Question

A tennis ball bouncing on a hard surface compresses and then rebounds. The details of the rebound are specified in tennis regulations. Tennis balls, to be acceptable for tournament play, must have a mass of 57.5 { \rm g}. When dropped from a height of 2.5 {\rm m} onto a concrete surface, a ball must rebound to a height of 1.4 {\rm m}. During impact, the ball compresses by approximately 6 {\rm mm}. How fast is the ball moving when it hits the concrete surface? (Ignore air resistance.) A. 5, B. 7, C. 25, D. 50m/s? If the ball accelerates uniformly when it hits the floor, what is its approximate acceleration as it comes to rest before rebounding? A. 1000, B. 2000, C. 3000, D. 4000m/s? By what percent does the kinetic energy decrease? When a tennis ball bounces from a racket, the ball loses approximately 30% of its kinetic energy to thermal energy. A ball that hits a racket at a speed of 10m/s will rebound with approximately what speed?

Answer 1

Answer:

B. 7 m/s

D. 4000 m/$s^{2}$

It loses 44% of its KE.

Its velocity after rebounding is 8.36 m/s

Explanation:

The equation of motion are as follows-

$v = u + at$

$s = ut +\frac{1}{2} at^{2}$

$v^{2} =u^{2} + 2as$

where, v= final velocity

            u= initial velocity

            s= displacement

             t= time

             a= acceleration  

The body falls from a height a of 2.5 m, so we will apply third law of motion,

u=0 ( as it falls from rest)

s=2.5 m

a= 9.8 m/$s^{2}$

$v^{2}=2 * a * s$

$v^{2} =2 * 9.8 * 2.5$

$v^{2}=49$

v=7 m/s

When the ball hits the ground it comes to rest after compressing 6mm hence bringing its velocity completely to rest. At that moment the ball unifromly accelerates to rest as given in the question.

Applying third equation again,

u=7 m/s

v=0

s=6mm= 0.006 m

a=??

$0=49 + 2* a *s$

$-49= 2 * a * 0.006$

$a=-\frac{49}{2*0.006}$

a=-4083.33 m/$s^{2}$≈-4000 m/$s^{2}$

It can go to the height of 1.4 m only ie. the potential energy it can gain is

m*g*h where h=1.4 m against the earlier height 2.5 m

As PE + KE = Total Energy

The difference in the above energy is the energy that the ball has lost.

Therefore the energy lost= mg$h_{1}$ - mg$h_{2}$

                                            =0.0575 × 9.8 × 1.1=0.61985 J

KE lost %= $\frac{mgh_{1} - mgh_{2}}{mgh_{1}} *100$

             = $\frac{1.1}{2.5} *100$

             = 44 %

It has lost 44 % of its kinetic energy.

KE= $\frac{1}{2} mv^{2}$

    Let the velocity after rebounding from the racket be x.

$\frac{1}{2} mx^{2}$=0.7 × $\frac{1}{2} mv^{2}$

As given in the question.

$x^{2} =0.7 v^{2}$

$x=0.836v$

x=8.36 m/s

The velocity of ball after rebounding is 8.36 m/s