Inelastic collision and Kinetic Energy Conservation

  • Thread starter stonnn
  • Start date
  • #1
8
0
1)
A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. What is the speed of the wreckage just after the collision?

Relevant equations: m1v1 - m2v2 = (m1+ m2) vfinal
My attempt: This is an inelastic collision because the cars collide and stick together. So, momentum is conserved but kinetic energy is not.



2)A 1.80-kg block slides on a rough horizontal surface. The block hits a spring with a speed
of 2.00 m/s and compresses it a distance of 11.0 cm before coming to rest. If the coefficient
of kinetic friction between the block and the surface is μk = 0.560, what is the force
constant of the spring?

My attempt:

1/2mv2 - Mgd - 1/2kx = 0
mv2 -Mg
[(1.80kg)(2m/s)(2m/s) - (0.560)(10)]/0.1meters
k = 16 n x m
 

Answers and Replies

  • #2
73
0
For #1, momentum is conserved in its components.

m1v1x + m2v2x = (m1+ m2) v'x
m1v1y + m2v2y = (m1+ m2) v'y

The car travelling north has no velocity in the x-direction and similarly, the car heading east has no velocity in the y-direction.

Solve both equations to get the components on the velocity after the collision and use pythagoream theorem to solve for the final velocity and the direction
 
  • #3
8
0
thank you so much!!
 

Related Threads on Inelastic collision and Kinetic Energy Conservation

Replies
3
Views
1K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
3
Views
21K
Replies
2
Views
8K
Replies
9
Views
10K
Replies
3
Views
3K
Replies
3
Views
3K
Replies
3
Views
913
Replies
1
Views
2K
Replies
1
Views
2K
Top