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A tennis player receives a shot with the ball (0.060 0 kg) traveling horizontally at 20.0 m/s and returns the shot with the ball traveling horizontally at 40.0 m/s in the opposite direction. (a) What is the impulse delivered to the ball by the tennis racket? (b) Some work is done on the system of the ball and some energy appears in the ball as an increase in internal energy during the collision between the ball and the racket. What is the sum W 2 DEint for the ball?

Respuesta :

Answer:

a) 3.6 N.s

b) 36 J

Explanation:

Given:

The mass of ball, m = 0.06 kg

The initial velocity v₁ = -20 m/s

final velocity v₂ = 40 m/s  

Now,

The impulse  is given as:

I= change in momentum

or

I = m( v₂ - v₁)

or

I = 0.06 × (40 - (-20))

or

I = 3.6 N.s

(b) work done = change in kinetic energy

W.D = ΔK.E

W.D = [tex]\frac{1}{2}\times m\times(v_2^2-v_1^2)[/tex]

on substituting the values

W.D =  [tex]\frac{1}{2}\times 0.06\times(40^2-(-20)^2)[/tex]

W.D = 36 J

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