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Collision of two balls where some kinetic energy is lost

  • Thread starter hms.tech
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1. Homework Statement [/b]

Two smooth uniform spheres, A and B, are of the same size, but A has twice the mass of B. They are placed at rest on a smooth horizontal table and A is projected directly towards B with speed U. The spheres collide, and then B strikes a barrier perpendicular to its path, rebounds, and hits A again. Find the velocities of A and B immediately after this second collision between A and B, given that the coefficient of restitution is 1/2 at all three impacts, and neglecting air resistance.

2. Homework Equations

Conservation of linear momentum
law of restitution

3. The Attempt at a Solution

Defining some variables :

[itex]U_{0}[/itex]= velocity of A before collision
[itex]U_{1}[/itex]= velocity of A after collision with B(which was stationary)
[itex]U_{3}[/itex]= velocity of A after collison with B for the second time (where B was moving)

[itex]V_{0}[/itex]=0= velocity of B initially
[itex]V_{1}[/itex]=velocity of B after collision with A for the first time
[itex]V_{2}[/itex]= Velocity of B after collision with the barrier
[itex]V_{3}[/itex]= Velocity of B after the collision with A for the 2nd time.

Now here is the solution :

Momentum is conserved in the First collision thus :

1) 2[itex]U_{0}[/itex] = 2[itex]U_{1}[/itex]+[itex]V_{1}[/itex]

by law of restitution for first collision :

2) 0.5 = -[itex](U_{1}-V_{1})/U_{0}[/itex]

3) by law of restitution in the collision between ball B and the barrier :
4) conservation of momentum between ball A and ball B
5)newton's law of restitution in the 2nd collision between ball A and B.


basically we have to find [itex]V_{3}[/itex] and [itex]U_{3}[/itex]
After solving all the five equations i get :

[itex]V_{3}[/itex]= 0.5[itex]U_{0}[/itex]
[itex]U_{3}[/itex]= 3/8[itex]U_{0}[/itex]

But the correct answers are :

[itex]V_{3}[/itex] same as mine
[itex]U_{3}[/itex]= 0

I have checked my working for any arithmetic mistakes many times, but i dont think its that kind of a mistake.
It seems to me that the error is related with +/- signs which i apply when forming these five equations specially when i use the velocity of B after colliding with the barrier, because it changes direction.

Equations 1 and 2 will always carry correct signs. (becuase i know they are correct)
its the equations 3,4 and 5 which involve [itex]V_{2}[/itex] and that is where i get confused .

can someone form equations 3 ,4 and 5 for me. For example equation 4 as i formed it was :

2[itex]U_{1}[/itex]-[itex]V_{2}[/itex] = 2[itex]U_{3}[/itex] + [itex]V_{3}[/itex]

Now in the above equation i used the -ve sign for [itex]V_{2}[/itex], is that right ?

And if use -[itex]V_{2}[/itex] instead of +[itex]V_{2}[/itex], i arrive at my result which i just quoted above (which is wrong )
 
Last edited:

TSny

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by law of restitution for first collision :

2) 0.5 = [itex](U_{1}-V_{1})/U_{0}[/itex]
Review the definition of coefficient of restitution to see if you have a sign error here.
 
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Review the definition of coefficient of restitution to see if you have a sign error here.
Yes, that was just type error, the problem still remain.
i have edited the first post and made it right .

Can u for equation four correctly for me
 

TSny

Homework Helper
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For example equation 4 as i formed it was :

2[itex]U_{1}[/itex]-[itex]V_{2}[/itex] = 2[itex]U_{3}[/itex] + [itex]V_{3}[/itex]

Now in the above equation i used the -ve sign for [itex]V_{2}[/itex], is that right ?

And if use -[itex]V_{2}[/itex] instead of +[itex]V_{2}[/itex], i arrive at my result which i just quoted above (which is wrong )
I believe you want to write 2[itex]U_{1}[/itex]+[itex]V_{2}[/itex] on the left side rather than with a minus sign. The direction of [itex]V_{2}[/itex] will be accounted for by the value of [itex]V_{2}[/itex] being positive or negative.
 
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I believe you want to write 2[itex]U_{1}[/itex]+[itex]V_{2}[/itex] on the left side rather than with a minus sign. The direction of [itex]V_{2}[/itex] will be accounted for by the value of [itex]V_{2}[/itex] being positive or negative.
I am confused about it, whether to write it as positive hence considering [itex]V_{2}[/itex] as a vector or -ve thus considering it as speed (not velocity)

But if i take it as positive (as a vector) the i must take all the values of [itex]V_{2}[/itex] as positive, right ?

Here is what happens if i take it as a vector (ie +ve) :

equation 3)

2([itex]V_{2}[/itex]-[itex]V_{1}[/itex]) = [itex]U_{1}[/itex]-[itex]V_{1}[/itex]

From Equations 1 and 2 :

[itex]U_{1}[/itex]= [itex]\frac{1}{2}[/itex][itex]U_{0}[/itex]
and
[itex]V_{1}[/itex]= [itex]U_{0}[/itex]

Substituting these values in equation 3 gives us :
[itex]V_{2}[/itex]= 0.25[itex]U_{0}[/itex]

Now this value doesn't make sense, because [itex]V_{2}[/itex] should be in the -x direction, because it rebounded from the wall, it cant be in the direction of its initial velocity, that is just wrong.

Can someone help me understand this ...
 
Last edited:

TSny

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equation 3)

2([itex]V_{2}[/itex]-[itex]V_{1}[/itex]) = [itex]U_{1}[/itex]-[itex]V_{1}[/itex]
I don't understand this equation. Aren't you trying to express what happens when the second ball collides with the barrier? Ball 1 doesn't take part in this collision. You should be able to get [itex]V_{2}[/itex] from just [itex]V_{1}[/itex] and the coefficient of restitution.
 
247
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I don't understand this equation. Aren't you trying to express what happens when the second ball collides with the barrier? Ball 1 doesn't take part in this collision. You should be able to get [itex]V_{2}[/itex] from just [itex]V_{1}[/itex] and the coefficient of restitution.
That is absolutely right and i think this is the reason i was getting a wrong answer.

Many thanks to you .

Since the collision is between B and the wall, the "correct" equation 3 should be :

3) [itex]V_{2}[/itex] = 0.5 [itex]V_{1}[/itex]

P.S : I am taking [itex]V_{2}[/itex] as a vector, so that i dont have to use -ve signs with the velocities .
and finally my answer matches !
 
Last edited:

TSny

Homework Helper
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3) [itex]V_{2}[/itex] = 0.5 [itex]V_{1}[/itex]

P.S : I am taking [itex]V_{2}[/itex] as a vector, so that i dont have to use -ve signs with the velocities .
I think this equation still needs a minus sign. (These signs are a nightmare, aren't they.) We know that ball B will change direction when it hits the wall. So, the [itex]V_{2}[/itex] vector must have the opposite direction of the [itex]V_{1}[/itex] vector. So, it must be [itex]V_{2}[/itex] = -0.5 [itex]V_{1}[/itex].

We can derive this by applying the coefficient of restitution equation to the collision with the wall. We would have

e = -[itex]\frac{V_{2}-W_{2}}{V_{1}-W_{1}}[/itex]

where [itex]W_{1}[/itex] and [itex]W_{2}[/itex] are the velocity vectors of the wall before and after the collision and e is the coefficient of restitution. Of course, [itex]W_{1}[/itex] = [itex]W_{2}[/itex] = 0. So, the equation will reduce to [itex]V_{2}[/itex] = -e [itex]V_{1}[/itex].
 

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