Revision Topic:              Momentum

 

Important Points:         Momentum is a vector Þ Setting a positive direction is important.

                                      Rate of change of momentum = Force (note: N2Law)

                                      Only momentum (of a system) is conserved in both Elastic and Inelastic Collision.

 

 

This is a common example. Assume collision is elastic.

 

First, take the direction to the right as positive (important).

 

Therefore,    initial momentum = p_i = + m v

final momentum = p_f = - m v

 

Change in momentum       = p_f - p_i

                                      = (- m v) - (+ m v)

                                      = - 2 m v

 

Notes:

~ Change in momentum is a vector.

~ The - ve sign for the change in momentum indicates “to the left”.

~ By convention, we always use final momentum minus the initial momentum.

 

 

This is the same example as the previous one.

 

First, take the horizontal direction to the right as positive (important).

 

Therefore,    initial momentum = p_i = + m v cos q

final momentum = p_f = - m v cos q

 

Change in momentum       = p_f - p_i

                                      = (- m v cos q) - (+ m v cos q)

                                      = - 2 m v cos q

 

Notes:

 

Note that the examples used above involves stationary walls.

 

Let’s consider a Thermodynamics problem where a ideal gas in

an insulated container is allowed to expand using a piston as shown.

 

Using the First Law of Thermodynamics (DU = DQ + DW), it

is easily shown that there will be a negative increase (DU = -ve)

in internal energy and therefore, the temperature of the gas

will increase (U µ T).

 

But when explaining in molecular terms we need to explain how the c_rms of the gas decreased (c_rms µ ÖT).

In this case, we need to point out that when the molecules collide with the moving wall (the red molecule in the diagram), they will bounce of with a slower velocity. There is no contradiction in conservation of momentum here, because relative to the moving wall, the red molecules’ velocity is still the same.

 

Let’s consider a TYS problem J96/III/1.

 

The compressed air in the toy rocket forces out the water through

the nozzle at the base of the rocket (because the compressed air has a

higher pressure than the surrounding atmospheric pressure).

 

At a particular instant, the velocity of the water being forced out is v.

 

To conserve the momentum of the system, the rocket will be thrusted

upwards (N2Law).

 

It’s also correct to use N3Law. Since the rocket exerts a force on the

water downwards, the action reaction pair to this force is the force

exerted by the water upwards. The latter is known as the thrust

experienced by the rocket.

 

At this instant, the change of momentum of the ejected water per second is

= p_f - p_i

= mv - 0

= r_w Av²

where m is mass of water ejected per second (m = r_wV = r_w Av).

 

Sine this the change of momentum per second (N2Law), this is the force exerted by the rocket on the water. And in turn, this is the magnitude of the force by the water on the rocket, the thrust.

 

To find the instantaneous acceleration, use

 

Thrust - M_{(rocket+leftover water)}g = M_{(rocket+leftover water)}a

 

Notes:

~ In cases involving a continuous flow of matter (water, sand, air etc), it’s always a good idea to consider how much mass is moving for every one second.

 

More examples will be included in future updates if time permits.