Calculating Final Velocities for Identical Plates on Ice

[n_h0987]

Homework Statement

Identical plates are placed close together on ice. A 0.742 kg rabbit that is initially sitting on the plate on the left jumps to the other plate and immediately back to the original plate. The rabbit leaves both plates with a horizontal component of velocity of magnitude 3.00 m/s with respect to the ice. If the mass of each plate is 12.8 kg, calculate the final velocities of both plates.

Homework Equations

p = mv
Ek = 1/2mv^2
m1v1i + m2v2i = m1v1f + m2v2f
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2 m2v2f^2

The Attempt at a Solution

I don't even know where to begin with this one...

 

[Doc Al]

m1v1i + m2v2i = m1v1f + m2v2f

This is all you need.

Just do it step by step:

When the rabbit jumps off of plate 1, figure out the speed of that plate.

When the rabbit lands on plate 2, figure out the speed of 'rabbit + plate 2'. (Assume they end up with the same speed.)

And so on until you've got it all.

 

[huynhtn2]

This is all you need.

Just do it step by step:

When the rabbit jumps off of plate 1, figure out the speed of that plate.

When the rabbit lands on plate 2, figure out the speed of 'rabbit + plate 2'. (Assume they end up with the same speed.)

And so on until you've got it all.

This is what i tried:
m1v1 +m2v2 = m1v1 F + m2v2 F
m1v1 + 0 = (m1+m2) vF
(0.742)(3.00) = (0.742 + 12.8) vF
vF= 0.164 m/s

then i added the velocity added when the rabbit jumps back to the left plate
p=mv
= (0.742)(3)
= 2.226 m/s
p=mv
1.47 =12.8v
v= 0.11

so the total velocity is 0.274m/s
but this is incorrect, what did do wrong,

 

[Doc Al]

This is what i tried:
m1v1 +m2v2 = m1v1 F + m2v2 F
m1v1 + 0 = (m1+m2) vF
(0.742)(3.00) = (0.742 + 12.8) vF
vF= 0.164 m/s

This is the speed of 'rabbit + plate 2'.

then i added the velocity added when the rabbit jumps back to the left plate
p=mv
= (0.742)(3)
= 2.226 m/s
p=mv
1.47 =12.8v
v= 0.11

so the total velocity is 0.274m/s

I don't understand what you're doing here.

 

[huynhtn2]

This is the speed of 'rabbit + plate 2'.


I don't understand what you're doing here.

When the rabbit jumps back to the left plate doesn't it increase the plate 2 velocity when it pushes off from it? or is that wrong?

 

[Doc Al]

When the rabbit jumps back to the left plate doesn't it increase the plate 2 velocity when it pushes off from it? or is that wrong?

Sure it does. Is that what you were trying to calculate?

 

[huynhtn2]

Sure it does. Is that what you were trying to calculate?

Yeah... was it a fail?

 

[Doc Al]

Yeah... was it a fail?

I think so. Explain what you were doing.

 

[huynhtn2]

I think so. Explain what you were doing.

momentum of rabbit as it leaves plate 2
p=mv
= (0.742)(3)
= 2.226

backward momentum of plate
p=mv
2.226 =12.8v
v= 0.173

then take this v and add it to the v from before to get final v?

 

[Doc Al]

momentum of rabbit as it leaves plate 2
p=mv
= (0.742)(3)
= 2.226

OK.

backward momentum of plate
p=mv
2.226 =12.8v
v= 0.173

You neglected the initial velocity of the rabbit.

 

[huynhtn2]

OK.


You neglected the initial velocity of the rabbit.

momentum of rabbit as it leaves plate 2
p=mv
= (0.742)(3-0.164)
= 2.104

backward momentum of plate
p=mv
2.104 =12.8v
v= 0.164

I should subtract the velocity from the first equation right?

 

[Doc Al]

momentum of rabbit as it leaves plate 2
p=mv
= (0.742)(3-0.164)
= 2.104

Why did you change this? You had this part correct before.

Just apply conservation of momentum in one step.

What's the momentum of 'rabbit + plate 2'? Set that equal to the momentum of the rabbit after he jumps + the momentum of plate 2.

 

[huynhtn2]

Why did you change this? You had this part correct before.

Just apply conservation of momentum in one step.

What's the momentum of 'rabbit + plate 2'? Set that equal to the momentum of the rabbit after he jumps + the momentum of plate 2.

mv = mv
(0.742)(3)=(0.742 +12.8)v?

I am confused out of my mind now, please help.

