What is the velocity of the cars right after the collision?

  • Thread starter Bob Loblaw
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  • #1
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Homework Statement



A 1500 kg car moving east at 17 m/s collides with a 1610 kg car moving south at 15 m/s and the two cars stick together.

(a) What is the velocity of the cars right after the collision?

(b) How much kinetic energy was converted to another form during the collision?



The attempt at a solution

Solving A was simply a matter of:
x: m1v1=(m1+m2)vx
y: m2v2=(m1+m2)vy

vx=m1/(m1+m2)=8.2 m/s^2
vy=m2/(m1+m2)=7.8 m/s^2

using pythagorean's theorum I reolved the velocity as 11.29 and by using the arctan function I found the angle was 43.442 south of east.

I am at a loss on how to compute how much kinetic energy was transformed. I tried comparing m1v1-m2v2 but that did not work.

Any ideas?
 
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Answers and Replies

  • #2
nrqed
Science Advisor
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Homework Statement



A 1500 kg car moving east at 17 m/s collides with a 1610 kg car moving south at 15 m/s and the two cars stick together.

(a) What is the velocity of the cars right after the collision?

(b) How much kinetic energy was converted to another form during the collision?



The attempt at a solution

Solving A was simply a matter of:
x: m1v1=(m1+m2)vx
y: m2v2=(m1+m2)vy

vx=m1/(m1+m2)=8.2 m/s^2
vy=m2/(m1+m2)=7.8 m/s^2

using pythagorean's theorum I reolved the velocity as 11.29 and by using the arctan function I found the angle was 43.442 south of east.

I am at a loss on how to compute how much kinetic energy was transformed. I tried comparing m1v1-m2v2 but that did not work.

Any ideas?

calculate [tex] \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 [/tex] before and after
 
  • #3
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Thanks for the help.
 

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