Question

during the football match the ball shot towards the goal struck the defenders foot at the speed of 10 metre per second and it bounce back at 20 metre per second if the time of impact was 0.2 second and mass the ball is half kg then the average force exerted by the defender on ball is​

Answer 1

Answer:

in the question given that

v = 10 m/s

final v= 20m/s

time of impact T = 0.2$

To calculate the force applied by defender on football

1. We used the newton second law
2. F=ma
3. using equation of motion
4. v= u+at
5. start calculating, we get answer

by equation we get a = 50 m/s^2

as mass given half kg

so

applying

formula

we get

Force = 0.5×50=25 newton

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Answer 2

Answer:  

The average force exerted by the defender on ball is 75N

Explanation:

Given:

the initial velocity of the ball (u) = 10 $m/s$

the final velocity of the ball (v) = -20 $m/s$

('-' sign because its bounced back)

mass of the ball (m) = $\frac{1}{2} kg$

the time of impact (t) = 0.2 s

to find:

The average force exerted by the defender on ball

procedure:

force is equal to the change in momentum change in time i.e,

the momentum of the ball is P = mv

(where m is mass and v is velocity)

F = ΔP/ ΔT

change in momentum ΔP= $mv-mu$

change in time period ΔT= $t$

F =  $\frac{mv-mu}{t}$

F = $\frac{m(v-u)}{t}$ --- (1)

now substitute the given values in the equation (1)

F = $0.5 \frac{(-20-10)}{0.2}$

F = $0.5\frac{(-30)}{0.2}$

F = $\frac{-15}{0.2}$

F = - 75N

the average force exerted by the defender on ball is​ 75N

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