2 bumper cars collide (check my work please?)

[Rijad Hadzic]

Homework Statement

So my book doesn't have an answer for this problem at the back, was just wondering if someone could check my work..

Two bumper cars at the county fair are sliding toward one another. Initially bumper car 1 is traveling to the east at 5.62 m/s and bumper car 2 is traveling 60 degrees south of west at 10 m/s. After they collide, bumper car 1 is observed to be traveling west with a speed of 3.14 m/s. Fiction is negligible between the cars and the ground. If the masses of bumper cars 1 and 2 are 596 kg and 625 kg respectively, what is the velocity of bumper car 2 immediately after the collision?

Homework Equations

K_i = K_f
$V_f = ((m_1 - m_2)/(m_1 + m_2)) (V_{im_1}) + ((2m_2)/(m_1 + m_2))(V_{im_2})$

The Attempt at a Solution

So the x velocity of bumper car 2 according the the eq ^ up there.

$V_x = ((625 -596) / (625 + 596)) 10cos(240) + \frac {2(596)}{(625 + 596) }(5.62)$

x velocity = 5.62 m/s

$V_y = ( (625-596)/(625+596) ) 10sin(240) + 0$

y velocity = -.2057 m/s

I put 0 here because the y velocity of car 1 doesn't exist

Now I find magnitude:

$( (5.3682)^2 + (-.2057)^2 )^{1/2}$

= 5.3721 m/s at 357 degrees (using arctan)

Can anyone check my work and make sure its correct? Am I doing the right thing here?

I also got another answer around 50 m/s using Kinetic energy initial = Ke final but that doesn't seem right so I will use this one since 5.37 m/s seems about right to me.

 

[TomHart]

I haven't gone through the math, but it does not seem reasonable to me that one of the bumper cars at the county fair would be traveling in excess of 100 miles per hour after the impact. Call me a stick in the mud, but that may be a bit too extreme for children on bumper cars.

 

[Rijad Hadzic]

I haven't gone through the math, but it does not seem reasonable to me that one of the bumper cars at the county fair would be traveling in excess of 100 miles per hour after the impact. Call me a stick in the mud, but that may be a bit too extreme for children on bumper cars.

I know lol, that's what my answer was using the Kinetic energy initial = Ke final approach, but I dismissed it since it seemed too large to me.

Instead I got v = 5.3721 m/s at 357 degrees which seems reasonable to me but I don't have the answer for this one in the back of my book unfortunately. :(

 

[TomHart]

I'll try to work through it, but some smarter person will probably beat me to it.

 

[Rijad Hadzic]

I'll try to work through it, but some smarter person will probably beat me to it.

I have faith in you and your abilities TomHart.

 

[TomHart]

I got a different answer than you did. Let me ask you a question. What is the y component of car 2's velocity before impact? And, since car 1 has no y component of velocity, either before or after the impact, what does that tell you about car 2's y component of velocity after impact?

 

[Rijad Hadzic]

I got a different answer than you did. Let me ask you a question. What is the y component of car 2's velocity before impact? And, since car 1 has no y component of velocity, either before or after the impact, what does that tell you about car 2's y component of velocity after impact?

Car 2's y component = 10sin(240) degrees, right? That means car 2's y component of velocity after impact remains the same?

 

[TomHart]

Yes, that's right. So that is a magnitude of 8.66 m/s. And since your answer of 5.3721 m/s for the magnitude of the final velocity is less than the y component of 8.66, that means your answer has to be wrong.

But since you now know the final y component of the velocity of car 2, it should now be a simple matter of equating before and after total x component momentum to find the x component of car 2's velocity.

 

[Rijad Hadzic]

Yes, that's right. So that is a magnitude of 8.66 m/s. And since your answer of 5.3721 m/s for the magnitude of the final velocity is less than the y component of 8.66, that means your answer has to be wrong.

But since you now know the final y component of the velocity of car 2, it should now be a simple matter of equating before and after total x component momentum to find the x component of car 2's velocity.

Sorry one question: Y initial was first 10sin(240) then it was the absolute value of 10sin(240), correct??

 

[TomHart]

Personally, I don't like the angle 240. It is kind of confusing to me. When the problem says, "60 degrees south of west", it means, from the west direction you rotate 60 degrees toward the south. It seems natural to me to define the north direction as positive y, and east direction as positive x. So starting from the +y direction, the initial angle of car 2 would actually be 210 degrees if you rotate clockwise from the starting point.
So that means the initial y component of car 2's velocity would be (10)(cos30) = 8.66 m/s in the negative y (or south) direction. And because car 1 has no y component of velocity, either before or after impact, due to conservation of the y component of momentum, the final y component of car 2's velocity has to remain unchanged - specifically, 8.66 m/s in the negative y (or south) direction. In other words, the y component of car 2's velocity is unaffected by the collision.

 

[Rijad Hadzic]

Personally, I don't like the angle 240. It is kind of confusing to me. When the problem says, "60 degrees south of west", it means, from the west direction you rotate 60 degrees toward the south. It seems natural to me to define the north direction as positive y, and east direction as positive x. So starting from the +y direction, the initial angle of car 2 would actually be 210 degrees if you rotate clockwise from the starting point.
So that means the initial y component of car 2's velocity would be (10)(cos30) = 8.66 m/s in the negative y (or south) direction. And because car 1 has no y component of velocity, either before or after impact, due to conservation of the y component of momentum, the final y component of car 2's velocity has to remain unchanged - specifically, 8.66 m/s in the negative y (or south) direction. In other words, the y component of car 2's velocity is unaffected by the collision.

https://imgur.com/a/4wNiS

That is the picture that came along with the question. I thought y component was suppose to be using sin not cos??

 

[TomHart]

For the angle defined the way it is in that picture, yes, it should be sin instead of cos.

 

[Rijad Hadzic]

For the angle defined the way it is in that picture, yes, it should be sin instead of cos.

Okay, so going by the picture I can assume the cars initial Y velocity to be 10sin(240), right?

And after the collision its still 10sin(240) because car 1 doesn't have a y component, right?

 

[TomHart]

Exactly . . . But I guess to be more accurate we should really say, car 2's y component of velocity has to remain the same because we know that car 1's y component of velocity is unchanged.

 

[Rijad Hadzic]

Exactly . . . But I guess to be more accurate we should really say, car 2's y component of velocity has to remain the same because we know that car 1's y component of velocity is unchanged.

I see.

Did you get $v_{Car2} = 3.35 ihat m/s + 10sin(240) jhat m/s$?

Sorry I forgot how to do the hat symbol again -_-

 

[TomHart]

Yes, that looks right. I got v_x = +3.354 m/s and v_y = -8.66 m/s

 

[Rijad Hadzic]

Yes, that looks right. I got v_x = +3.354 m/s and v_y = -8.66 m/s

Cool thanks TomHart