Question #248231

8. A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when

it is struck by a club. If the club and ball are in contact for 2.00 ms, what

average force acts on the ball? Is the effect of the ball’s weight during

the time of contact significant? Why or why not?


1
Expert's answer
2021-10-10T16:00:28-0400

Given:

m=0.0450kgm=0.0450\:\rm kg

vi=0v_i=0

vf=25.0m/sv_f=25.0\:\rm m/s

Δt=0.00200m/s\Delta t=0.00200\:\rm m/s


The average force

F=mΔvΔt=0.045025.00.00200=563NF=m\frac{\Delta v}{\Delta t}=0.0450*\frac{25.0}{0.00200}=563\:\rm N

The weight of the ball

W=mg=0.04509.81=0.44NW=mg=0.0450*9.81=0.44\:\rm N

Since WFW\ll F, so the weight isn't significant.


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