Understanding the Relationship between Kinetic Energy and Momentum in Collisions

[stevo1]

I am having problems with the kinetic energy formula KE = 1/2 mv^2.
If an object of 1kg travels at a speed of 1ms its kinetic energy is 1/2 * 1 * 1^2 = 0.5J.
But if it collides with an object of 0.5kg which is stationary, to conserve momentum which is equal to mass * velocity = 1 * 1 = 1kgms the new speed of the 0.5kg object = momentum / mass = 1/0.5 = 2ms. The kinetic energy of this new object is equal to 1/2 * 0.5 * 2^2 = 1/4 * 4 = 1J. This is double the energy of the initial object which is impossible. It does not make sense to me that the faster an object travels a disproportionate amount of energy is required. I think that the origin of KE (1/2 mv^2), the integration of F=ma, is something that cannot be integrated in reality, only in theory, if momentum is to be conserved.
Can anyone shed any light on this?

Thanks
Stephen Lewis

 

[jcsd]

What you're doigng is describing what is called an inelastic collsion, in an inelsatic collsion both kinetic energy and momentum cannot be conserved here's the one-diemsnional proof

Imagine two objects of masses m_1 and m_2, traveling with velocities of v_1 and v_2 which after collding inelastically form a new object of m_3, traveling at velocity m_3, which due to the conservations of mass must equal m_1 + m_2

We can say this due to the conservation of momentum:

m_3v_3 = m_1v_1 + m_2v_2

therefore:

v_3 = \frac{m_1v_1 + m_2v_2}{m_3}

We can also say that due to the conservation of energy:

\frac{1}{2}m_3{v_3}^2 = \frac{1}{2}m_1{v_1}^2 + \frac{1}{2}m_2{v_2}^2

therefore:

v_3 = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}}

combing the equations we get:

\frac{m_1v_1 + m_2v_2}{m_3} = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}}

square and mutiply {m_3}^2 byboth sides,substitue in m_3 = m_1 + m_2 and mutiply out:

{m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + 2m_1m_2v_1v_2 = {m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + m_1m_2{v_1}^2 + m_1m_2v_2^2

Simply eliminate and you get:

2v_1v_2 = v_1^2 + v_2^2

Which can be re-arranged as:

v_1^2 - 2v_1v_2 + v_2^2 = 0

using the quadratic formula we can solve for v_1

And we find that:

v_1 = v_2


So for an inelastic collision the intial velcoties of the two colliding objects must be the same, hence no collision.

Therfore the nergy must take some other form rtaher than kinetic enrgy.

 

[Doc Al]

I am having problems with the kinetic energy formula KE = 1/2 mv^2.
If an object of 1kg travels at a speed of 1ms its kinetic energy is 1/2 * 1 * 1^2 = 0.5J.

OK.

But if it collides with an object of 0.5kg which is stationary, to conserve momentum which is equal to mass * velocity = 1 * 1 = 1kgms the new speed of the 0.5kg object = momentum / mass = 1/0.5 = 2ms.

Says who? During the collision the total momentum is conserved. In this case, the total momentum equals 1 kg-m/s. The situation you describe, where the struck object gets all the momentum, cannot happen--it would violate conservation of energy (assuming there is no energy source--like an explosive--involved). It could happen, but only if the two objects were the same mass.

The kinetic energy of this new object is equal to 1/2 * 0.5 * 2^2 = 1/4 * 4 = 1J. This is double the energy of the initial object which is impossible.

Right. It won't happen.

It does not make sense to me that the faster an object travels a disproportionate amount of energy is required. I think that the origin of KE (1/2 mv^2), the integration of F=ma, is something that cannot be integrated in reality, only in theory, if momentum is to be conserved.

It works just fine, done correctly.

 

[Doc Al]

super-elastic collision

One more note. Collisions are often classified as elastic or inelastic. In a purely elastic collision, the KE is conserved. In a more realistic inelastic collision, some of that energy is lost to thermal energy and deformation of the objects: so the total KE after the collision is less than what it was before the collision.

The example you gave--in which the total KE increased--would be a super-elastic collision. This would require an additional source of energy.

 

[chris_tams]

You need to consider Newton's coefficient of Restitution

e = Relative Speed of separation/Relative Speed of Approach

Where e is the the coefficient of resitution. This is used when there is a energy loss. i.e. the collision is inelastic.

Yes youth!

 

[stevo1]

Thanks to all those who replied.
Sorry for the long delay in this reply.

One final thing, could two objects possesses the same momentum but hold different kinetic energies, for example an object of 1kg mass traveling at 1ms, KE = 1/2 * 1 * 1^2 = 0.5J and momentum = 1 * 1 = 1kgms and an object of 0.5kg mass traveling at 2ms KE = 1/2 * 0.5 * 2^2 = 1J and momentum = 0.5 * 2 = 1kgms.
Is this possible? If so, can you explain why?

Many thanks

Stephen Lewis

 

[Doc Al]

One final thing, could two objects possesses the same momentum but hold different kinetic energies,...
...
Is this possible? If so, can you explain why?

It's not only possible, but if the objects have different masses it's inevitable. Momentum and kinetic energy are two entirely different concepts.

 

[stevo1]

Thanks Doc Al, very interesting.