# Momentum and conservation of energy

1. Oct 21, 2005

### Werg22

I two object collide, and there is no loss of kinetic energy, the following relationship is true:

$${v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}$$

But also by conservation of energy,

$${v'_1^{2}}{m'}/2 + {v_1^{2}}{m}/2= {v'_1^{2}}{m'}/2 + {v_1^{2}}{m}/2$$

$${v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}$$

That means for any two mass, and velocities that share the relationship

$${v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}$$

$${v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}$$

which is obviously not true. Why is there a paradox?

Last edited: Oct 21, 2005
2. Oct 21, 2005

### Tide

Why do you think it is not true?

3. Oct 21, 2005

### Werg22

$${v'_1}{m'} + {v_1}{m}= {v'_2}{m'} + {v_2}{m}$$

$${v'_1}{m'} - {v'_2}{m'} = {v_2}{m} - {v_1}{m}$$

$$m'(v'_1 - v'_2) = m(v_2 - v_1)$$

Now with this relationship

$${v'_1^{2}}{m'} + {v_1^{2}}{m} = {v'_2^{2}}{m'} + {v_2^{2}}{m}$$

$$m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})$$

Let m' = 4, m = 4, v'_1 = 4, v_1 = -3

$$4(4 - v'_2) = 4(v_2 + 3)$$

$$4 - v'_2 = v_2 + 4$$

Let v'_2 be -2, so that v_2 = 3

$$m'(v'_1^{2} - v'_2^{2}) = m(v_2^{2} - v_1^{2})$$

$$4(4^{2} - (-2)^{2}) = 4(3^{2} - ((-3)^{2})$$

$$12 = 0$$

So it is not true

Last edited: Oct 21, 2005
4. Oct 21, 2005

### HallsofIvy

Staff Emeritus
No, you can't do that. If you have an elastic collision between two bodies, you can't select the final speed of either. Given your initial information:
both masses= 4, v'_1= 4, v_1= -3,
Then:
conservation of momentum: 4(4)+4(-3)= 4v'_2+ 4(v_2)
conservation of kinetic energy: 4(16)+ 4(9)= 4(v'_2)^2+ 4(v_2)^2.
Of course, you can just cancel all of the masses (4s):
v'_2+ v_2= 1 and v'_2^2+ v_2^2= 25.

Now, you have two equations in two unknowns.

5. Oct 21, 2005

### Werg22

Humm yes... they are consequent one of another so this makes more sense. Thank you!