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# Physics/Proving the law of conservation of momentum

Question
Hi there. How are you? I was shown on TV once how to prove the law of conservation of momentum I.e. momentum before = momentum after. The written proof involved trolleys, steps and explanations and equations. It also used Newton's Third Law of Motion. I wish I still had the darn thing. Would you be able to help me prove it in the same fashion at grade 12 level if its not too much trouble? If possible a hand written proof would do wonders. Its not a homework question. Its just me going back in time to grade 12 and keeping a printed document for my collections. My email address is petergoodwill(at)telkomsa.net if it needs to be hand written. This would be sincerely appreciated.

Regards

Peter

PS Have a wonderful festive season.

Hello Peter,

Sorry for the delay in answering. I can do that using this study of trolley A and trolley B. Let them have masses Ma & Mb and original velocities Ua and Ub. When they collide, trolley A exerts average force Fa on trolley B while trolley B exerts average force Fb on trolley A. The time of contact for both trolleys is time t.

Using Newton's 2nd law, we know that trolley A experiences acceleration
aA = Fb/Ma
(Because acceleration is usually a lower case a, I'll make the subscript, designating which trolley's acceleration it is, an upper case.)
That causes the velocity of trolley A to change to
Va = Ua + aA*t = Ua + Fb*t/Ma

Likewise, we know that trolley B experiences acceleration
aB = Fa/Mb
That causes the velocity of trolley B to change to
Vb = Ub + aB*t = Ub + Fa*t/Mb

The total original momentum of the 2 trolleys is Ma*Ua + Mb*Ub.
Their final momentum, after the collision, is Ma*(Ua + Fb*t/Ma) + Mb*(Ub + Fa*t/Mb).

If momentum is conserved, then we must be able to set the original and final momentum expressions equal to each other and show that is truly an equality. So
Ma*Ua + Mb*Ub =? Ma*(Ua + Fb*t/Ma) + Mb*(Ub + Fa*t/Mb)
Expanding the right side
Ma*Ua + Mb*Ub =? Ma*Ua + Ma*Fb*t/Ma + Mb*Ub + Mb*Fa*t/Mb
Transferring the Ma*Ua and Mb*Ub terms from right to left and simplifying the remainder of the right side
0 =? Fb*t + Fa*t

Obeying Newton's 3rd law, we realize that Fb = -Fa. Therefore -Fa can be substituted for Fb giving
0 =? -Fa*t + Fa*t
and indeed, 0 does truly equal -Fa*t + Fa*t

So using the basic kinematic formula Vf = Vo + a*t and Newton's 2nd and 3rd laws, we have shown that expressions for momentum before and after the collision that are equivalent, proving that momentum is conserved.

I hope this helps,
Steve

Physics

Volunteer

#### Steve Johnson

##### Expertise

I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

##### Experience

I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

Education/Credentials
BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University