- #1
hms.tech
- 247
- 0
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.
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 )
Last edited: