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a 60.0 grams tennis ball approaches a racket at 30.0m/s and is in contact with the racket strings for 5.0 milliseconds then rebounds at 30m/s what was the average force exerted on the ball ?

Sagot :

Indrawboyidn

The average force exerted on the ball : 720 N

Further explanation

Impulse: the product of the average net force on an object and time interval

Can be symbolized: J

The formula:

[tex]\rm J=F_{average}.\Delta T[/tex]

J = F average  (t2-t1)

Impulses also show changes in momentum , so they can be combined into

F • Δ t = m • Δ v.

So the forces acting on objects are:

[tex]\rm F=\dfrac{m.\Delta v}{\Delta t}[/tex]

Given

m tennis ball = 60 gr = 0.06 kg

Δt = 5.0 ms = 5 x 10⁻³ s

since force is a vector and time is a scalar, then we assume the direction of the racket that hits the tennis ball is positive, then the direction v1 = vi = negative and v2 = vf = positive

v1 = v approaching = vi(initial velocity) = -30 m / s

v2 = v away = vf(final velocity) = + 30 m / s

Δ v = vf-vi = 30 - (-30) = 60 m / s

The average force :

[tex]\rm F=\dfrac{0.06.60}{5.10^{-3}}\\\\F=\boxed{\bold{720~N}}[/tex]

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