What are the equations for solving the collision problem in momentum?

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SUMMARY

The collision problem involves a car with a mass of 2000 kg moving at 40 m/s colliding with a stationary car of mass 1500 kg. The conservation of momentum equation, represented as m1v1 = m2v2, is used to determine the speed of the second car prior to the collision, yielding a speed of 53.3 m/s. The direction of the second car can be analyzed using vector equations, where the first car's direction is defined as +x. The discussion emphasizes the importance of considering both momentum and kinetic energy in collision problems.

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Violagirl
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Homework Statement


A car moving to the right at 40 m/s has a mass of 2000 kg. It strikes a car of mass 1500
kg and both cars immediately come to rest. What was the speed and direction of the
second car just prior to the collision?

Homework Equations



pinitial=pfinal

p=mv

K=1/2mv2



The Attempt at a Solution



In determining the speed of the second car, I set up my formula to try to resemble what the problem provides in how the first car is moving:

If pinitial=pfinal, then:

m1v1=m2v2

I set it up like this since we're only told that the first car is traveling in a given direction towards the second car and we we want it so that we can find v2. So:

(2000 g) (40 m/s) = (1500 g)v2

v2=53.3 m/s

Otherwise, I'm not sure what approach to take the find the direction. I thought maybe in the x direction with sin equaling 180 degrees with car one in applying the kinetic energy equation but I am not exactly sure if this is correct.
 
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Violagirl said:
If pinitial=pfinal, then:

m1v1=m2v2
What does that imply about the direction of the second car?
I'm not sure what approach to take the find the direction. I thought maybe in the x direction with sin equaling 180 degrees with car one in applying the kinetic energy equation but I am not exactly sure if this is correct.
If "to the right" (the direction of the 1st car) is the +x direction, then 180 degrees from the +x direction is what?

You could try doing it formally - call the direction of the first car +x and write out vector equations:

before collision:
##\vec{p}_{b}=p_1\hat{\imath}+\vec{p}_{2}=(m_1v_1+m_2v_2\cos{\theta})\hat{\imath}+m_2v_2\sin(\theta)\hat{\jmath}##

after collision:
##\vec{p}_{a}=0##

conservation of momentum:
##\vec{p}_a = \vec{p}_b##

... gives you two equations and two unknowns.

If ##\theta=0## then car-2 is traveling in the same direction as car-1; which is to say: "to the right".
To get full marks, you should always try to provide final answers using the same terms as used in the question.

In general: if you lay out all conservation-of-stuff problems like this, you'll find them easier to think your way through.

Note: write the expression for total kinetic energy before and after the collisions.
Is there any value for ##v_2## (your only unknown) which will make KE before equal KE after? i.e. is kinetic energy conserved in the collision?
 
Last edited:
Both cars are moving initially and "finally" both are stationary.
So pinitial need to have both components and pfinal = ...
 

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