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

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The discussion centers on solving a one-dimensional elastic collision problem involving three blocks with varying masses. Block 1, moving at 2.40 m/s, collides with stationary Block 2, which is twice its mass, followed by Block 2 colliding with stationary Block 3, also twice the mass of Block 2. The correct speed of Block 3 after the collisions is calculated to be 1.78 m/s. Participants express confusion over the calculations and the relationships between the speeds, kinetic energy, and momentum of the blocks, noting that Block 3's speed and kinetic energy are less than Block 1's, while its momentum is greater. The thread emphasizes the importance of following the correct equations for elastic collisions to arrive at the right solution.
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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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