Introduction to Basic Physics (Mechanics) Pt. 6

We now proceed to Newton's 3rd law of motion, which is stated simply as:

If 2 bodies interact, the force that body 1 exerts on body 2 is equal and opposite to the force that body 2 exerts on body 1. E.g. (F(12) = F(21) = - F(12)).

As a simple static demonstration consider the diagram and an object resting on a surface (Fig. 1). This could be a simple block (A) of mass m resting on a surface, B. The forces are equal and opposite so that we must have (for the third law to hold):

F(AB) = - F(AB) = F (BA)

Here: F(AB) = m(A)g (the weight of block A) and F(BA) is the normal force, N acting against it.

Exercise:

Re-design the linear air track experiment (Pt.1) to show a kinetic (as opposed to static) demonstration of Newton’s 3rd law. You can use any additional equipment you want, simply show or explain how it is incorporated into the existing experiment, and what results you will expect and how they can be measured. (Hint: you will want two different sets of timing tape, two timers.)

Based on at least 10 trials, prepare a data table and show that Newton's Third law holds.


Problem:

The diagram provided (lower graphic) is to be used to arrive at a validation of Newton’s 2nd and 3rd laws from an experiment in which two trolleys travel toward each other on a frictionless surface. Trolley m1 (with mass m1) travels at velocity u1, and trolley m2 travels at velocity u2:

a) Using Newton’s 2nd law show the applied resultant forces for each mass, m1a1 and m2a2 if m1 accelerates to velocity v1, and m2 accelerates to velocity v2. Assume each acceleration or increase in velocity occurs in the same time interval, delta t.

b)Show by Newton’s 3rd law that on collision, the forces exerted by the masses are equal and opposite.

c)Hence or otherwise, show the total momentum before the collision is the same as the total momentum after collision (take the momentum to be the product of the mass times the velocity).


Solution for part (a):

a) The respective resultant forces are:

F1 = m1(v1 - u1)/(delta t) and F2 = m2(v2 - u2)/(delta t)

Then: a1 = (v1 - u1)/(delta t) and a2 = (v2 - u2)/(delta t)

Therefore F1 = m1a1 and F2 = m2a2


Readers are invited to work out parts (b) and (c).