Elastic collisions of a sliding block

AI Thread Summary
In a discussion on elastic collisions, Block 1 with a mass m1 and initial speed of 3.6 m/s collides elastically with stationary Block 2, which has a mass of 0.69m1. After the collision, the final velocities calculated are v1f = 0.6604 m/s for Block 1 and v2f = 4.260 m/s for Block 2. To determine how far each block slides in a region with a coefficient of kinetic friction of 0.39, the deceleration due to friction can be calculated using the formula F = m*a = μ*m*g, leading to a = μ*g. The discussion emphasizes the importance of applying these principles to find the sliding distance for both blocks.
intenzxboi
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1. Homework Statement [/b]
Block 1, with mass m1 and speed 3.6 m/s, slides along an x-axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass m2 = 0.69m1. The two blocks then slide into a region where the coefficient of kinetic friction is 0.39; there they stop. How far into that region do (a) block 1 and (b) block 2 slide?



I used the elastic collision formula and got v1f= .6604m/s and v2f=4.260 m/s

can i use conservation of ke?? to find mass?
 
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intenzxboi said:
1. Homework Statement [/b]
Block 1, with mass m1 and speed 3.6 m/s, slides along an x-axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass m2 = 0.69m1. The two blocks then slide into a region where the coefficient of kinetic friction is 0.39; there they stop. How far into that region do (a) block 1 and (b) block 2 slide?

I used the elastic collision formula and got v1f= .6604m/s and v2f=4.260 m/s

can i use conservation of ke?? to find mass?

You can determine the distance by observing the deceleration of the friction for each block and applying it to the velocity.

F = m*a = μ*m*g

a = μ*g
 

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