# 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.

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

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

## 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)

## 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.

## To find the final speed of the first 4-kilogram mass, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is given by the product of its mass and velocity.

Before the collision, the first 4-kilogram mass has a speed of 6 m/s, so its momentum is:

Momentum = mass × velocity
= 4 kg × 6 m/s
= 24 kg·m/s

The second 4-kilogram mass is initially at rest, so its momentum is zero.

After the head-on elastic collision, the momentum of the two masses is redistributed. However, since the system is isolated and no external forces are involved, the total momentum remains the same. Therefore, the total momentum after the collision is also 24 kg·m/s.

Now, let's consider the two masses that stick together after the collision.

Since both masses are identical and have the same mass of 4 kilograms, we can represent their combined mass as 8 kilograms.

Let the final speed of the two masses sticking together be v.

The momentum of the combined masses is equal to the product of the total mass and the final velocity:

Momentum = mass × velocity
= 8 kg × v
= 8v kg·m/s

According to the principle of conservation of momentum, the total momentum before the collision (24 kg·m/s) is equal to the total momentum after the collision (8v kg·m/s):

24 kg·m/s = 8v kg·m/s

Now, we can solve for v:

8v = 24
v = 24 / 8
v = 3 m/s

Therefore, the final speed of the first 4-kilogram mass is 3 m/s.

And the final speed of the two 4-kilogram masses that stick together is also 3 m/s.

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