A ball of mass m = 0.500 kg was dropped from an initial height 3.00 m above a ha

Question

A ball of mass m = 0.500 kg was dropped from an initial height 3.00 m above a hard floor. It rebounds to a maximum height h (less than 3.00 m). During the inelastic bounce, 11.6 J of energy are dissipated as heat and sound. The time the ball is in contact with the floor is 10.0 ms.

a) Find the upward speed of the ball just after it rebounds from the floor (i.e. at the height of the floor).
b) Obtain the numeric value of the height, h.
c)Determine the magnitude and the direction of the impulse delivered to the ball by the floor.

d) Determine the average force on the ball during the collision.

Explanation / Answer

velocity before collision u = (2gh) = (2*9.8*3) = 7.668 m/s

velocity after collision = v

(1/2)m(v^2 - u^2) = -11.6 J

0.5 * 0.5 * (v^2 - 7.668^2) = -11.6

v = 3.521 m/s

h = v^2/2g = 3.521^2 / (2*9.8) = 0.632 m/s

Impulse I= m(v-(-u)) = m(v+u) = 0.5*(3.521+7.668) = 5.5945 kg-m/s in upward direction

Ft = I

F = I/t = 5.5945/0.01 = 559.45 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.