 

[Doc Al]

mv = mv
(0.742)(3)=(0.742 +12.8)v?

I am confused out of my mind now, please help.

That's the momentum of 'rabbit + plate 2', which happens to just equal the momentum of the rabbit. If you want the speed of plate 2 after the rabbit jumps off, continue as I suggested before:

Set that equal to the momentum of the rabbit after he jumps + the momentum of plate 2.

Hint: Pay attentions to signs.

 

[huynhtn2]

That's the momentum of 'rabbit + plate 2', which happens to just equal the momentum of the rabbit. If you want the speed of plate 2 after the rabbit jumps off, continue as I suggested before:

Hint: Pay attentions to signs.

I finally got it, thanks so much for your help. I know i was annoying so thanks for sticking with helping me.

 

[ghostanime2001]

Identical plates are placed close together on ice as shown. A 0.645 kg rabbit that is initially sitting on the plate on the left jumps to the other plate and immediately back to the original plate. The rabbit leaves both plates with a horizontal component of velocity of magnitude 2.87 m/s with respect to the ice.

(a) If the mass of each plate is 12.0 kg, calculate the final velocity of the plate on the right. (Ans = 3.09e-01 m/s)
(b) Calculate the final velocity of the plate on the left. (Ans = -2.93e-01 m/s)

I have a similar question. I get the right numerical answer to the second jump (when the rabbit jumps from right plate back to left plate) except the sign is reversed. I followed the sign convention in the conservation of momentum equations, but I do not understand why it isn't working out. The right plate works out perfectly when following sign conventions but it gives trouble for the left plate. Anyway, here is my process for the left plate ONLY:

m = 12 kg
m[itex]_{R}[/itex] = 0.645 kg
v[itex]_{R}[/itex] = 2.87 m/s
v[itex]_{i}[/itex] = 0 = initial velocity of left plate before jump (rabbit sitting initially on the left plate)
v[itex]_{f}[/itex] = final velocity of left plate (rabbit's first jump from left plate to right plate)
v = final velocity of left plate (rabbit's second jump from right plate back to left plate)

There are 2 unknown, so 2 equations are required.
Also following the sign conventions, if the rabbit were to jump to the right, it's momentum is positive. The left plate moves left because the rabbit pushes back on the left plate, so it's momentum is negative. These are shown in the conservation of momentum equations below:

Before Jump = After jump (rabbit jumps from left plate to right plate)
[itex]\left(m+m_{R}\right)v_{i}=m_{R}v_{R}-mv_{f}[/itex]

[itex]0=m_{R}v_{R}-mv_{f}[/itex]

[itex]mv_{f}=m_{R}v_{R}[/itex]

I don't understand why this equation if solved for v[itex]_{f}[/itex] gives a positive value. I have followed the sign conventions in the first line, but I do not understand why it works out to give a positive velocity instead of a negative velocity.

The second conservation of momentum equation of left plate (second jump from right plate back to left plate):

Before landing = After landing
[itex]-mv_{f}-m_{R}v_{R}=-(m+m_{R})v[/itex]

looking back to the first equation, substitute [itex]mv_{f}[/itex] for [itex]m_{R}v_{R}[/itex] and solve for [itex]v[/itex]

[itex]-m_{R}v_{R}-m_{R}v_{R}=-(m+m_{R})v[/itex]

[itex]-2m_{R}v_{R}=-(m+m_{R})v[/itex]

[itex]\dfrac{-2m_{R}v_{R}}{-(m+m_{R})}=v[/itex]

[itex]0.29278 \text{ m/s} = v[/itex]

I know the problem stems somewhere from the first conservation of momentum equation, but when I followed the sign conventions for the right plate, everything worked out fine, the answer came out positive and correct. However, when following sign conventions for the left plate the answer still comes out positive which is wrong.
Thank you to anyone's help ! =D

 

[Doc Al]

Before Jump = After jump (rabbit jumps from left plate to right plate)
[itex]\left(m+m_{R}\right)v_{i}=m_{R}v_{R}-mv_{f}[/itex]

[itex]0=m_{R}v_{R}-mv_{f}[/itex]

[itex]mv_{f}=m_{R}v_{R}[/itex]

I don't understand why this equation if solved for v[itex]_{f}[/itex] gives a positive value. I have followed the sign conventions in the first line, but I do not understand why it works out to give a positive velocity instead of a negative velocity.

