Question

A tennis ball of mass 0.0600 kg is initially moving horizontally with a speed of 25.0 m/s. A tennis player hits it with his racket with an average force of 66 N, sending the ball back in the opposite direction with a speed of 35.0 m/s. For how much time was the racket and ball in contact, in units of seconds?

Answer 1

Let racket and ball are in contact for t time .

Mass of the ball m=0.0600 kg.

Initial velocity u=25.0 m/s.

The final velocity of the ball is v = - 35.0 m/s. [v is negative because the direction of v is in the negative x-axis and we took negative x-axis as negative in magnitude and positive x-axis as positive in magnitude]

We know from newton's law of motion force

  $F=\frac{mv-mu}{t}$

   $or\: \: \: -66=\frac{0.0600\times (-35.0)-0.0600\times (25.0)}{t}$

  $or\: \: \: -66=-\frac{0.0600\times (35.0)+0.0600\times (25.0)}{t}$

  $or\: \: \: t=\frac{0.0600\times (35.0)+0.0600\times (25.0)}{66}\: \: \: s$

  
s t = 0.0545

$\therefore$ For 0.0545 s was the racket and ball in contact.