How can kinetic energy and momentum be conserved in an elastic collision?

[Hypercubes]

Homework Statement

I must be overlooking something here. In an elastic collision, both momentum and kinetic energy are conserved, right? The math is not making sense...

Homework Equations

Momentum is conserved:
$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1} + m_{2}v_{2}$$

and kinetic energy is conserved:
$$\frac{m_1u_1^2}2+\frac{m_2u_2^2}2=\frac{m_1v_1^2}2+\frac{m_2v_2^2}2$$

The Attempt at a Solution

How is this possible? The two equations are not equivalent. If, for example, I solve for v2:
$$v_{2} = (m_{1}*u_{1} - m_{1}*v_{1} + m_{2}*u_{2})/m_{2}$$
with the first formula, but
$$v_{2} = \sqrt{(m_{1}*u_{1}^2/m_{2} - m_{1}*v_{1}^2/m_{2} + u_{2}^2)}$$ with the second.

Thanks in advance!

 

[cupid.callin]

obvoiudly the 2 eqn's are not same ... if they were same when what will be the use of 2 eqn's ? ...
if they were same ... you'll get(suppose) v = a+ 1 ... from eqn 1
and v = a+1 from 2 again ...

then how will they help solve your problem

like in math you get 2 eqn's ...
2a + 3b = 5
18a + 17b = 25 ... they will not give identical expression for a ... that's why you can solve for a,b

 

[Hypercubes]

obvoiudly the 2 eqn's are not same ... if they were same when what will be the use of 2 eqn's ? ...
if they were same ... you'll get(suppose) v = a+ 1 ... from eqn 1
and v = a+1 from 2 again ...

then how will they help solve your problem

like in math you get 2 eqn's ...
2a + 3b = 5
18a + 17b = 25 ... they will not give identical expression for a ... that's why you can solve for a,b

Well, often in physics equivalent equations are used for solving for different things.

Then which one should be used for elastic collisions?

 

[cupid.callin]

Well, often in physics equivalent equations are used for solving for different things.

like?

Then which one should be used for elastic collisions?

both ... in some cases one might be enough but in complex questions you need to use both

 

[Hypercubes]

I figured out the problem. There is only one value for which they are both equal, and this would only occur in a perfectly elastic collision. So for realistic collisions, the momentum one should be used.

 

[Riad]

-Simplify the problem by taking unit masses and cancelling the 1/2 factor.
v1+v2=u1+u2; v1^2+v2^2=u1^2+u2^2. Square the two sides of the first equation and subtract from the second and get; v1*v2=u1*u2 as a solution satisfying both conditions. This remains correct if the masses are not equal or the velocities are general vectors.
Of course we assume there is no change of energy into potential energy due to changes in position. This condition is satisfied only if the two come from far away and go back far away after their elastic collision.

 

[PeterO]

I figured out the problem. There is only one value for which they are both equal, and this would only occur in a perfectly elastic collision. So for realistic collisions, the momentum one should be used.

That is the way we work.

In EVERY collision, momentum is conserved. In elastic collisions energy is also conserved. If you are unsure of whether a collision is elastic - stick with the momentum equation.

Note, while the equations involved the same variables, The first one used the Vector velocities, where as the Energy equation only uses the speed - the magnitude of the velocity.

I one case, a change in direction is important - in the other it is not.

 

[azizlwl]

Homework Statement

I must be overlooking something here. In an elastic collision, both momentum and kinetic energy are conserved, right? The math is not making sense...

Homework Equations

Momentum is conserved:
$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1} + m_{2}v_{2}$$

and kinetic energy is conserved:
$$\frac{m_1u_1^2}2+\frac{m_2u_2^2}2=\frac{m_1v_1^2}2+\frac{m_2v_2^2}2$$

The Attempt at a Solution

How is this possible? The two equations are not equivalent. If, for example, I solve for v2:
$$v_{2} = (m_{1}*u_{1} - m_{1}*v_{1} + m_{2}*u_{2})/m_{2}$$
with the first formula, but
$$v_{2} = \sqrt{(m_{1}*u_{1}^2/m_{2} - m_{1}*v_{1}^2/m_{2} + u_{2}^2)}$$ with the second.

Thanks in advance!

$$m_{1}\vec u_{1}+m_{2}\vec u_{2}=m_{1}\vec v_{1} + m_{2}\vec v_{2}$$

$$\frac{m_1|u_1|^2}2+\frac{m_2|u_2|^2}2=\frac{m_1|v_1|^2}2+\frac{m_2|v_2|^2}2$$

One is vector and the other is scalar.
Velocity 5m/s going north is not equal to 5m/s going south.
For scalar both are identical.

$$\vec v_{2} \ne |v_{2}|$$