The figure above shows the velocity graphs for a \"sideswipe\"collision between

Question

The figure above shows the velocity graphs for a "sideswipe"collision between two carts. In this collision, one cart comes frombehind at a high velocity, scrapes the second cart, and then goeson past it. The initial and final velocities are:v1i = 0.65 m/s, v1f = 0.25 m/s,v2i = 0.04 m/s, v2f = 0.2 m/s. Theinertia of cart 1 is M1 = 0.50 kg. How much kineticenergy is lost during the collision?

(Hint: Think carefully about what you know. You needM2.)

2) You are on arollercoaster with a circular loop of radius R. Assuming therollercoaster is frictionless, what is the minimum height the carhas to start at to make it around the loop without falling off thetracks? There is no initial velocity or acceleration, so the entireenergy of the car is due to its initial potentialenergy. 1) 3 R 2) R 3) 2 R 4) 5/2 R You are on arollercoaster with a circular loop of radius R. Assuming therollercoaster is frictionless, what is the minimum height the carhas to start at to make it around the loop without falling off thetracks? There is no initial velocity or acceleration, so the entireenergy of the car is due to its initial potentialenergy. You are on arollercoaster with a circular loop of radius R. Assuming therollercoaster is frictionless, what is the minimum height the carhas to start at to make it around the loop without falling off thetracks? There is no initial velocity or acceleration, so the entireenergy of the car is due to its initial potentialenergy. 3 R R 2 R 5/2 R 1) 3 R 2) R 3) 2 R 4) 5/2 R

Explanation / Answer

1) Let's calculate the K separately for each car: M1 K = 1/2mv2 KM1 = [(1/2)(0.50)(.25)2] -[(1/2)(0.50)(0.6502] KM1 = -0.09 J M2 K = 1/2mv2KM2 = [(1/2)m2 (.02)2] - [(1/2) m2(0.04)2] KM2= 0.02m2 -0.0008m2KM2 =0.0192m2 Ktotal = KM1 +KM2 Ktotal = -0.09 J + 0.0192m2 Therefore, the total loss in kinetic energy is: 0.09 J -0.0192m22) In order for the roller coaster cart to make it all the way to thetop, the initial potential energy has to equal the final kineticenergy at the top. The only way for that to happen is if theroller coaster cart started at the very top. That is, soUi = mg(2R).   Thus, your answer is (3)2R.

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