Question #191160

In a ball game a 70 kg player is running at 10 m/s when he is hit by the other player he bounces off in the opposite direction at 2 m/s. What is the players change in momentum?


1
Expert's answer
2021-05-12T11:28:19-0400

before collision:\text{before collision:}

m=70;V1=10;m = 70;\vec{V_1}=10;

p1=mV1=1070=700\vec p_1=m*\vec{V_1} = 10*70 = 700

after collision:\text{after collision:}

m=70;V2=2;m = 70;\vec{V_2}=-2;

p2=mV2=270=140\vec p_2=m*\vec{V_2} = -2*70 = -140

Δp=p2p1=140700=840kgms\Delta p= \vec p_2-\vec p_1= -140-700 =-840kg*\frac{m}{s}

Answer: Δp=840kgms\text{Answer: }\Delta p=-840kg*\frac{m}{s}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS