Speed of Toy Car on Curved Track: Exploring Kinetic Energy

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SUMMARY

The discussion focuses on determining the speed of a toy car at the highest point of its trajectory after leaving a curved track. The car has an initial speed VB at point B and leaves at an angle θ. The key conclusion is that the problem can be simplified to projectile motion, where the horizontal component of velocity remains constant, and the vertical component is zero at the highest point. Thus, the speed at the highest point is equal to the horizontal component of the velocity, which can be expressed in terms of VB and θ without needing additional variables such as time or acceleration.

PREREQUISITES
  • Understanding of projectile motion principles
  • Knowledge of kinetic and potential energy concepts
  • Familiarity with trigonometric functions related to angles
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the principles of projectile motion in-depth
  • Learn about energy conservation in mechanical systems
  • Explore the effects of angle θ on projectile trajectories
  • Investigate the equations of motion for objects under gravity
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of projectile motion and energy conservation.

Michele Nunes
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Homework Statement


A toy car coasts along the curved track shown. The car has initial speed VA when it is at point A at the top of the track, and the car leaves the track at point B with speed VB at an angle θ above the horizontal. Assume that energy loss due to friction is negligible.

Determine the speed of the car when it is at the highest point in its trajectory after leaving the track, in terms of VB and θ. Briefly explain how you arrived at your answer.

Homework Equations

The Attempt at a Solution


Okay so conceptually I think I understand how to do the problem. At point A, all potential energy, at point B, almost all kinetic energy. Then when the car leaves point B, energy is lost due to the downward force of the car's weight, so I want to find the work done by the car's weight and then subtract that from the original amount of energy the car possessed and then go from there. But in terms of how this plays out in the actual equations, I have no idea. I'm not sure where to start in terms of the setting equations equal to each other and how exactly to set that up.
 

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Michele Nunes said:
Determine the speed of the car when it is at the highest point in its trajectory after leaving the track, in terms of VB and θ. Briefly explain how you arrived at your answer.

If this is all you need to find then there's no need for considerations of energy, work or whatever goes on at the curved track. It is simply a projectile motion question. What is the speed of a projectile at its highest point? Think in terms of the vertical and horizontal speeds
 
JeremyG said:
If this is all you need to find then there's no need for considerations of energy, work or whatever goes on at the curved track. It is simply a projectile motion question. What is the speed of a projectile at its highest point? Think in terms of the vertical and horizontal speeds
But how would I do that if I don't know any other variable? Like I don't know time or acceleration or displacement so how would I get the speed only in terms of VB and θ?
 
1. For projectile motion, neglecting air resistance, the horizontal component of the velocity remains constant. The only force acting on the projectile during this parabolic motion is its own weight, in the vertical direction.

2. At its highest point, what is its vertical and horizontal speed? You do not need time, acceleration nor displacement to come up with an expression of the velocity at the highest point.
 
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JeremyG said:
1. For projectile motion, neglecting air resistance, the horizontal component of the velocity remains constant. The only force acting on the projectile during this parabolic motion is its own weight, in the vertical direction.

2. At its highest point, what is its vertical and horizontal speed? You do not need time, acceleration nor displacement to come up with an expression of the velocity at the highest point.
Oh yes okay, I was definitely over thinking this, thank you!
 
No problem! Glad to help! :)
 

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