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In a little league baseball game, the 145 grams ball enters the strike zone with a speed of 11.0 meters per second. The batter hits the ball and it leaves his bat with a speed of 25.0 meters per second in exactly the opposite direction. If the bat is in contact with the ball for 1.0 m/s, what is the magnitude of the average force exerted by the bat on the ball?​

Sagot :

hamzaahmeds

Answer:

Force = 5.22 N

Explanation:

According to Newton's Second Law of motion:

[tex]Force = Rate\ of\ Change\ of\ Momentum\\\\Force = \frac{mv_f-mv_i}{t}\\[/tex]

where,

m = mass of ball = 145 g = 0.145 kg

vf = final speed of ball after hit = 25 m/s

vi = initial speed of ball before hit = -  11 m/s (negative sign due to opposite direction)

t = time of contact = 1 s

Therefore,

[tex]Force = \frac{(0.145\ kg)(25\ m/s)-(0.145\ kg)(-11\ m/s)}{1\ s} \\\\[/tex]

Force = 5.22 N

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