Collision of two balls where some kinetic energy is lost

In summary: 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...But the question is in 1 dimension, so the velocity of the wall is only in the y-direction.So we should only consider the y-components of the velocities.So, e = -\frac{V_{2,y}-W_{2,y}}{V_{1,y}-W_{1,y}}But W_{1,y} and W_{2,y} are both 0, so we end up withe = -\frac
  • #1
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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 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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  • #2
hms.tech said:
by law of restitution for first collision :

2) 0.5 = [itex](U_{1}-V_{1})/U_{0}[/itex]

Review the definition of coefficient of restitution to see if you have a sign error here.
 
  • #3
TSny said:
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
 
  • #4
hms.tech said:
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 )

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.
 
  • #5
TSny said:
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.

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]
and
[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 can't be in the direction of its initial velocity, that is just wrong.

Can someone help me understand this ...
 
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  • #6
hms.tech said:
equation 3)

2([itex]V_{2}[/itex]-[itex]V_{1}[/itex]) = [itex]U_{1}[/itex]-[itex]V_{1}[/itex]

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.
 
  • #7
TSny said:
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.

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 don't have to use -ve signs with the velocities .
and finally my answer matches !
 
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  • #8
hms.tech said:
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 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 [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].
 

1. What is the collision of two balls where some kinetic energy is lost?

The collision of two balls where some kinetic energy is lost is a physical phenomenon in which two objects, in this case balls, collide with each other and some of their kinetic energy is converted into other forms of energy such as heat, sound, or deformation.

2. Why does some kinetic energy get lost in a collision of two balls?

In an ideal scenario, where there is no friction or external forces, the total kinetic energy of the two balls before and after the collision will be the same. However, in real-life situations, there is always some friction present which causes the kinetic energy to be lost.

3. How does the mass and velocity of the balls affect the amount of kinetic energy lost in a collision?

The mass and velocity of the balls play a crucial role in determining the amount of kinetic energy lost in a collision. Generally, the higher the mass and velocity of the balls, the more kinetic energy will be lost due to the larger impact force between them.

4. Is it possible for two balls to collide without any kinetic energy being lost?

In theory, it is possible for two balls to collide without any kinetic energy being lost if the collision is perfectly elastic, meaning there is no deformation or loss of energy. However, in real-life scenarios, it is nearly impossible to achieve a perfectly elastic collision due to the presence of external forces and friction.

5. How does the elasticity of the balls affect the amount of kinetic energy lost in a collision?

The elasticity of the balls, or their ability to deform and return to their original shape, plays a significant role in determining the amount of kinetic energy lost in a collision. In a perfectly elastic collision, where there is no deformation, no kinetic energy is lost. However, in a partially elastic collision, some kinetic energy will be lost due to the deformation of the balls.

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