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Question
A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J K−1.
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Solution
Height of the floor from which ball is dropped, h1 = 2.0 m
Height to which the ball rises after collision, h2 = 1.5 m
Let the mass of ball be m kg.
Let the speed of the ball when it falls from h1 and h2 be v1 and v2, respectively.
`nu_1=sqrt(2gh_1)=sqrt(2xx10xx2)=sqrt40"m/s"`
`nu_2=sqrt(2gh_2)=sqrt(2xx10xx1.5)=sqrt30"m/s"`
Change in kinetic energy is given by
`DeltaK=1/2xxmxx40-(1/2m)xx30=(10/2)m`
`rArrDeltaK=5"m"`
If the position of the ball is considered just before hitting the ground and after its first collision, then 40% of the change in its KE will give the change in thermal energy of the ball. At these positions, the PE of the ball is same. Thus,
Loss in PE = 0
The change in kinetic energy is utilised in increasing the temperature of the ball.
Let the change in temperature be ΔT. Then,
`(40/100)xxDeltaK=mxx800xxDeltaT`
`(40/100)xx10/2m=mxx800xxDeltaT`
`rArrDeltaT=1/400=0.0025`
`=2.5xx10^-3°C`
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