Newton’s second law in terms of momentum 

• Linear momentum, p, is defined to be the product of an object’s mass m with its velocity v.

• p = mv

  linear momentum

• Its units are obtained directly from the formula and are kg m s-1.
  

  EXAMPLE: What is the linear momentum of a 4.0-gram NATO SS 109 bullet traveling at 950 m/s?

  SOLUTION:

  ·Convert grams to kg (jump 3 decimal places left) to get m = .004 kg.

  ·Then p = mv = (.004)(950) = 3.8 kg m s-1.

Newton’s second law in terms of momentum

• Fnet = ma = m((Dv)/(Dt)) = (mDv)/(Dt) = (Dp)/(Dt)
• This last is Newton’s second law in terms of change in momentum rather than mass and acceleration.
  

  Fnet=(Dp)/(Dt)

  Newton’s second law ( p-form)

• EXAMPLE: A 6kg object increases its speed from 5 ms-1 to 25 m s-1 in 30 s. What is the net force acting on it?

  SOLUTION:  

  Fnet=(Dp)/(Dt)  =m(v–u)/(Dt) 

  =6(25–5)/30  = 4 N.

Kinetic energy in terms of momentum

EXAMPLE: Show that kinetic energy can be calculated directly from the momentum using the following:

SOLUTION:  

• From p=mv we obtain v=p/m. Then

EK=1/2mv 2 =1/2 m(p/m)^2

=mp2/2m2

=p2/2m

Kinetic energy in terms of momentum

PRACTICE: What is the kinetic energy of a 4.0-gram NATO SS 109 bullet traveling at 950 m/s

and having a momentum of  3.8 kg m s-1?

SOLUTION: Start from scratch using  EK=1/2 mv 2 or you can use EK=p2/2m 

• Let’s use the new formula…

EK=p2/2m

                        =3.82/((2)(0.004) )

                        = 1800 J

Collisions

• A collision is an event in which a relatively strong force acts on two or more bodies for a relatively short time.
• The Meteor Crater in the state of Arizona was the first crater to be identified as an impact crater.
• Between 20,000 to 50,000 years ago, a small asteroid about 80 feet in diameter impacted the Earth and formed the crater.

• Consider two colliding pool balls…
  

  FYI    A system boundary is the “area of interest” used by physicists in the study of complex processes. ·A closed system has no work done on its parts by external forces

• If we take a close-up look at a collision between two bodies, we can plot the force acting on each mass during the collision vs. the time :

Impulse and force – time graphs

• • Although the force varies with time, we can simplify it by “averaging it out” as follows: 

• Imagine an ant farm (two sheets of glass with sand in between) filled with the sand in the shape of the above force curve:
• We now let the sand level itself out (by tapping or shaking the ant farm):
• The area of the rectangle is the same as the area under the original force vs. time curve.
• The average force Fis the height of this rectangle.
• We define a new quantity called impulseJ as the average force times the time.
• This amounts to the area under the force vs. time graph.
  

  J=F∆t             area under F vs. t graph

  impulse

  

• Since F=(Dp)/(Dt) we see that F∆t=∆p and so we can interpret the impulse as the change in momentum of the object during the collision.
  

  J=F∆t=∆p        = area under F vs. t graph

  impulse

  • It is well to point out here that during a collision there are two objects interacting with one another.
  • Because of Newton’s third law, the forces are equal but opposite so that F = – F.
• Thus for one object, the area (impulse or momentum change) is positive, while for the other object the area (impulse or momentum change) is negative
  

  FYI    ·Thus impulse can be positive or negative

• EXAMPLE:  A 0.140-kg baseball comes in at 40.0 m/s, strikes the bat, and goes back out at 50.0 m/s.  If the collision lasts 1.20 ms (a typical value),

  find the impulse imparted to the ball from the bat during the collision.

  SOLUTION: