A ball of mass 0.400 kg is dropped from a height of 9.80 m to a concrete floor.

Question

A ball of mass 0.400 kg is dropped from a height of 9.80 m to a concrete floor. The ball rebounds to a height of 7.35 m. (a) Calculate the initial gravitational potential energy of the ball relative to the concrete floor. (b) State the kinetic energy of the ball as it reaches the concrete floor. (C) Calculate the total mechanical energy of the ball as it rebounds to 7.35 m. (d) Calculate the difference between the gravitational potential energy at this time and the kinetic energy you stated at (b) and account for this difference.

Explanation / Answer

(a) P =mgH

=0.4 * 9.81 * 9.8 =38.455 J

*(B) By conservation fo enrgy

total enrgy at height 9.8 = total enrgy at just before floor

K1 + P1 = K2 + P2

0 + mgH = K2 + 0

K.E. = mgH = 38.455 J

(C) Total mechanical enrgy at H1= 7.35

here V=0 m/s

Total mech enrgy = K.E. + P

=1/2 mV2 + mgH1   

= 0 = .4 * 9.81 * 7.35

=28.84 J

(d) Difference in enrgy = K.E. before collison to floor - mgH1

=38.455- 28.84 = 9.613 J

this differnce is loss of enrgy which is due to the inelastic collison at the floor .

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