Understanding Momentum and Kinetic Energy Conservation in Elastic Collisions

[joe789]

Hi, here's my problem: A 5 kg ball moving right at a velocity of 2 m/s on a frictionless table, collides head-on with a stationary 7.5 kg ball. Find the velocities of the balls if the collision is elastic.
I know you have to use momentum and kinetic energy conservation, then subsitute for one variable, but I don't understand how the equations should be rearranged.
Also, I don't understand how a system can lose kinetic energy without losing momentum, don't they both use the same units (the mass and velocity of the object), so how can one be reduced while the other stays the same?
Thanks for any help

 

[Diane_]

To your first question: find the initial momentum and total kinetic energy for the system. You'll end up with numbers, as you have all the requisite values. Then set up the equations for the final momentum and kinetic energy of the system. You'll have two variables - the final velocity of the first ball and the final velocity of the second ball. You'll also have two equations - the one for momentum and the one for kinetic energy. Two equations, two unknowns - the rest is just algebra.

As for the second - remember that momentum is linear with velocity but kinetic energy is quadratic. If you halve the velocity of one component, you'll halve its momentum, but you'll cut its kinetic energy to one-quarter. It's relatively easy to conserve momentum while not conserving kinetic energy.

 

[joe789]

To your first question: find the initial momentum and total kinetic energy for the system. You'll end up with numbers, as you have all the requisite values. Then set up the equations for the final momentum and kinetic energy of the system. You'll have two variables - the final velocity of the first ball and the final velocity of the second ball. You'll also have two equations - the one for momentum and the one for kinetic energy. Two equations, two unknowns - the rest is just algebra.
As for the second - remember that momentum is linear with velocity but kinetic energy is quadratic. If you halve the velocity of one component, you'll halve its momentum, but you'll cut its kinetic energy to one-quarter. It's relatively easy to conserve momentum while not conserving kinetic energy.

The algebra part is what I don't get


My textbook has a similar problem which they solve for you and the momentum and kinetic energy conservation equations end up being:


m1vf1 + m2vf2 = m1vo1 + 0


(1/2)m1vf1^2 + (1/2)m2vf2^2 = (1/2)m1vo1^2 + 0


Then without explaining how they did it, they rearrange the equations into:


vf1 = (m1-m2/m1+m2)vo1


vf2 = (2m1/m1+m2)vo1


I can solve the problem using these last two equations, but I don't understand how they rearranged the original momentum and kinetic energy equations into these last two equations.


About your answer to my second question, are you saying that there is some momentum loss, just that it's very small compared to the kinetic energy loss?

 

[Diane_]

If you're not sure where the equations come from, it's best not to use them. You never know when an assumption made for the derivation doesn't apply to the specific problem you're using.

The problem is complicated by the fact that your two equations are not of the same order, but that can be gotten around. Consider the following two equations:

x + y = 7

x^2 + y^2 = 25

Geometrically, you're looking for the intersection of a line with a circle. Algebraically, probably the easiest way to solve this is to solve the linear equation for one of the variables and substitute the result into the quadratic, viz:

x = 7 - y

x^2 + y^2 = 25

(7 - y)^2 + y^2 = 25

49 - 14y + y^2 + y^2 = 25

2y^2 - 14y + 49 = 25

2y^2 - 14y + 24 = 0

y^2 - 7y + 12 = 0

(y - 3)(y-4) = 0

y = 3 or y = 4

Therefore, x = 4 or x = 3. Note that the two solutions are exactly the same, but with the variables reversed. The same process should work for your problem, too.

As for the second part: almost. Remember - any momentum lost by one object is going to have to be gained by the other to preserve momentum conservation. Because of the way that velocity affects the two quantities differently, it's very easy to set things up so that corresponding changes in momentum lead to very different changes in kinetic energy. The vector nature of momentum verses the scalar nature of kinetic energy also comes into play.

This is not the best example, but perhaps it will do: consider the case of two lumps of clay, each of the same mass, heading directly towards each other at the same speed (but, obviously, different directions). The momentum of one will be mv. The momentum of the other will be -mv. Total momentum: 0. Total kinetic energy: mv^2. Now: suppose when they hit, they "mush" together. You end up with one object, and a perfectly inelastic collision. The new total momentum must still be 0, which means the new object (mass 2m) will have velocity 0. Change in momentum: 0. The new kinetic energy of the object will, obviously, also be 0. Change in kinetic energy: -mv^2.

Does that help at all?