Show that v < 5.5. How can we prove that?

[lioric]

1.
Two particles P and Q are moving in opposite directions along the same horizontal straight ine. Particle P is moving due east and particle Q is moving due west. Particle P has mass 3m and particle Q has mass 2m. The particles collide directly. Immediately before the collision, the speed of P is 7u and the speed of Q is v. As a result of collision the direction of motion of P unchanged and is moving speed 2u.

Homework Equations

[/B]
Total momentum initial = Total momentum final
For any collision
Total Kinetic energy initial > total Kinetic energy final

The Attempt at a Solution

[/B]
Total momentum initial = Total momentum final
3m x 7u + 2m x v = 3m x 2u + 2m x ?

M canceled out from both sides

? = (15u - 2v)/2

This gives us the final velocity of Q in terms of u and v

Total KE initial = Total KE final

1/2 x 3m x (7u)^2 + 1/2 x 2m x v^2 = 1/2 x 3m x (2u)^2 + 1/2 x 2m x ((15u-2v)/2)^2

The halves and 3m cancels and we are left with(7u)^2 + v^2 = (2u)^2 + ((15u-2v)/2)^2

Now I'm stuck
I'm left with

49u^2 +v^2 = 4u^2 + (225u^2 - 60uv + 4v^2)/4

Please help

 

[PeroK]

Try solving for $v_f - v$.

 

[lioric]

Try solving for $v_f - v$.

Could you elaborate a little bit more
I mean Vf - Vi is = to what?

 

[PeroK]

Your final momentum equation can be written as:

$v_f - v = 7.5u$

Perhaps leave it like that. Now, what do get from the energy equation?

Hint: $v_f^2 - v^2 = (v_f - v)(v_f + v)$

PS I assume you are trying to find the maximum final velocity of Q? That's what I've called $v_f$.

 

[lioric]

Your final momentum equation can be written as:

$v_f - v = 7.5u$

Perhaps leave it like that. Now, what do get from the energy equation?

Hint: $v_f^2 - v^2 = (v_f - v)(v_f + v)$

PS I assume you are trying to find the maximum final velocity of Q? That's what I've called $v_f$.

No I'm trying to prove that initial velocity of Q is < 5.5u

 

[PeroK]

No I'm trying to prove that initial velocity of Q is < 5.5u

Try $v = u$ and see what you get.

Note: I think the question might be wrong.

PS or try $v = 5u$.

Definitely wrong. I think I can see what the question should be.

 

[lioric]

Try $v = u$ and see what you get.

Note: I think the question might be wrong.

PS or try $v = 5u$.

Definitely wrong. I think I can see what the question should be.

No the question is right
May show a snapshot of the question?

 

[PeroK]

No the question is right
May show a snapshot of the question?

View attachment 216687

Okay, sorry, I was taking $v$ as a velocity. And, it's got nothing to do with energy!

Do you want to completely rethink the problem? Or do you want another hint?

 

[lioric]

Okay, sorry, I was taking $v$ as a velocity. And, it's got nothing to do with energy!

Do you want to completely rethink the problem? Or do you want another hint?

Please give me another hint

 

[PeroK]

Please give me another hint

Think about the possible range of values for $v_f$. Hint: The two objects can't pass through each other!

 

[lioric]

Think about the possible range of values for $v_f$. Hint: The two objects can't pass through each other!

Opposite direction values?

 

[PeroK]

Opposite direction values?

Another hint:

Why can't $v_f =0$?

 

[lioric]

Another hint:

Why can't $v_f =0$?

It says that both objects were moving

 

[PeroK]

It says that both objects were moving

That's at the start. Maybe P hit Q and stopped it dead?

 

[lioric]

That's at the start. Maybe P hit Q and stopped it dead?

But it cannot cause the other particle kept moving
So unless P goes through Q, Q has to move move out of P's way cause P kept going in the same direction

 

[PeroK]

But it cannot cause the other particle kept moving
So unless P goes through Q, Q has to move move out of P's way cause P kept going in the same direction

Yes, but Q can't get out the way, can it?

So, after the collision Q has a minimum speed of ?

 

[lioric]

Yes, but Q can't get out the way, can it?

So, after the collision Q has a minimum speed of ?

