A 4-kilogram mass has a speed of 6 m/s on a horizontal frictionless surface. The mass collides head-on and elastically with an identical 4-kilogram mass initially at rest. The second 4-kilogram mass then collides head-on and sticks to a third 4-kilogram mass initially at rest.

Question

The final speed of the first 4-kilogram mass is...

The final speed of the two 4-kilogram masses that stick together is...

Answer 1

If they collide elastically the sum of 1/2 m v^2 after has to equal the sum of 1/2 m v^2 s before and the momentums (m1v1+m2v2) must be the same before and after.
This can only happen if the first one stops and the second one proceeds at 6 m/s.
I will leave you to prove that to yourself using those energy and momentum equalities.
Now the second part.
We have no energy conservation now because it is not elastic.
momentum is still conserved though and the two stick so they become one 8 kg mass after
so
4 (6) + 4(0) = 8 (v final)

Answer 2

so for the first part... the initial m1v1 is (4kg)(6m/s). the final is what i can't get. is the final 12kg? arent we only concerned with the first mass in that question?

Answer 3

In the first part of the problem
a 4 kg mass moving at 6 m/s hits another 4 kg mass
that means
4 kg * 6 m/s + 4 kg* 0 m/s = initial momentum
the final momentum is
4 kg * v1 + 4 kg*v2
so

24 + 0 = 4v1 + 4 v2
or
v1 = 6 - v2

then energy
initial ke = (1/2) m v^2 = 2*36 = 72 Joules
final ke is also 72 joules
so
72 = 2 v1^2 + 2 v2^2
36 = (6-v2)^2 + v2^2
36 = 36 - 12 v2 + 2 v2^2
v2^2 - 6 v2 = 0
v2(v2-6) = 0
so v2 = 0 or v2 = 6
v2 may not equal zero unless mass 1 goes straight through mass 2
so
v2 = 6
(like I tried to tell you :) The first one stops and the second one takes over at the original speed.

Answer 4

In the second part of the problem, the 4 kg mass suddenly doubles in mass, so it has to go half as fast to have the same momentum

Answer 5

That is it, ten before 11, I am turning into a pumpkin ! Or at least I am turning into sleep.

Answer 6

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