http://www.ck12.org Chapter 7. Momentum
pf=mvf=m(−vi) = ( 0 .145 kg)(− 46 .9 m/s) =− 6. 81
kg·m
s
The change in momentum of the ball is therefore:
∆p=pf−pi=− 6. 808 −(+ 6. 808 ) =− 13. 6 kgs·mor a magnitude of 13. 6 kgs·m
http://www.ck12.org Chapter 7. Momentum
pf=mvf=m(−vi) = ( 0 .145 kg)(− 46 .9 m/s) =− 6. 81
kg·m
s
The change in momentum of the ball is therefore:
∆p=pf−pi=− 6. 808 −(+ 6. 808 ) =− 13. 6 kgs·mor a magnitude of 13. 6 kgs·m