Question

A soft tennis ball is dropped onto a hard floor from a height of 1.95 m and rebounds to a height of 1.55 m. (Assume that the positive direction is upward.) (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms. (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Answer 1

Answer:

a)$v=6.19m/s$

b)$v=5.51m/s$

c)$a=3.3*10^{3}m/s^{2}$

d)$x=5.78*10^{-3}m$

Explanation:

h1=195m

h2=1.55

a) Velocity just before the ball strikes the floor:

Conservation of the energy law

$E_{o}=E_{f}$

$E_{o}=mgh_{1}$

$E_{f}=1/2*mv^{2}$

so:

$v=\sqrt{2gh_{1}}=\sqrt{2*9.81*1.95}=6.19m/s$

b) Velocity just after the ball leaves the floor:

$E_{o}=E_{f}$

$E_{o}=1/2*mv^{2}$

$E_{f}=mgh_{2}$

so:

$v=\sqrt{2gh_{2}}=\sqrt{2*9.81*1.55}=5.51m/s$

c) Relation between Impulse, I, and momentum, p:

$I=\Delta p\\ F*t=m(v_{f}-v{o})\\ (ma)*t=m(v_{f}-v{o})\\\\ a=\frac{ v_{f}-v{o}}{t}=\frac{ 5.51- (-6.19)}{3.5*10^{-3}}=3.3*10^{3}m/s^{2}$

d) The compression of the ball:

The time elapsed between the ball touching the ground and it is fully compressed, is half the time the ball is in contact with the ground.

$t_{2}=t/2=3.5/2=1.75ms$

Kinematics equation:

$x(t)=v_{o}t+1/2*a*t_{2}^{2}$

Vo is the velocity when the ball strike the floor, we found it at a) 6.19m/s.

a, is the acceleration found at c) but we should to use it with a negative sense, because its direction is negative a Vo, a=-3.3*10^3

So:

$x=6.19*1.75*10^{-3}-1/2*3.3*10^{3}*(1.75*10^{-3})^2=5.78*10^{-3}m$