Bouncing Balls and Rising Temperatures

Classical Mechanics Level 4

A steel ball is dropped from a height of 1m to a horizontal non-conducting surface. Every time it bounces, it reaches 80% of its previous height. Nearly by how much will the temperature of the ball rise after 4 bounces?

Details and Assumptions

  1. Specific heat capacity of the ball is 0.1 cal/g C 0.1 \text{cal/g C} .

  2. Neglect loss in heat to the surroundings and the floor.

1.4 0.59 0.059 0.014

Rajdeep Dhingra by Rajdeep Dhingra , 20, India

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1 solution

Jimmy Qin
Jan 18, 2016

The answer is 0.0014 C ∘ C^{\circ} , not 0.014 C ∘ C^{\circ} .

The loss in mechanical energy after four bounces is Δ U = m g ( h 0 − h f ) = m g h 0 ( 1 − ( 0.8 ) 4 ) . \Delta U = mg (h_0 - h_f) = mgh_0 (1-(0.8)^4). This mechanical energy is converted into heat: Δ U = m g h 0 ( 1 − ( 0.8 ) 4 ) = Q = m c Δ T . \Delta U = mgh_0 (1-(0.8)^4) = Q = mc\Delta T. Therefore Δ T = g h 0 ( 1 − 0. 8 4 ) c = ( 9.8 m / s 2 ) ( 1 m ) ( 1 − 0. 8 4 ) 4184 J / k g C ∘ = 0.0014 C ∘ . \Delta T = \frac{gh_0(1-0.8^4)}{c} = \frac{(9.8 m/s^2)(1 m)(1-0.8^4)}{4184 J/kg C^{\circ}} = 0.0014 C^{\circ}.

1 Helpful 1 Interesting 0 Brilliant 0 Confused

No value turns out be 0.014 after calculating . please recheck

Prakhar Bindal - 5 years, 4 months ago

I think you forgot to multiply g(taken as 10)

Shreyash Rai - 5 years, 4 months ago

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