How Do Carts Behave and Springs Compress in an Elastic Collision?

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SUMMARY

In an elastic collision involving a 0.60 kg cart moving at 5.0 m/s (W) and a 0.80 kg cart moving at 2.0 m/s (E), the post-collision velocities were calculated using the formulas for elastic collisions. The first cart's velocity after the collision is -1 m/s (E), while the second cart's velocity is 4 m/s (W). The maximum compression of the spring, with a spring constant of 1200 N/m, requires further analysis of the potential energy stored in the spring, which is given by U = 0.5 * k * X^2.

PREREQUISITES
  • Understanding of elastic collisions and conservation of momentum
  • Familiarity with the equations for post-collision velocities
  • Knowledge of spring mechanics and Hooke's Law
  • Ability to manipulate algebraic equations for energy calculations
NEXT STEPS
  • Study the principles of conservation of momentum in elastic collisions
  • Learn how to derive post-collision velocities for two-body collisions
  • Explore the relationship between kinetic energy and potential energy in spring systems
  • Investigate the effects of varying spring constants on compression and energy storage
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Physics students, educators, and anyone interested in understanding the dynamics of elastic collisions and spring mechanics.

fa08ti
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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s (W) collides with a 0.80 kg cart moving at 2.0m/s (E). the collision is cushioned by a spring (k=1200 N/m)

a find the velocity of each cart after the collision

b find the maximum compression of the spring

ATTEMPT:

cart 1:
v1=( m1-m2/m1+m2)(v1)
v1= (0.6-0.8/0.6+0.8)(7)
v1= -1

then i did -1 + -1 = -3 m/s (E)

cart 2:

v2= (2m1/m1+m2) (v1)
v2= 6
6 + -2= 4 m/s (W)

i'm totally lost for part b
 
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It may help to recall the potential energy stored in a compressed spring is equal to

U = 0.5 * K * X^2
 
you might want to check (a) again
i am getting different answers
 

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