25. (30 points) A 0.3 5kg ball Is dropped from rest at a height of 8.30m. Every

Question

25. (30 points) A 0.3 5kg ball Is dropped from rest at a height of 8.30m. Every time It hits the floor, it loses 14% of Its energy. a.) How fast Is the ball going Just before It hits the ground the first time? b.) How fast does it rebound the first time? c.) What is the impulse delivered to the ball on the first bounce? d.) If the duration of this first bounce is 3.6ms, what is the average force on the ball during the first bounce? e.) How high does the ball get (max) after the third bounce? f.) Let?s call the force that robs the ball of its energy at each bounce the unelastic force. How much work has the unelastic force done after 10 bounces?

Explanation / Answer

(a) velocity of the ball before its the ground

      0.5 m v^2 = mgh

      v = sqrt ( 2gh) = sqrt ( 2 * 9.8 * 8.3) = 12.755 m/s

(b) It loses 14 % of energy therefore

     0.5 * m * vf^2 = 0.86 * 0.5 * m * v^2

     vf = sqrt(0.86) * v = 11.83 m/s

(c) Impulse = m* (v - vf) = 0.35 * (12.755 - 11.83) = 0.3243 kgm/s

(d) force = impulse / time = 0.3243 / 0.0036 = 90.07 N

(e) after 3 bounces

      vf = sqrt(0.86*0.86*0.86) * v = 10.172 m/s

      Height = vf^2 / 2g = 10.172^2 / 19.6 = 5.28 m

(f) work done by the unelastic force = loss in energy of the ball after 10 bounces

    Height after 10 bounces = 0.86^10 * 12.755 ^2 / 19.6 = 1.837 m

    loss in energy = work done = 0.35 * 9.8 * (8.3 - 1.837 ) = 22.168 J

    

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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