Collision of two balls where some kinetic energy is lost

[hms.tech]

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.

Homework Equations

Conservation of linear momentum
law of restitution

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) 2U_{0} = 2U_{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}= 0.5U_{0}
U_{3}= 3/8U_{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 don't 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 :

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

 

[TSny]

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.

 

[hms.tech]

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]

For example equation 4 as i formed it was :

2U_{1}-V_{2} = 2U_{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 2U_{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.

 

[hms.tech]

I believe you want to write 2U_{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}= 0.25U_{0}

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

Can someone help me understand this ...

 

[TSny]

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.

 

[hms.tech]

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 don't have to use -ve signs with the velocities .
and finally my answer matches !

 

[TSny]

3) V_{2} = 0.5 V_{1}

P.S : I am taking V_{2} as a vector, so that i don't 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}.