and the final momentum is Gpf=⋅7400kg m s i.bg#The impulse on it equals the change in its momentum: ()3ˆˆ7.4 10 N s ij .fiJpp=−=×⋅−GGG(b) The initial momentum of the car is Gpi7400kg m s ibg#and the final momentum is Gpf=0. The impulse acting on it is 3ˆ( 7.4 10 N s)i.p=−×⋅G(c) The average force on the car is GGGFptJtavgkg m s ij4.6sNi j===⋅−ΔΔΔ74001600bgejbgej##and its magnitude is 3avg1600N22.3 10 N.F==×(d) The average force is GGFJtavgkg m s isNi−⋅×=− ×−Δ7400350 1021 1034b
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.