Speed of Toy Car on Curved Track: Exploring Kinetic Energy

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Homework Help Overview

The problem involves a toy car moving along a curved track, transitioning to projectile motion after leaving the track. The original poster seeks to determine the car's speed at the highest point of its trajectory, expressed in terms of its speed at point B and the angle of projection.

Discussion Character

  • Conceptual clarification, Problem interpretation, Mixed

Approaches and Questions Raised

  • The original poster attempts to relate potential and kinetic energy to find the speed at the highest point, expressing uncertainty about setting up the equations. Some participants suggest that the problem can be simplified to projectile motion, questioning the need for energy considerations.

Discussion Status

Participants are exploring different interpretations of the problem, with some suggesting that the focus should shift to the components of velocity in projectile motion. Guidance has been offered regarding the constancy of horizontal velocity and the role of vertical speed at the highest point.

Contextual Notes

There is a noted lack of information regarding time, acceleration, or displacement, which some participants mention as constraints in determining the speed solely in terms of VB and θ.

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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