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SUMMARY

The discussion focuses on solving a physics problem involving two air track gliders with masses of 300 g and 200 g, moving towards each other at speeds of 50 cm/s and 100 cm/s, respectively. For part (a), the velocities after an elastic collision can be determined using the conservation of momentum and kinetic energy equations. For part (b), the most inelastic collision scenario, where the gliders stick together, results in a final velocity of -1.0 m/s and a kinetic energy of 0.25 J. The problem is framed within a one-dimensional context, emphasizing the importance of both momentum and kinetic energy conservation in elastic collisions.

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Homework Statement


Two air track gliders of a mass 300 g and 200 g are moving towards each other in opposite directions with speeds of 50 cm/s and 100cm/s respectively. Take the direction of the more massive glider as positive.

A.) determine the velocity of each glider after the collision if the collision is elastic.

b.) The most inelastic collision would occur if the two gliders stuck together on impact. If this was the case, find the velocity of the pair after the collision, and the kinetic energy in standard units of "J".


Homework Equations


m1v1i +m2v2i = m1v1f + m2v2f


The Attempt at a Solution



a.) in order to find the unknown velocites you need to know at least one of the rebound velocities do you not?

0.3(0.50) + 0.2(-1.00) = 0.3v1f + 0.2v2f

b.)0.3(0.50) + 0.2(-1.00) = 0.5vf

vf = (0.15-0.2)/0.5
= -1.0 m/s

KE = 1/2mv^2

= 1/2(0.5)(-1.0)^2
= 0.25 J
 
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Is this elastic collision only one-dimensional?
 
physicskid72 said:

Homework Equations


m1v1i +m2v2i = m1v1f + m2v2f

The Attempt at a Solution



a.) in order to find the unknown velocites you need to know at least one of the rebound velocities do you not?

0.3(0.50) + 0.2(-1.00) = 0.3v1f + 0.2v2f

You have enough information to solve it (given that this is a 1 dimensional problem -- things can get more complicated if the objects can freely move in more than 1 dimension).

You have the conservation of momentum equation that you listed above. But since you are assuming it is a perfectly elastic collision, you can also use conservation of kinetic energy (conservation of kinetic energy only applies if the collision is perfectly elastic). So you have 2 equations and 2 unknowns, which is solvable.

b.)0.3(0.50) + 0.2(-1.00) = 0.5vf

vf = (0.15-0.2)/0.5
= -1.0 m/s

Try that last calculation again. I think you missed a decimal point somewhere.
 

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