1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Collision of two balls where some kinetic energy is lost

  1. Aug 31, 2012 #1
    1. The problem statement, all variables and given/known data[/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. Relevant 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: Aug 31, 2012
  2. jcsd
  3. Aug 31, 2012 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Review the definition of coefficient of restitution to see if you have a sign error here.
  4. Aug 31, 2012 #3
    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
  5. Aug 31, 2012 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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.
  6. Aug 31, 2012 #5
    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]
    [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: Aug 31, 2012
  7. Aug 31, 2012 #6


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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.
  8. Sep 1, 2012 #7
    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: Sep 1, 2012
  9. Sep 1, 2012 #8


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    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].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook