Correct statement about perfectly elastic collision

[songoku]

Homework Statement

In perfectly elastic collision between two atoms, it is always true to say that
a. the initial speed of one atom will be the same as the final speed of the other atom
b. the relative speed of approach between two atoms equals their relative speed of separation
c. the total momentum must be conserved, but a small amount of the total kinetic energy lost
d. whatever their initial state of motion, neither atom can be stationary after the collision
e. none of the above

Homework Equations

conservation of momentum
coefficient of restitution

The Attempt at a Solution

I think the answer is B. Or maybe it is E because it should be relative velocity instead of relative speed?

Thanks

 

[ehild]

Homework Statement

In perfectly elastic collision between two atoms, it is always true to say that
b. the relative speed of approach between two atoms equals their relative speed of separation

Homework Equations

conservation of momentum
coefficient of restitution

The Attempt at a Solution

I think the answer is B. Or maybe it is E because it should be relative velocity instead of relative speed?

Thanks

How do you define velocity of approach and velocity of separation? And how is speed of approach and speed of separation defined?
Can you find the relation between the relative velocities before and after collision?

 

[songoku]

How do you define velocity of approach and velocity of separation? And how is speed of approach and speed of separation defined?

I am not sure. Velocity is vector so the direction is included, while speed is scalar.

If object A moves to right with 5 m/s and B with 3 m/s, I would say velocity of A = speed of A = 5 m/s while velocity of B is -3 m/s and speed of B = 3 m/s

Can you find the relation between the relative velocities before and after collision?

u₁ - u₂ = v₂ - v₁

 

[PeroK]

I am not sure. Velocity is vector so the direction is included, while speed is scalar.

What about a rubber ball being dropped and bouncing elastically on the floor?

 

[songoku]

What about a rubber ball being dropped and bouncing elastically on the floor?

The speed will be the same but velocities will have different sign

So this means the answer should be E?

Thanks

 

[ehild]

I am not sure. Velocity is vector so the direction is included, while speed is scalar.

If object A moves to right with 5 m/s and B with 3 m/s, I would say velocity of A = speed of A = 5 m/s while velocity of B is -3 m/s and speed of B = 3 m/s

Does B move to right or to left? What about the velocity of approach and the velocity of separation? How do you define them?

u₁ - u₂ = v₂ - v₁

Yes, you can consider the difference of the velocities, the relative velocities, as velocity of approach and velocity of separation. You derived correctly that the relative velocity changes sign after the collision, but the magnitude does not change. And you said, that speed is magnitude of velocity. Is the statement "b. the relative speed of approach between two atoms equals their relative speed of separation" true?

 

[songoku]

Does B move to right or to left? What about the velocity of approach and the velocity of separation? How do you define them?

Sorry, B moves to left.
Velocity of approach = 8 m/s
Velocity of separation will depend on calculation after they collide.

Yes, you can consider the difference of the velocities, the relative velocities, as velocity of approach and velocity of separation. You derived correctly that the relative velocity changes sign after the collision, but the magnitude does not change. And you said, that speed is magnitude of velocity. Is the statement "b. the relative speed of approach between two atoms equals their relative speed of separation" true?

It is wrong then because we need to include direction so the correct statement will be relative velocity of approach = relative velocity of separation. The answer is E

 

[ehild]

It is wrong then because we need to include direction so the correct statement will be relative velocity of approach = relative velocity of separation. The answer is E

I do not understand your argument.
The statement is about speed, where direction does not count. More, the relative velocities are not the same before and after collision.

 

[songoku]

I do not understand your argument.
The statement is about speed, where direction does not count. More, the relative velocities are not the same before and after collision.

Let me try again:
A moves to right 5 m/s and B moves to left 3 m/s. They both have same masses and the collision is elastic so after collision A will move to left 3 m/s and B moves to right 5 m/s.
Speed of A before collision = 5 m/s
Velocity of A before collison = 5 m/s
Speed of B before collision = 3 m/s
Velocity of B before collision = -3 m/s
Speed of A after collision = 3 m/s
Velocity of A after collision = -3 m/s
Speed of B after collision = 5 m/s
Velocity of B after collision = 5 m/s

(i)
Relative speed of approach = 5 - 3 = 2 m/s
Relative speed of separation = 5 - 3 = 2 m/s
(Is this correct?)