It comes out positive because you have already included the minus sign in your equation. Your v_f is the magnitude of the velocity. If you want the sign of v_f to be negative (letting the solution provide the proper sign), you would have just done:

[itex]\left(m+m_{R}\right)v_{i}=m_{R}v_{R} + mv_{f}[/itex]

[itex]0=m_{R}v_{R} + mv_{f}[/itex]

[itex]mv_{f}= -m_{R}v_{R}[/itex]

Which would give you a negative value.

Personally, I would have done it the way you did, letting v_f in your equation just stand for the magnitude. (It's easier to work with variables that you know will be positive when you know the direction.)

Even though your variable is positive, the actual momentum is of course negative.

The second conservation of momentum equation of left plate (second jump from right plate back to left plate):

Before landing = After landing
[itex]-mv_{f}-m_{R}v_{R}=-(m+m_{R})v[/itex]

Once again, you have incorporated the direction of the various momenta by adding minus signs in your equation, thus your values for v_f, v_R, and v are all positive.

 

[ghostanime2001]

I need 2 equations so the first equation should work out such that the answer gives the correct magnitude and direction. When the first equation is inserted into the second equation, the mathematics should automatically work out to give the correct magnitude and direction, otherwise the two equations are in conflict with each other (as is the case here).

1. if I DON'T take into account the signs, the second equation collapses to 0=combined mass momentum

2. If I DO take into account the signs, the second equation gives positive answers (using intuition for directions can give out correct answer, but I don't understand why using sign conventions for the right plate works but not with the left plate).

the math should automatically work out to give the negative value, I know something is missing.

 

[Doc Al]

I need 2 equations so the first equation should work out such that the answer gives the correct magnitude and direction. When the first equation is inserted into the second equation, the mathematics should automatically work out to give the correct magnitude and direction, otherwise the two equations are in conflict with each other (as is the case here).

1. if I DON'T take into account the signs, the second equation collapses to 0=combined mass momentum

2. If I DO take into account the signs, the second equation gives positive answers (using intuition for directions can give out correct answer, but I don't understand why using sign conventions for the right plate works but not with the left plate).

the math should automatically work out to give the negative value, I know something is missing.

You only used the sign convention half-way. You were lucky with the right plate, because it always moves in the positive direction.

Let's pretend we have no idea how the left plate will move. We'll use the sign convention and let the equations give us the answers.

Let's call the speed of the plate after the first jump v₁ and after the second jump v₂.

First jump:
[tex]0 = m_Rv_R + mv_1[/tex]
So:
[tex]v_1 = - \frac{m_Rv_R}{m}[/tex]
Well we know that v_R is positive, so that means v₁ is negative. Which makes sense, since the plate must move opposite to the rabbit.

Second jump:
[tex]mv_1 - m_Rv_R = (m + m_R)v_2[/tex]

Now since we know v₁ is negative, if you crank out v₂ you'll find that to be negative also.

Works just fine! (Let me know if that clears things up.)

 

[ghostanime2001]

That clears it up but it's like the sign convention switched for the left plate. Is there a reason behind doing so ?

 

[Doc Al]

That clears it up but it's like the sign convention switched for the left plate. Is there a reason behind doing so ?

I don't understand. The sign convention is the same for both: To the right is positive.

 

[ghostanime2001]

In your previous post, the sign convention is positive to the left for the left plate

 

[Doc Al]

In your previous post, the sign convention is positive to the left for the left plate

That's because I was taking advantage of my knowing that the left plate moved left. That way I can let my unknowns be positive quantities.

 

[ghostanime2001]

So then shouldn't the momentum of the rabbit be negative for the first jump ? since it is moving right ?

 

[Doc Al]

So then shouldn't the momentum of the rabbit be negative for the first jump ? since it is moving right ?

Actually, I stuck with the usual sign convention. I just let v_f stand for the magnitude of the plate's velocity, thus it was positive. The momentum of the plate is now -mv_f, not mv_f, since the plate's final velocity is -v_f.

As I said before, I would have done exactly what I thought you started out doing:

[itex]\left(m+m_{R}\right)v_{i}=m_{R}v_{R}-mv_{f}[/itex]

[itex]0=m_{R}v_{R}-mv_{f}[/itex]

[itex]mv_{f}=m_{R}v_{R}[/itex]

v_f is of course positive, but the plate moves in the negative direction.

 

[ghostanime2001]

I understand now but I am now having second thoughts about this question. Now I am confused about how the rabbit lands on the plates. My first question is the rabbit's velocity is 2.87 m/s after it jumps AND also 2.87 m/s before it lands on their respective plates ?

