Question

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 52.0 m/s and returns the shot with the ball traveling horizontally at 32.0 m/s in the opposite direction. (Assume the initial direction of the ball is in the -x direction.)

(a) What is the impulse delivered to the ball by the racquet?
?

(b) What work does the racquet do on the ball?

How many Joules?

Answer 1

mass of the ball m = 0.06 Kg

v1 = -52 m/s

v2 = 32 m/s

(a)

The impulse delivered to the ball = change in momentum

                                                 = m(v2 - v1)

                                                 = 0.06(32-(-52))

                                                 = 0.06(32+52)

                                                 = 0.06*84

                                                 = 5.04 Kgm/s   

(b)

work done on the ball = change in it's kinetic energy

                                 = (m/2)(v1² - v2²)  

                                = 0.03*[2704 - 1024]

                                 = 50.4 J  

Answer 2

given that

      mass of the ball m = 0.0600kg

   intial velocity v_i = -52 m/s

final velocity v_f = 32 m/s

impulse

             FΔt = Δp

                   = m( v_f - ( -v_i)

                    = 5.04 N.s

(b) wor done is change in kinetic energy

               w = ΔKE

                   = ( 1/2) m ( v_f² - v_i²)

                  = -50.4J

negative sign means rocket is supplying work to the ball

Answer 3

Impulse = Change in momentum.

J = Δp
J = mΔv
J = .06(-32 - 52)
J = -5.04 N*s

Work = ΔKE
= (m/2)(v12 - v22)
= 0.03*[2704 - 1024]
= 50.4 J