. A 0.145 kg baseball pitched at 35.0 m/s is hit on a horizontal line drive straight back at the pitcher at 56.0 m/s. If the contact time between the bat and ball is 5.00 × 10−3 ????, calculate the force (assumed to be constant) between the ball and bat.

Respuesta :

Answer:

2639 N

Explanation:

The impulse on the baseball is given by:

[tex]I=F \Delta t[/tex] (1)

where F is the force exerted on the ball and [tex]\Delta t[/tex] the contact time between the bat and the ball.

The impulse is also equal to the change in momentum of the ball:

[tex]I=\Delta p = m (v-u)[/tex] (2)

where m is the ball's mass, v is its final velocity, u is its initial velocity.

By using (1) and (2) simultaneously we can write an expression for F, the force exerted on the ball:

[tex]F=\frac{m(v-u)}{\Delta t}[/tex]

where:

m = 0.145 kg

u = 35.0 m/s

v = -56.0 m/s

[tex]\Delta t=5.00 \cdot 10^{-3}s[/tex]

Substituting,

[tex]F=\frac{(0.145 kg)((-56.0 m/s)-35.0 m/s)}{5.00\cdot 10^{-3} s}=-2639 N[/tex]

and the negative sign means that the force is simply in the opposite direction to the ball's initial direction.

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