In the first plate (left plate) i know the rabbit jumps to the right and the plate moves to the left.
In the second plate (right plate) i know the rabbit moves to the right and the plate moves to the right also.
In the second plate (right plate) when the rabbit is jumping to the left to the first plate the second plate must move right.

I calculated the final velocities of the left plate in the first jump and it is -0.154 m/s. In the second plate the velocity is 0.146 m/s. When the rabbit jumps from the right plate back to the left plate the velocity should be greater than 0.146 however I do not get a larger velocity instead I get zero. I do not understand I have used conservation of momentum before and after the jump and before and after the land for both plates. Please help !

 

[Doc Al]

I understand now but I am now having second thoughts about this question. Now I am confused about how the rabbit lands on the plates. My first question is the rabbit's velocity is 2.87 m/s after it jumps AND also 2.87 m/s before it lands on their respective plates ?

Yes. That's given in the problem statement.

In the first plate (left plate) i know the rabbit jumps to the right and the plate moves to the left.
In the second plate (right plate) i know the rabbit moves to the right and the plate moves to the right also.
In the second plate (right plate) when the rabbit is jumping to the left to the first plate the second plate must move right.

OK.

I calculated the final velocities of the left plate in the first jump and it is -0.154 m/s. In the second plate the velocity is 0.146 m/s.

Good.

When the rabbit jumps from the right plate back to the left plate the velocity should be greater than 0.146 however I do not get a larger velocity instead I get zero. I do not understand I have used conservation of momentum before and after the jump and before and after the land for both plates.

Show how you did that last calculation as you must have made an error somewhere. (Careful with signs.)

 

[ghostanime2001]

it says the rabbit's velocity is 2.87 m/s when it leaves the left plate. What is the velocity when it is landing on the right plate ? greater than 2.87 m/s right ? So then would I have to use kinetic energy conservation ?

 

[ghostanime2001]

for part (a) here is my method (when the rabbit jumps ON the right plate):

Before landing = After landing

[itex]m_{R}\left(v_{i}\right)_{R}+m_{p}\left(v_{i}\right)_{p}=m_{R}\left(v_{f}\right)_{R}+m_{p}\left(v_{f}\right)_{p}[/itex]

[itex]m_{R}\left(v_{i}\right)_{R}=\left(m_{R} + m_{p}\right)v_{f}[/itex]

[itex]\dfrac{m_{R}\left(v_{i}\right)_{R}}{m_{R} + m_{p}}=v_{f}[/itex]

[itex]\dfrac{\left(0.645\right)\left(2.87\right)}{0.645+12}=v_{f}[/itex]

[itex]0.146 m/s=v_{f}[/itex]

(when the rabbit jumps FROM the right plate back to the left plate):

Before jump = After jump

[itex]m_{R}\left(v_{i}\right)_{R}+m_{p}\left(v_{i}\right)_{p}=m_{R}\left(v_{f}\right)_{R}+m_{p}\left(v_{f}\right)_{p}[/itex]

[itex]\left(m_{R} + m_{p}\right)v_{f}=m_{R}\left(v_{f}\right)_{R}+m_{p}\left(v_{f}\right)_{p}[/itex]

[itex]\left(m_{R} + m_{p}\right)v_{f}=m_{R}\left(v_{f}\right)_{R}+m_{p}\left(v_{f}\right)_{p}[/itex]

 

[Doc Al]

it says the rabbit's velocity is 2.87 m/s when it leaves the left plate. What is the velocity when it is landing on the right plate ? greater than 2.87 m/s right ?

No. The speed of the rabbit when it's in the air is the same.

So then would I have to use kinetic energy conservation ?

Kinetic energy is not conserved in this problem.

 

[Doc Al]

(when the rabbit jumps FROM the right plate back to the left plate):

Before jump = After jump

[itex]m_{R}\left(v_{i}\right)_{R}+m_{p}\left(v_{i}\right)_{p}=m_{R}\left(v_{f}\right)_{R}+m_{p}\left(v_{f}\right)_{p}[/itex]

[itex]\left(m_{R} + m_{p}\right)v_{f}=m_{R}\left(v_{f}\right)_{R}+m_{p}\left(v_{f}\right)_{p}[/itex]

[itex]\left(m_{R} + m_{p}\right)v_{f}=m_{R}\left(v_{f}\right)_{R}+m_{p}\left(v_{f}\right)_{p}[/itex]

Good. And you know (are given) the final speed of the rabbit after it jumps. So use that to figure out the final speed of the plate.

 

[ghostanime2001]

I think I got it but let me send you a PDF that I produced in LaTeX to see if my work make sense or not. Please check out the attached PDF file.