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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

Conservation of linear momentum

law of restitution

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 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 [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 )

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 :

[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 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 [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 )

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