Collisions and Momentum

  • #1
A 3 kg mass moving laterally at 5 m/s collides with a 5 kg mass at rest. As a result the 5 kg mass travels at 2 m/s at a 26 angle counterclockwise from the direction of the motion of the initial mass. The 5 kg mass then collides with a 4 kg mass at rest. The 4 kg mass then leaves at 1 m/s at a 30 angle counterclockwise from the direction of motion of the 5 kg mass.

a. Find the final magnitude and direction for the velocity of each of the first 2 two masses.

Does this apply to only the 3 kg and 5 kg mass?

Let us say m1 = 3 kg, m2 = 5 kg, m3 = 4 kg.

Now I am a bit confused since 3 masses and not 2 are involved.
This is how I started:

1st scene
x component: m1v1 = m1v1'*cos(0) + m2v2'*cos(26)
y component: 0 = m1v1'*sin(0) + m2v2'*sin(26)

But I am not sure if my signs are correct. Then from here I was going to add the 2 equations together to determine v1' and then use substitution for v2' . . .

The 2nd scene with m2 and m3 has stumped me. Do I use similar equations like those above but just replace it with the new velocity of m2 (v2' will be v2 now?) and angles (26 for m2 and 30 for m3)?

b. Are these collisions elastic? Why or why not?

Do I count all the collisions separately or will they all be elastic or inelastic?

Basically I have to see if kinetic energy is conserved by calculating the total KE after the collision(s) and comparing it with before the collision(s). But I don't know what to do from here because a collsion follows another collision, and I am totally lost.

Thanks for the help.
 

Answers and Replies

  • #2
futb0l
Try to do a drawing and put the collisions drawing up the angles, this may help you.

The 2nd scene with m2 and m3 has stumped me. Do I use similar equations like those above but just replace it with the new velocity of m2 (v2' will be v2 now?) and angles (26 for m2 and 30 for m3)?

Yes - you should change the velocity - v2 = v2' do not change the angle for mass 2 to 26 - treat it as another collision.
 
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  • #3
Are there any hints to help me out on the kinetic energy portion?
 
  • #4
Doc Al
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Soaring Crane said:
a. Find the final magnitude and direction for the velocity of each of the first 2 two masses.

Does this apply to only the 3 kg and 5 kg mass?
Since you are given the final velocity of the 4 kg mass--no point in asking for that! :smile:

Let us say m1 = 3 kg, m2 = 5 kg, m3 = 4 kg.

Now I am a bit confused since 3 masses and not 2 are involved.
Just do it step by step.
This is how I started:

1st scene
x component: m1v1 = m1v1'*cos(0) + m2v2'*cos(26)
y component: 0 = m1v1'*sin(0) + m2v2'*sin(26)
I assume you mean that m1 moves at some angle [itex]\theta[/itex], not 0!

But I am not sure if my signs are correct. Then from here I was going to add the 2 equations together to determine v1' and then use substitution for v2' . . .
You can work with the signs you chose. (Your angle will come out negative.) Hint: Isolate the m1v1' terms so you can solve for the angle.

The 2nd scene with m2 and m3 has stumped me. Do I use similar equations like those above but just replace it with the new velocity of m2 (v2' will be v2 now?) and angles (26 for m2 and 30 for m3)?
Do the same analysis for the 2nd collision. (Draw a picture to keep the angles straight.)

b. Are these collisions elastic? Why or why not?

Do I count all the collisions separately or will they all be elastic or inelastic?
Each collision must be analyzed separately.

Basically I have to see if kinetic energy is conserved by calculating the total KE after the collision(s) and comparing it with before the collision(s).
Right. That's all there is to it. Since you already know the speeds before and after each collision (from part a), this part should be easy.
 
  • #5
For angle 1, is it around 19? For the second collision, do I solve for angle 2 (so this is why I do not use 26)?

As for the actual equation is y-comp: 0 = m2v2'*cos(theta_2) + m3v3'*cos(30)?

Tell me if I'm wrong so I can amend it. Thanks.
 
  • #6
Doc Al
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Soaring Crane said:
For angle 1, is it around 19?
Not according to my calculation.
For the second collision, do I solve for angle 2 (so this is why I do not use 26)?
Since you need to find the direction, of course you need to find angle 2. I would measure the angle from the direction of the incoming particle (just like you did for the first collision), so I would not use 26 degrees in the momentum equation. (Although in presenting your final answer for the direction of the velocity, you may need to make use of that 26 degrees.)

As for the actual equation is y-comp: 0 = m2v2'*cos(theta_2) + m3v3'*cos(30)?
Looks fine to me.
 
  • #7
For the 1st part:

0 = m1v1'*sin(theta_1) + m2v2'*sin(26)
-[(m2v2'*sin(26)]/[m1v1'] = sin(theta_1)
theta_1 = -17

Now v1' = [m1v1 - m2v2'*cos(26)]/[m1cos(-17)] = 2.1 m/s


2nd part:
0 = m2v2'*sin(theta_2) + m3v3'*sin(30)
-[m3v3'*sin(30)]/[m2v2'] = sin(theta_2)
theta_2 = -11.54 = -12

v2' = [m2v2 - m3v3'*cos(30)]/[m2cos(-12)] = 1.3 m/s

Wrong or right? Please check. Thanks.
 
  • #8
Doc Al
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Soaring Crane said:
For the 1st part:

0 = m1v1'*sin(theta_1) + m2v2'*sin(26)
-[(m2v2'*sin(26)]/[m1v1'] = sin(theta_1)
theta_1 = -17
Incorrect. Since v1' is unknown, this equation not allow you to solve for theta_1. (I assume you got your answer by using v1 instead of v1'.)

To solve for the angle you need to combine both equations. See my hint in post #4.
 

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