I have an image but I can't post it because it is a website.

block 1 of mass m1 slides along an x-axis on a frictionless floor with speed of v1i=4.00 m/s. Then it undergoes a one dimentional elastic collision with stationary block 2 of mass m2=.500m1. Next, block 2 undergoes a one dimentional elastic collision with stationary block 3 of mass m3=.500m2. (a) What then is the speed of block 3? Are (b) the speed, the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block1?

The lines, the "x" and numbers are kinda croocked, sorry.

a)We use the principle of conservation of momentum here.

Initially momentum is (m1)(4)
After collision with m2, it will be (v)(m1+0.5m1) = (v)(1.5m1)
After collision with m3, it will be (u)(1.5m1+0.25m1) = (u)(1.75m1)
this is bevause m3 = 0.5m2 = 0.5(0.5m1) = 0.25m1.

They must be equal so
(m1)(4) = (u)(1.75m1)
u = 2.3m/s roughly!

(b) Calculate the KE of m3. It is 1/2(2.3)(0.25m1)2
The KE of m1 is 1/2(m1)(42)

See which one is greater!

(d) They are asking the momentum of m3 and not total momentum. Right?This will be (0.25m1)(2.3) where as that of m1 is (m1)(4) so of course that of m1 is greater

Lets examine the math: 4m1=m1V1' + .5m1*v2'
1/2 16 m1=1/2 m1 v1'^2 + 1/2 .5 v2'^2
we are intersted in v2'.
v1'=4-.5v2'
V2'^2 =32-2v1^2= 16-2 (4-.5v2')^2= 16-2(16 + 1/4 v2'^2 -4v2')
V2'^2 =32-3 2 - 1/2 v2'^2+8v2'
1.5v2'^2=8v2' or v2'= 16/3 ..
Hmmm the second velocity has a higher velocity in the forward direction than the original ball.
But this is true. Experiment: drop two balls, one smaller. THe larger on the bottom, the smaller on top. The smaller will bounce very high.

Work this problem thru. The next collisioon will do the same. THe third ball will have much higher velocity.

Huh?

You need to redo your math. The third ball will have much more velocity than the first.

Explain Bot Explain Bot answered
8 months ago

To determine the speed of block 3, we can use the principle of conservation of momentum. The initial momentum of block 1 is given as (m1)(v1i), where m1 is the mass of block 1 and v1i is its initial speed.

After the collision with block 2, the total momentum will be (v)(m1 + 0.5m1), where v is the speed of block 2 after the collision. Since block 2 is stationary initially, its velocity is 0.

After the collision with block 3, the total momentum will be (u)(1.5m1 + 0.25m1), where u is the speed of block 3. We know that m3 = 0.5m2 = 0.5(0.5m1) = 0.25m1.

Setting the initial momentum equal to the final momentum, we have (m1)(v1i) = (u)(1.75m1).

Solving for u, we find that u = 2.3 m/s.

For part (b), we can compare the kinetic energy (KE) of block 3 to the KE of block 1. The KE of block 3 is given by 1/2(mu^2), where m is the mass of block 3 and u is its speed. The KE of block 1 is given by 1/2(m1)(v1i^2). We can calculate these values using the known velocities.

Finally, for part (d), we compare the momentum of block 3 to the momentum of block 1. The momentum of block 3 is given by (0.25m1)(u), while the momentum of block 1 is given by (m1)(v1i). We can compare these values to see if they are greater than, less than, or the same as the initial momentum of block 1.

It seems there was confusion in the previous response about the second calculation. The correct calculation should show that block 3 has a higher velocity than block 1. This can be demonstrated with experiments, such as dropping two balls of different sizes and observing the smaller one bouncing higher.

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