Question

A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are released from rest at the same moment, to fall through a distance of 1.08 m, (a) Find the magnitude of the downward velocity with which the basketball reaches the ground. Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. (b) Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?

Answer 1

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

$u^2-u'^2 = 2as$

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

$u^2-0^2 = 2\times9.8\tiimes1.08$

or

$u = \sqrt{21.168}$

or

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed ($v_f$) = v = 4.6 m/s

Then conservation of energy says  

$\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2$  

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s  

u₂ = velocity of the basketball before collision= 4.6 m/s  

v₁ =  velocity of the tennis ball after collision  

v₂ = velocity of the basketball  after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

$v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2$

substituting the values in the above equation we get

$v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6$

or

$v = -8.565-3.965$

or

$v = -12.53m/s$

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

$K.E = \frac{1}{2}mv^2$

or

$K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2$

or

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

or

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m