1. CHAPTER 6

2. Momentum- the product of the mass and velocity of an object. It is equal to In general the momentum of an object can be conceptually thought of as the tendency for an object to continue to move in its direction of travel. As such, it is a natural consequence of Newton's first law. Momentum is a conserved quantity, meaning that the total momentum of any closed system (one not affected by external forces) cannot be changed.

3. The amount of momentum that an object has depends on two physical quantities: the mass and the velocity of the moving object in the frame of reference. In physics, the symbol for momentum is usually denoted by a small p (bolded because it is a vector), so this can be written: where p is the momentum, m is the mass, and v the velocity. The velocity of an object is given by its speed and its direction. Because momentum depends on velocity, it too has a magnitude and a direction and is a vector quantity. For example the momentum of a 5-kg bowling ball would have to be described by the statement that it was moving westward at 2 m/s. It is insufficient to say that the ball has 10 kg m/s of momentum because momentum is not fully described unless its direction is given.

4. Practice Problem A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck? m= 2250 kg v= 25 m/s to the east p=? p=mv=(2250 kg)(25 m/s)= 56,000 kgm/s to the east Practice 6 a 1-3

5. A change in momentum takes force and time According to Newton's second law, force is equal to the change in momentum with respect to time We see F=ma which comes from the original form F=  p/  t = the change in momentum/change in time So with algebra we rearrange the equation to be F  t=  p this is the impulse momentum theorem The left side is known as the impulse. It is the force applied to a object in a given time interval that changes the momentum of the object. The common equation relating to force can be applied if the mass of the object is constant. Luckily, that scenario is extremely common. If a system is in equilibrium, then the change in momentum with respect to time is equal to 0

6. Practice Problem A 1400 kg car is moving west with a velocity of 15 m/s collides with a pole and stops in.30 s. Find the magnitude of the force exerted on the car during collision? F  t =  p = mvf – mvi F= mvf – mvi /  t = (1400kg)(0) – (1400kg)(15m/s) /.30 s =70,000 N to the east Problems 6 b 1-4

7. Stopping times and distances also depend on the impulse-momentum theorem Something that has more mass than it originally did must undergo a greater change in momentum in order to stop than it would with less mass. Also a change in momentum over a longer time requires less force Coming to a red light the fast your going the more time your going to need to slow down if you apply the brakes way in advanced you apply them easier and it takes less force. If you did it almost right at the light you have to slam on your brakes and apply a bigger force to stop.

8. Practice problem A 2250 kg car traveling west slows down uniformly from 20 m/s to 5 m/s. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car go during the deceleration? F  t =  p = mvf – m  t = mvf – mvi / F = (2250 kg)(-5m/s) – (2250kg)(20m/s)/8450kgm^2/s^2 = 4s d= ½(vi + vf)  t =1/2(-20 m/s-5 m/s)(4s) = -50 m = 50 m west Problems 6c 1-3