Dynamics Questions // Elastic Collision with different Vo

[sunnnystrong]

Homework Statement

[/B]
A 2.3 kg object traveling at 6.1 m/s collides head-on with a 3.5 kg object traveling in the opposite direction at 4.8 m/s. If the collisions is perfectly elastic, what is the final speed of the 2.3 kg object?

Homework Equations

Conservation of Energy:
1/2 m*vi^2 = 1/2 m*vf^2

Conservation of Momentum:
Pix = Pfx

The Attempt at a Solution

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Pix For 2.3 kg Object
(2.3kg)(6.1m/s) = (2.3kg)(Vf2) + (3.5kg)(Vf1)

Solving for Vf1 -->
[14.03 Ns - (2.3kg)(Vf2)]/3.5kg = Vf1

KE
1/2(2.3kg)(6.1m/s)^2 = 1/2(2.3kg)(Vf1)^2 + 1/2(3.8kg)(Vf2)^2
42.7915 = 1.15(Vf1)^2 + 1.9(Vf2)^2

Plugging in Vf1
42.7915 = 1.15(0.431837(Vf2)^2-5.26841Vf2 + 16.0686) + 1.9(Vf2)^2
Simplify and you get a quadratic:
0=2.39661 (Vf2)^2 - 6.05867 (Vf2) -24.3126
Solve for Vf2 = 4.69 m/s

Plug back into:
[14.03 Ns - (2.3kg)(Vf2)]/3.5kg = Vf1

Solve for Vf1 = 0.96m/s (which is definitely wrong)

I'm really confused & not sure how to solve for Vf1?

 

[Doc Al]

Conservation of Momentum:
Pix = Pfx

The Attempt at a Solution

[/B]
Pix For 2.3 kg Object
(2.3kg)(6.1m/s) = (2.3kg)(Vf2) + (3.5kg)(Vf1)

Total momentum of the system is conserved. Don't forget the initial momentum of the other mass. (Careful with signs!)

 

[sunnnystrong]

Total momentum of the system is conserved. Don't forget the initial momentum of the other mass. (Careful with signs!)

So going back -->
(2.3kg)(6.1m/s) + (3.5kg)(-4.8m/s) = (2.3kg)(Vf2) + (3.5kg)(Vf1)
Solve for Vf1 :
Vf1 = (-2.77Ns -2.3kg(Vf2))/(3.5kg)
Square it for later use:
0.431837(Vf2)^2 + 1.04016(Vf2) + .6264 = (Vf1)^2

Looking at total system -->
1/2(2.3kg)(6.1m/s)^2 + 1/2(3.5kg)(-4.8m/s)^2 = 1/2(2.3kg)(Vf1)^2 + 1/2(3.8kg)(Vf2)^2
Simplify this:
42.7915+40.32 = 1.15(Vf1)^2 + 1.9(Vf2)^2
Plug in again you get a quadratic:
83.1115 = 1.15(0.431837(Vf2)^2 + 1.04016(Vf2) + .6264) + 1.9(Vf2)^2
Solve
Vf2 = 5.62 m/s :((
I'm confused because the answer is 6.6m/s

 

[sunnnystrong]

Total momentum of the system is conserved. Don't forget the initial momentum of the other mass. (Careful with signs!)

Do you think you could show me how to set it up?

 

[Doc Al]

Do you think you could show me how to set it up?

Your set up looks good to me. (Now that you've corrected the momentum equation.) I didn't check all the steps you did, but if you're getting a different answer than expected double check your arithmetic.

 

[sunnnystrong]

Your set up looks good to me. (Now that you've corrected the momentum equation.) I didn't check all the steps you did, but if you're getting a different answer than expected double check your arithmetic.

I'm not sure anymore. i went through and combed through my work & can't find where i went wrong.

If you have time, could you look at my last response? I'd really appreciate it as my teacher is useless haha

 

[TomHart]

1/2(2.3kg)(6.1m/s)^2 + 1/2(3.5kg)(-4.8m/s)^2 = 1/2(2.3kg)(Vf1)^2 + 1/2(3.8kg)(Vf2)^2

You used 3.8 kg instead of 3.5 kg. I don't know if that will solve all of your problem, but it is at least part of it.

