/3 from the bottom

 

[nasu]

So there is only one initial speed that satisfies the conditions. Nice.

 

[haruspex]

I got it. Thank you haruspex!

@Nasu: If you care the answer is

Spoiler

1/3 from the top or 2/3 from the bottom

Hmm... I got 1/6 from the top.

 

[Lunippa]

How do I know that a ball falling in free fall, which collides with another object in free fall, is experiencing 0 gravity?

 

[neilparker62]

How do I know that a ball falling in free fall, which collides with another object in free fall, is experiencing 0 gravity?

Presumably because both are in free fall and therefore not accelerating relative to each other.

 

[neilparker62]

I think there are a number of assumptions one must make with this problem:

1) Masses are equal
2) The collision is elastic
3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for $\vec{v_a}=2\vec{v_b}$ since they are in opposite directions.

Given the above, the collision is actually something of a red herring since all it does is swap the respective velocity values and change their sign. This means that if $\lvert v_a \rvert = \lvert 2v_b \rvert$ after collision then $\lvert v_b \rvert = \lvert 2v_a \rvert$ before collision. So we are left with the relatively simple task of solving:

$$ v - gt = 2gt $$ leading to the answer given by Haruspex above. I think the OP's answer comes from mixing up which is twice which - he/she appears to have rather solved (effectively):

$$ 2(v - gt) = gt $$

 

[Lunippa]

I think there are a number of assumptions one must make with this problem:

1) Masses are equal

I think there are a number of assumptions one must make with this problem:

1) Masses are equal
2) The collision is elastic
3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for $\vec{v_a}=2\vec{v_b}$ since they are in opposite directions.

Given the above, the collision is actually something of a red herring since all it does is swap the respective velocity values and change their sign. This means that if $\lvert v_a \rvert = \lvert 2v_b \rvert$ after collision then $\lvert v_b \rvert = \lvert 2v_a \rvert$ before collision. So we are left with the relatively simple task of solving:

$$ v - gt = 2gt $$ leading to the answer given by Haruspex above. I think the OP's answer comes from mixing up which is twice which - he/she appears to have rather solved (effectively):

$$ 2(v - gt) = gt $$

2) The collision is elastic
3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for $\vec{v_a}=2\vec{v_b}$ since they are in opposite directions.

Given the above, the collision is actually something of a red herring since all it does is swap the respective velocity values and change their sign. This means that if $\lvert v_a \rvert = \lvert 2v_b \rvert$ after collision then $\lvert v_b \rvert = \lvert 2v_a \rvert$ before collision. So we are left with the relatively simple task of solving:

$$ v - gt = 2gt $$ leading to the answer given by Haruspex above. I think the OP's answer comes from mixing up which is twice which - he/she appears to have rather solved (effectively):

$$ 2(v - gt) = gt $$

Thank you!
But what if the mass of the ball is a lot smaller and the 2 objects collide falling in the same direction, towards earth?

 

[neilparker62]

But what if the mass of the ball is a lot smaller and the 2 objects collide falling in the same direction, towards earth?

Are you still referring to the situation in which one object is dropped and the other thrown or fired upwards? And the dropped object is the lighter one ?

 

[jbriggs444]

3) Magnitude of A's velocity is twice the magnitude of B's velocity. Not possible for $\vec{v_a}=2\vec{v_b}$ since they are in opposite directions.

Whoah there. What guarantees that they are in opposite directions?

 

[neilparker62]

Whoah there. What guarantees that they are in opposite directions?

Assuming elastic collision between two equal masses. Before collision:

$$p_1=mv_1 $$ $$p_2=-mv_2$$ where the - sign indicates the two velocities are in opposite directions - this will be the case per OP's post.

Collision impulse (elastic collision): $$Δp = 2μΔv=m(v_1+v_2)$$ where reduced mass μ=m/2 and relative velocity Δv=v₁+v₂

Immediately after collision: $$p_1=mv_1-Δp=mv_1-m(v_1+v_2)=-mv_2$$ $$p_2=-mv_2+Δp=-mv_2+m(v_1+v_2)=mv_1$$
The respective momenta simply exchange during collision and so if the velocities were opposite before collision, they remain so after collision. Since both objects are in free fall, I am assuming that g does not affect the normal collision dynamics at all.