Inelastic collision and Kinetic Energy Conservation

AI Thread Summary
In an inelastic collision, a 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s, resulting in the cars sticking together. Momentum is conserved in both the x and y components, allowing for the calculation of the wreckage's speed post-collision using the equation m1v1 + m2v2 = (m1 + m2)vfinal. For the second scenario, a 1.80-kg block compresses a spring after sliding on a rough surface, with kinetic friction affecting its motion. The force constant of the spring can be determined using energy conservation principles, factoring in the work done against friction. Overall, both scenarios illustrate the principles of momentum conservation and energy transformation in physics.
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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
 
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For #1, momentum is conserved in its components.

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

The car traveling 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
 
thank you so much!
 
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