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Irina-Kira [14]
3 years ago
13

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

es he exert on a golf ball if the club head and golf ball are in contact for 0.030 seconds?
Physics
1 answer:
Semenov [28]3 years ago
7 0

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

F=\frac{\Delta p}{\Delta t}

where

F is the force

\Delta p is the change in momentum

\Delta t=0.030 s is the time interval

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

v=281 km/h =78.1 m/s is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N

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