Solve 1D Elastic Collision of Blocks: Speed of Block 3

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SUMMARY

The discussion focuses on solving a one-dimensional elastic collision problem involving three blocks with specified masses and initial velocities. Block 1, with mass m1, moves at 2.40 m/s and collides elastically with stationary Block 2 (mass m2 = 2.00m1), which then collides with Block 3 (mass m3 = 2.00m2). The calculated speed of Block 3 after the collisions is 1.78 m/s. The user struggles with the equations of motion and the relationships between speed, kinetic energy, and momentum across the blocks.

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elastic collision in one dimension
1. Homework Statement

Block 1 of mass m1 slides along an x-axis on a frictionless floor with a speed of 2.40 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 2.00m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 2.00m2.

what is the speed of block 3? the answer is 1.78.

i just can't figure it out.


more questions that relate to problem...are the speed, kinetic energy, and the momentum of block 3 greater than or less than, or th same as the initial values for block one?

the answers are less, less, greater...

but i don't need help with them because i haven't attempted them yet.

2. Homework Equations

v1f = m1 - m2 / m1 + m2 * v1i

v2f = 2m1 / m1 + m2 *v1i

3. The Attempt at a Solution

v1f = v2i = m1 - m2 / m1 + m2 * v1i
= m - 2m/ m + 2m *4
= -m/3m * 4
= -4/3

v2f = v3i = 2m1 / m1 + m2 * v1i
= 2m / m +2m *4
=2/3 *4
=8/3

v3f = m2 - m3 / m2 + m3 * v2i
= 2 - 4 / 2 + 4 * (-4/3)
= -2/6 (-4/3)
=-8/18

clearly this is wrong. i really did try a lot of different things. what am i doing wrong?

when i multiply (4/3) by (8/3) and divide by 2, i get the right answer! but that's just luck.
 
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