 

[sunnnystrong]

You used 3.8 kg instead of 3.5 kg. I don't know if that will solve all of your problem, but it is at least part of it.

I went back and did it carefully in my notebook but this time got 9m/s which is also wrong XD
at this point i just don't know what went wrong... do you have any ideas? does my set up look right?

 

[TomHart]

When I work these perfectly elastic collision problems, it is easy for me to make a mistake. But what I got for an answer was as follows:
The 2.3 kg object initially had a velocity of +6.1 m/s. Its final velocity is -7.055 m/s.
The 3.5 kg object initially had a velocity of -4.8 m/s. Its final velocity is 3.845 m/s.

But like I said, it is easy for me to make mistakes because of the numerous math steps involved.

does my set up look right?

I thought your set-up looked right. I will try to go through your solution to see if I can find the problem.

 

[sunnnystrong]

When I work these perfectly elastic collision problems, it is easy for me to make a mistake. But what I got for an answer was as follows:
The 2.3 kg object initially had a velocity of +6.1 m/s. Its final velocity is -7.055 m/s.
The 3.5 kg object initially had a velocity of -4.8 m/s. Its final velocity is 3.845 m/s.

But like I said, it is easy for me to make mistakes because of the numerous math steps involved.I thought your set-up looked right. I will try to go through your solution to see if I can find the problem.

I'd really appreciate it. this is for my exam on tuesday. This problem & another one has been causing me troubles haha.
I used the same process throughout ^^ However if some of my numbers are weird above ignore those as i went through and did it again and got an answer around 9m/s for the car.

 

[Doc Al]

When I work these perfectly elastic collision problems, it is easy for me to make a mistake. But what I got for an answer was as follows:
The 2.3 kg object initially had a velocity of +6.1 m/s. Its final velocity is -7.055 m/s.
The 3.5 kg object initially had a velocity of -4.8 m/s. Its final velocity is 3.845 m/s.

But like I said, it is easy for me to make mistakes because of the numerous math steps involved.

I worked through it as well and agree with your answers.

 

[TomHart]

(2.3kg)(6.1m/s) + (3.5kg)(-4.8m/s) = (2.3kg)(Vf2) + (3.5kg)(Vf1)

1/2(2.3kg)(6.1m/s)^2 + 1/2(3.5kg)(-4.8m/s)^2 = 1/2(2.3kg)(Vf1)^2 + 1/2(3.8kg)(Vf2)^2

It looks like in the first equation you have the 3.5 kg object's velocity defined as V_f1, and in the second equation you have the 2.3 kg object defined as V_f1. That could be causing a problem.

 

[sunnnystrong]

I worked through it as well and agree with your answers.

When I work these perfectly elastic collision problems, it is easy for me to make a mistake. But what I got for an answer was as follows:
The 2.3 kg object initially had a velocity of +6.1 m/s. Its final velocity is -7.055 m/s.
The 3.5 kg object initially had a velocity of -4.8 m/s. Its final velocity is 3.845 m/s.

But like I said, it is easy for me to make mistakes because of the numerous math steps involved.I thought your set-up looked right. I will try to go through your solution to see if I can find the problem.

Thanks for your help guys. I actually misread the answer key & the answer is 7 m/s for the speed.
I will go back and find my math error because I set it up right ^^^ but made a mistake

 

[haruspex]

you get a quadratic:

In perfectly elastic collisions there is a shortcut that can avoid quadratics.
Newton's experimental law says that for two given materials (not necessarily perfectly elastic) in a straight line collision the relative velocity afterwards is -R times the initial relative velocity, where R is a constant (coefficient of restitution) depending only on the materials. For a perfectly elastic collision R=1.
Thus, in the present problem, you could write down that the relative velocity after the collision is 6.1+4.8=10.9 m/s (but reversed). Combining that with conservation of momentum gives you the answer without any quadratics.
The law, for the perfectly elastic case, can be deduced fromconservation of energy and momentum.

 

[Doc Al]

To add to what haruspex just said, you may want to look at our formulay, especially this post: Linear Momentum and Collisions