Chapter 9 section1

Physics of Football-momentum

p =mv•If 2 objects have the same

velocity, the more massive object has greater momentum.•I f 2 objects have the same mass, the object moving faster has greater momentum.•If an object is at rest its momentum is zero.

Physics of football - impulse

F∆t = ∆p or F∆t =m(vf -vi)•Impulse = change in momentum

•Force needed to change the momentum can be decreased by increasing the time over which it acts.•Force can be reduced by increasing the area over which it acts.• A negative value for impulse means a loss in momentum. (The force is in the opposite direction of the motion.)

Impulse demonstration

a. p=mv v=100 km/hr *1000m/1km*1h/3600s = 27.8

m/sm=725 kg

p=(27.8 m/s)(725kg) = 2.01*104 kgm/s

b. 2.01*104 kgm/s= (2175kg) v v= 9.24 m/s

FΔt = Δpa. Δp= (5000N)(2.0s)= 1.0 *104 kgm/sb. 1.0 *104 N·s (or kgm/s) West (or put – in

front of answer)c. FΔt = Δp so FΔt = m(vf -vi) - 1.0 *104 kgm/s = 725 kg (vf – 27.8 m/s) - 1.0 *104 kgm/s = 725 kg vf -20155

kgm/s10155 kgm/s = 725 kg vf

vf= 14m/s (50km/h)

a. Δp = m(vf -vi)Δp = 240kg (28m/s - 6m/s)Δp =240 kg (22 m/s)Δp =5280kgm/s

b. Δp = impulse =5280kgm/s (or N·s)

c. impulse=FΔt and FΔt =ΔpF(60.0 s) =5280kgm/sF=88.0 N

FΔt =Δp so FΔt = m(vf -vi)(30.0 N) (0.16 s) = 0.115 kg (vf - 0)4.8 N·s = 0.115 kg vf

vf = 42m/s