(15u-2v)/2

 

[PeroK]

(15u-2v)/2

Okay, let me tell you. P is moving east at $2u$, so Q must be moving east at $\ge 2u$

That's all there was to it!

 

[lioric]

Okay, let me tell you. P is moving east at $2u$, so Q must be moving east at $\ge 2u$

That's all there was to it!

Ya so meaning that Q has to go faster than P
Understood
How am I going to prove that? I mean that V is < 5.5

 

[PeroK]

Ya so meaning that Q has to go faster than P
Understood
How am I going to prove that? I mean that V is < 5.5

just go to your momentum equation:

$v = 7.5u - v_f$

 

[lioric]

Ya so meaning that Q has to go faster than P
Understood
How am I going to prove that? I mean that V is < 5.5

Is it by equating (15u-2v)/2 > 2u ?

 

[lioric]

Thank you very much

 

[PeroK]

Is it by equating (15u-2v)/2 > 2u ?

Yes, but your reluctance to introduce the variable $v_f$ is going to hurt you in these sorts of problems. At this level, you need to introduce variables yourself without the question telling you which ones you need.

 

[PeroK]

Thank you very much

Since I messed up, let me show you the trick for solving energy-momentum equations using the factorisation of the difference of squares.

From momentum we had:

$v_f + v = 7.5u$ (1)

Note that normally this is $-$ but since $v$ is a speed we have a plus here. That's where I went wrong.

From energy conservation we get:

$2m(v_f^2-v^2) = 3m(45)u^2$

Then we use the factorisation of the left hand side and equation (1) to get:

$15(v_f - v) = 3(45)u$

$v_f - v = 9u$ (2)

Now, equations (1) and (2) give you the solution for $v_f$.

Note that it is usually much neater to do this than to generate a quadratic like you did, so it's a trick worth remembering. This technique is useful all over physics.

 

[Ray Vickson]

1.
Two particles P and Q are moving in opposite directions along the same horizontal straight ine. Particle P is moving due east and particle Q is moving due west. Particle P has mass 3m and particle Q has mass 2m. The particles collide directly. Immediately before the collision, the speed of P is 7u and the speed of Q is v. As a result of collision the direction of motion of P unchanged and is moving speed 2u.

Homework Equations

[/B]
Total momentum initial = Total momentum final
For any collision
Total Kinetic energy initial > total Kinetic energy final

The Attempt at a Solution

[/B]
Total momentum initial = Total momentum final
3m x 7u + 2m x v = 3m x 2u + 2m x ?

M canceled out from both sides

? = (15u - 2v)/2

This gives us the final velocity of Q in terms of u and v

Total KE initial = Total KE final

1/2 x 3m x (7u)^2 + 1/2 x 2m x v^2 = 1/2 x 3m x (2u)^2 + 1/2 x 2m x ((15u-2v)/2)^2

The halves and 3m cancels and we are left with(7u)^2 + v^2 = (2u)^2 + ((15u-2v)/2)^2

Now I'm stuck
I'm left with

49u^2 +v^2 = 4u^2 + (225u^2 - 60uv + 4v^2)/4

Please help

(1) Saying that " For any collision Total Kinetic energy initial > total Kinetic energy final" is false, especially when you later say that
"Total KE initial = Total KE final". If you used a non-strict inequality ≥ in the first statement, that would have been OK. Be careful!

(2) It is usually easiest to analyze such collision problems in the CM (center-of-mass or center-of-momentum) frame. This is a coordinate system that moves with some constant velocity V in the "lab" frame (which is the one you start with). In the CM frame, both the initial and final total momenta are zero. Furthermore, if the collision is perfectly elastic (conserving total kinetic energy) it becomes particularly easy to get final velocities: in the CM the speed of each particle remains unchanged before and after the collision. Of course, the individual CM-frame velocities may change directions--as in this problem, where they would reverse their directions in a perfectly elastic collision. Transforming CM-frame final velocities back into the lab frame is easy.

Anyway, you have obtained a quadratic equation for v in terms of u---just go ahead and solve it, using the "quadratic" formula. (I have not checked the details of your work, so I don't know if your final quadratic equation is OK.)

