Confirming Conservation of Kinetic Energy: An Explanation

[hidemi]

Homework Statement

For a completely inelastic two-body collision the kinetic energy retained by the objects is the same as:

A) the total kinetic energy before the collision
B) the kinetic energy of the more massive body before the collision
C) the kinetic energy of the less massive body before the collision
D) 1/2 Mv^2, where M is the total mass and v_com is the velocity of the center of mass
E) the difference in the kinetic energies of the objects before the collision

The answer is D

Relevant Equations

KE = 1/2 *MV^2

D is correct, the reasoning is as follows:
1/2*(M1V1)^2 + 1/2*(M2V2)^2 = 1/2 * (M1 + M2) (Vcm)^2, since V1 =V2 =Vcm
KE retained = KE final = 1/2 *M(Vcm)^2
Let me know if reasoning is okay?

However, why A isn't correct?

 

[Doc Al]

D is correct, the reasoning is as follows:
1/2*(M1V1)^2 + 1/2*(M2V2)^2 = 1/2 * (M1 + M2) (Vcm)^2, since V1 =V2 =Vcm
KE retained = KE final = 1/2 *M(Vcm)^2
Let me know if reasoning is okay?

That reasoning is not OK. You are assuming that the initial KE equals the final KE. But translational KE is not conserved. (Which is also why A is incorrect.)

Here's a simple example to show why that is not the case. Imagine two balls of clay heading toward each other with the same speed and mass. When they collide, assuming a perfectly inelastic collision, what's their final combined speed?

 

[hidemi]

That reasoning is not OK. You are assuming that the initial KE equals the final KE. But translational KE is not conserved. (Which is also why A is incorrect.)

Here's a simple example to show why that is not the case. Imagine two balls of clay heading toward each other with the same speed and mass. When they collide, assuming a perfectly inelastic collision, what's their final combined speed?

Are the following what you mean?

(A): mv + m (-v) = ( m+m ) v'
v'= o
(D) (1/2 mv^2)*2 = 1/2 (2mv^2)

 

[Doc Al]

(A): mv + m (-v) = ( m+m ) v'
v'= o

Assuming you are talking about the example I gave, then yes: Conservation of momentum (which applies in any collision) shows that the final speed of the clay is zero. (Not sure why you labeled it "A".)

(D) (1/2 mv^2)*2 = 1/2 (2mv^2)

Not sure what you mean here. In my example, the final KE must be zero, since the final speed is zero. The initial KE is certainly not zero. KE is clearly not conserved in that collision.

 

[hidemi]

Assuming you are talking about the example I gave, then yes: Conservation of momentum (which applies in any collision) shows that the final speed of the clay is zero. (Not sure why you labeled it "A".)Not sure what you mean here. In my example, the final KE must be zero, since the final speed is zero. The initial KE is certainly not zero. KE is clearly not conserved in that collision.

Sorry for the confusion. Thanks for confirming Option (A).
As for Option (D) being the correct answer, does that contradict with the given answer that the KEs of initial and final are not conserved?

 

[Doc Al]

Thanks for confirming Option (A).

Just to be clear, I confirmed that Option (A) is incorrect.

As for Option (D) being the correct answer, does that contradict with the given answer that the KEs of initial and final are not conserved?

Option (D) is perfectly consistent with KE not being conserved. Note that they are just asking for the final KE, which you can get from conservation of momentum. And that final KE is given in Option (D).

 

[hidemi]

Here's a simple example to show why that is not the case. Imagine two balls of clay heading toward each other with the same speed and mass. When they collide, assuming a perfectly inelastic collision, what's their final combined speed?

If I use the example you gave previously for option (D),
then KE initial = 1/2*mv^2 + 1/2*m(-v)^2, while KE final (or, retained) = 1/2 * (2m)v'^2, where v'=0.
Therefore, the retained kinetic energy isn't equal to 1/2*mv^2, but 0.

I'm not sure if I misused your given example for Option (D). However, I now understand Option (A) and thanks again for the confirmation.

 

[Doc Al]

If I use the example you gave previously for option (D),
then KE initial = 1/2*mv^2 + 1/2*m(-v)^2, while KE final (or, retained) = 1/2 * (2m)v'^2, where v'=0.
Therefore, the retained kinetic energy isn't equal to 1/2*mv^2, but 0.

Good.

 

[hidemi]

Good.

With what I said in post #7, does that make Option (D) the given answer incorrect?

 

[Doc Al]

With what I said in post #7, does that make Option (D) the given answer incorrect?

No. Why would you think that?

In post #7, you are discussing my example. But even in that example, Option (D) is still correct. In my example, V_com = 0, so the final KE = 1/2(M)(V_com)^2 = 0. But Option (D) still applies, as it must for any completely inelastic collision.

 

[hidemi]

No. Why would you think that?

In post #7, you are discussing my example. But even in that example, Option (D) is still correct. In my example, V_com = 0, so the final KE = 1/2(M)(V_com)^2 = 0. But Option (D) still applies, as it must for any completely inelastic collision.

Ok I see. Thanks.

 

[haruspex]

None of the options is true in general.
Option D is correct if the collision is a coalescence. More generally, it could be a glancing blow, in which case it is $\frac 12Mv^2$, where v is the component of $v_{com}$ along the line of mass centres at collision.

 

[robphy]

None of the options is true in general.
Option D is correct if the collision is a coalescence. More generally, it could be a glancing blow, in which case it is $\frac 12Mv^2$, where v is the component of $v_{com}$ along the line of mass centres at collision.

But the problem says... "completely inelastic"...

Homework Statement:: For a completely inelastic two-body collision the kinetic energy retained by the objects is the same as:

 

[hmmm27]

'D' makes no allowance for RE, but it's the definition for (translational) kinetic energy.

f) none of the above.

 

[haruspex]

But the problem says... "completely inelastic"...

What I described involves no elasticity.
From https://en.m.wikipedia.org/wiki/Inelastic_collision:
"For two- and three-dimensional collisions the velocities in these formulas are the components perpendicular to the tangent line/plane at the point of contact."