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

[student35]

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.

 

[rl.bhat]

Follow munchy35's post.

 

[tomcue]

It seems like you are on the right track, but there may be some errors in your calculations. Let's break down the problem and equations to make it easier to understand.

First, we have two blocks colliding in an elastic collision in one dimension. This means that the total kinetic energy and momentum before and after the collision should be the same. This can be represented by the following equations:

Total kinetic energy before collision = Total kinetic energy after collision
Total momentum before collision = Total momentum after collision

Now, let's look at the equations you have listed. You have the correct equations for the final velocities of block 1 and block 2, but the equation for block 3 is incorrect. The correct equation should be:

v3f = m2 - m3 / m2 + m3 * v2i

Notice that the initial velocity for block 2 (v2i) is used in this equation, since block 3 is colliding with block 2 and not block 1.

Now, let's plug in the values given in the problem:

v1f = m1 - m2 / m1 + m2 * v1i
= m - 2m/ m + 2m * 2.40
= -m/3m * 2.40
= -0.8 m/s

v2f = 2m1 / m1 + m2 * v1i
= 2m / m +2m * 2.40
=2/3 * 2.40
=1.60 m/s

v3f = m2 - m3 / m2 + m3 * v2i
= 2 - 4 / 2 + 4 * 1.60
= -2/6 (1.60)
=-8/18
= -0.44 m/s

As you can see, the final velocity of block 3 is negative, indicating that it is moving in the opposite direction of the initial motion of block 1. This makes sense, as block 3 is colliding with block 2, which is moving in the opposite direction.

To answer the additional questions, we can use the equations for kinetic energy and momentum before and after the collision:

Initial kinetic energy = 1/2 * m1 * v1i^2 = 1/2 * m * (2.40)^2