 

[Ray Vickson]

1.
Two particles P and Q are moving in opposite directions along the same horizontal straight ine. Particle P is moving due east and particle Q is moving due west. Particle P has mass 3m and particle Q has mass 2m. The particles collide directly. Immediately before the collision, the speed of P is 7u and the speed of Q is v. As a result of collision the direction of motion of P unchanged and is moving speed 2u.

Homework Equations

[/B]
Total momentum initial = Total momentum final
For any collision
Total Kinetic energy initial > total Kinetic energy final

The Attempt at a Solution

[/B]
Total momentum initial = Total momentum final
3m x 7u + 2m x v = 3m x 2u + 2m x ?

M canceled out from both sides

? = (15u - 2v)/2

This gives us the final velocity of Q in terms of u and v

Total KE initial = Total KE final

Now I'm stuck
I'm left with

49u^2 +v^2 = 4u^2 + (225u^2 - 60uv + 4v^2)/4

Please help

If the collision is perfectly elastic---that is, total kinetic energy is conserved---the given problem conditions are impossible: we cannot have a particle P of mass 3m and velocity 7u > 0 colliding with a particle Q of mass 2m and velocity -v < 0 giving rise to a post-collision velocity of 2u for P. For this to happen, the collision must be inelastic, with some loss of kinetic energy.

It is easiest to see this by going into the CM frame, as suggested in Post #25. In the original (lab) frame the total initial momentum is
$$P_i = 3m \times 7u + 2m \times (-v) = 21 m u - 2 m v$$.
The velocity of the CM frame is
$$V_{cm} = \frac{P_i}{3m + 2m} = \frac{21}{5} u - \frac{2}{5} v.$$
Therefore, the initial velocity of P in the CM frame is
$$V(P)_{i ,cm}= 7u - V_{cm} = \frac{14}{5} u + \frac{2}{5} v.$$
The final velocity of P in the CM frame is
$$V(P)_{f,cm} = 2u - V_{cm} = -\frac{11}{5} u + \frac{2}{5} v.$$

Now comes the clincher: in the CM frame, both particles keep the same speeds in a perfectly elastic collision! (That is why the CM frame is so handy when dealing with perfectly elastic collisions: speeds remain the same before and after, and only the directions change.) So, if the collision is elastic, particle P bounces back in the CM frame at the same speed as it had initially; that is, the final velocity = -(initial velocity), so
$$-\frac{11}{5} u + \frac{2}{5} v = - \frac{14}{5} u - \frac{2}{5} v \; \Longrightarrow \; v = -\frac{3}{4} u.$$
We see that we would need $v < 0$, meaning that initially particle Q is actually moving due east with speed $|v| = (3/4)u$.

Conclusion: total kinetic energy cannot be conserved under the stated problem conditions.

So: how to proceed?

In the CM frame the particles do not "pass through" each other. Rather, particle P bounces back, changing from due-east travel to due-west; and particle Q also bounces back, changing from due-west travel to due-east. If you examine the implications of this fact for particle P you can get an inequality on $v$.

 

[PeroK]

@Ray Vickson Can you see yesterday's posts? I think you had this problem before.

This was all sorted out. The thread has been marked as "solved" in any case.

 

[Ray Vickson]

@Ray Vickson Can you see yesterday's posts? I think you had this problem before.

This was all sorted out. The thread has been marked as "solved" in any case.

Maybe it was marked "solved", but I don't think it really was solved at all. The fact is that kinetic energy cannot be conserved in this problem, so any solution that uses conservation of kinetic energy cannot be correct.

That is what my latest post is about.

Anyway: about my "problem" (of seeming to ignore previous posts?): for some reason, on my system I sometimes do not see several responses until after I have made a response and pressed the "enter" key. In this thread, about 15 posts did not appear on my screen until after I made a post as well. This has happened to me many times before in this forum, but never before to the extent in this thread.

 

[PeroK]

Maybe it was marked "solved", but I don't think it really was solved at all. The fact is that kinetic energy cannot be conserved in this problem, so any solution that uses conservation of kinetic energy cannot be correct.

That is what my latest post is about.
.

Looking at conservation of energy was a false start. The problem was solved more simply, without directly considering energy in fact.

 

[itfitmewelltoo]

As I recall, the energy and momentum equations give two simultaneous equations. Typically, problems have one of the two as the unknown.