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

Click For Summary
SUMMARY

The collision of a 1500 kg car moving east at 17 m/s and a 1610 kg car moving south at 15 m/s results in the two cars sticking together. The velocity of the combined mass immediately after the collision is calculated to be 11.29 m/s at an angle of 43.442 degrees south of east. The kinetic energy before the collision is computed using the formula K.E. = 1/2 m v^2, but the discussion highlights confusion regarding the calculation of kinetic energy converted to other forms during the collision.

PREREQUISITES
  • Understanding of momentum conservation principles
  • Familiarity with vector addition in physics
  • Knowledge of kinetic energy calculations
  • Proficiency in using trigonometric functions for angle determination
NEXT STEPS
  • Study momentum conservation in inelastic collisions
  • Learn how to calculate kinetic energy before and after a collision
  • Explore vector addition techniques for two-dimensional motion
  • Review the application of trigonometric functions in physics problems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of inelastic collisions and energy transformation concepts.

Bob Loblaw
Messages
67
Reaction score
0

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?
 
Last edited:
Physics news on Phys.org
Bob Loblaw said:

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
 
Thanks for the help.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
4
Views
2K