# Collision of two balls where some kinetic energy is lost

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 :

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

$V_{0}$=0= velocity of B initially
$V_{1}$=velocity of B after collision with A for the first time
$V_{2}$= Velocity of B after collision with the barrier
$V_{3}$= 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$U_{0}$ = 2$U_{1}$+$V_{1}$

by law of restitution for first collision :

2) 0.5 = -$(U_{1}-V_{1})/U_{0}$

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 $V_{3}$ and $U_{3}$
After solving all the five equations i get :

$V_{3}$= 0.5$U_{0}$
$U_{3}$= 3/8$U_{0}$

But the correct answers are :

$V_{3}$ same as mine
$U_{3}$= 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 $V_{2}$ 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$U_{1}$-$V_{2}$ = 2$U_{3}$ + $V_{3}$

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

And if use -$V_{2}$ instead of +$V_{2}$, i arrive at my result which i just quoted above (which is wrong )

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TSny
Homework Helper
Gold Member
by law of restitution for first collision :

2) 0.5 = $(U_{1}-V_{1})/U_{0}$
Review the definition of coefficient of restitution to see if you have a sign error here.

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
Gold Member
For example equation 4 as i formed it was :

2$U_{1}$-$V_{2}$ = 2$U_{3}$ + $V_{3}$

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

And if use -$V_{2}$ instead of +$V_{2}$, i arrive at my result which i just quoted above (which is wrong )
I believe you want to write 2$U_{1}$+$V_{2}$ on the left side rather than with a minus sign. The direction of $V_{2}$ will be accounted for by the value of $V_{2}$ being positive or negative.

I believe you want to write 2$U_{1}$+$V_{2}$ on the left side rather than with a minus sign. The direction of $V_{2}$ will be accounted for by the value of $V_{2}$ being positive or negative.
I am confused about it, whether to write it as positive hence considering $V_{2}$ 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 $V_{2}$ as positive, right ?

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

equation 3)

2($V_{2}$-$V_{1}$) = $U_{1}$-$V_{1}$

From Equations 1 and 2 :

$U_{1}$= $\frac{1}{2}$$U_{0}$
and
$V_{1}$= $U_{0}$

Substituting these values in equation 3 gives us :
$V_{2}$= 0.25$U_{0}$

Now this value doesn't make sense, because $V_{2}$ 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 ...

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TSny
Homework Helper
Gold Member
equation 3)

2($V_{2}$-$V_{1}$) = $U_{1}$-$V_{1}$
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 $V_{2}$ from just $V_{1}$ and the coefficient of restitution.

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 $V_{2}$ from just $V_{1}$ 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) $V_{2}$ = 0.5 $V_{1}$

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

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TSny
Homework Helper
Gold Member
3) $V_{2}$ = 0.5 $V_{1}$

P.S : I am taking $V_{2}$ 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 $V_{2}$ vector must have the opposite direction of the $V_{1}$ vector. So, it must be $V_{2}$ = -0.5 $V_{1}$.

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

e = -$\frac{V_{2}-W_{2}}{V_{1}-W_{1}}$

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