A tennis player hits a 0.1-kilogram tennis ball with her racket from the north end of a court. Her racket was traveling at 65 kilometers per hour. The ball accelerated at a rate of 10 meters per second squared. The player on the south end of the court hit the tennis ball with her racket at 75 kilometers per hour. How much force did the racket on the south end of the court exert on the ball?(1 point) A.1 N B.6.5 N C.7.5 N D.0.01 N

Explanation:

Given that:

Mass of ball = 0.1-kilogram

From the North End:

Acceleration of racket = 65 km/hour

Converting to m/seconds

= 65×1000/3600

65000/3600

= 18.055m/s

Ball acceleration= 10 meters per second squared

From the South End:

Acceleration of racket = 75 km/hour

Converting to m/seconds

75×1000/3600

= 75000/3600

= 20.833m/s

Now,

18.055m/s = 10m/s^2

20.833m/s = x

18.055m/s × x = 20.833m/s × 10m/s^2

18.055m/s × x = 208.33

x = 208.33/18.055m/s

X = 11.538m/s^2

Hence, the ball accelerated at the rate of 11.538m/s^2 in the South End.

F = ma

Where a = 11.538m/s^2

m = 0.1kg

F = 0.1×11.538m/s^2

F = 1.15N

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can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s