(ii)
Relative velocity of approach = 5 - (-3) = 8 m/s
Relative velocity of separation = 5 - (-3) = 8 m/s

In my oponion, (ii) is used in calculation so the correct statement is "relative velocity of approach = relative velocity of separation"

I do not see why relative velocity are not the same before and after collision

Thanks

Edit: I made mistake not writing information after collision

 

[PeroK]

I do not see why relative velocity are not the same before and after collision

Thanks

What about the bouncing ball?

 

[songoku]

What about the bouncing ball?

The collision is between ball and ground. I think it this way, let say the speed of the ball just before hitting the ground is 5 m/s.

Speed of ball before collision = 5 m/s
Speed of ground before collision = 0 m/s
Velocity of ball after collision = - 5m/s
Velocity of ball after collision = 0 m/s

Speed of ball after collision = 5 m/s
Speed of ground after collision = 0 m/s
Velocity of ball after collision = 5 m/s
Velocity of ground after collision = 0 m/s

Relative Speed of approach = 5 - 0 = 5 m/s
Relative Speed of separation = 0 - 5 = -5 m/s

Relative Velocity of approach = -5 - 0 = -5 m/s
Relative velocity of separation = 0 - 5 = - 5 m/s

 

[PeroK]

The collision is between ball and ground. I think it this way, let say the speed of the ball just before hitting the ground is 5 m/s.

Speed of ball before collision = 5 m/s
Speed of ground before collision = 0 m/s
Velocity of ball after collision = - 5m/s
Velocity of ball after collision = 0 m/s

Speed of ball after collision = 5 m/s
Speed of ground after collision = 0 m/s
Velocity of ball after collision = 5 m/s
Velocity of ground after collision = 0 m/s

Relative Speed of approach = 5 - 0 = 5 m/s
Relative Speed of separation = 0 - 5 = -5 m/s

Relative Velocity of approach = -5 - 0 = -5 m/s
Relative velocity of separation = 0 - 5 = - 5 m/s

I look at it very differently. When the ball is falling, the ball and the Earth are getting closer together. And, after the bounce the ball and the Earth are getting further apart. That's opposite direction of relative motion.

If you have a ball falling at the same time as another ball is rising, they cannot have the same separation velocity relative to the Earth.

Moreover, speed is a magnitude so it can't be negative.

 

[jbriggs444]

Relative speed of approach = 5 - 3 = 2 m/s
Relative speed of separation = 3 - 5 = -2 m/s

Relative velocity of approach = 5 - (-3) = 8 m/s
Relative velocity of separation = 5 - (-3) = 8 m/s

I do not think you have understood what the questioner intends by the phrasing "relative speed of separation" and "relative speed of approach". It is not the difference in speeds as measured in an arbitrary rest frame.

In my view, the [somewhat awkward] phrasing means "rate at which the distance between the two is increasing after the collision" and "rate at which the distance between the two is decreasing during approach". A useful approach would be to adopt a center-of-mass reference frame try to calculate those numbers from the perspective of such a frame.

 

[ehild]

Let me try again:
A moves to right 5 m/s and B moves to left 3 m/s. They both have same masses and the collision is elastic so after collision A will move to left 3 m/s and B moves to right 5 m/s.
Speed of A before collision = 5 m/s
Velocity of A before collison = 5 m/s
Speed of B before collision = 3 m/s
Velocity of B before collision = -3 m/s
Speed of A after collision = 3 m/s
Velocity of A after collision = -3 m/s
Speed of B after collision = 5 m/s
Velocity of B after collision = 5 m/s

Relative speed of approach/separation is meant as magnitude of relative velocities before/after collision.
Using "relative" is confusing. Speed of approach/separation means the magnitude of the relative velocities already.

 

[haruspex]

the [somewhat awkward] phrasing means

I feel the questioner is using "of" to mean "during". As you say, clumsy.

 

[songoku]

I am so sorry for really late reply.

Thank you very much for all the help perok, ehild, jbriggs444, haruspex