A ball of mass m is droped from a height h , losing x% of its energy(that it had before the respective drop) on each fall.Calculate how many times the ball will fall until it stops and how long it will take the ball to do that.(?)

(the ball bounces up and down)

1 Answer
A08
May 22, 2018

This is what I get

Explanation:

Initial potential energy of ball #=mgh#, where #g# is acceleration due to gravity
Energy lost in the first bounce#=mghx/100#
#=># Energy remaining after first bounce#=mgh(1-x/100)#
Therefore, for the second drop height #=h(1-x/100)# ......(1)

#=># for each subsequent drop height would be multiplied by a factor #=(1-x/100)#

To calculate time #t_1# taken for first drop we use of the kinematic expression

#h=ut+1/2g t^2#

Assuming that the ball is dropped with initial velocity #=0#, we get

#h=0xxt_1+1/2g t_1^2#
#=>t_1=sqrt((2h)/g)# .........(2)

From (1) we get time #t_2# for the second drop as

#t_2=sqrt((2h(1-x/100))/g)#
#=>t_2=t_1(1-x/100)^(1/2)# ........(3)

For third drop as #t_3#

#=>t_3=t_1(1-x/100)^(2/2)# .......(4)

Theoretically ball will bounce #oo# times before it comes to stop. Total time taken by the ball to come to stop is sum of infinite series as given below

#t_1(1+(1-x/100)^(1/2)+(1-x/100)^(2/2)+(1-x/100)^(3/2).....)#

Now sum of infinite GP is given by the expression as

#S_(oo)=a_1/(1-r)#
where #a_1# is the first term and #r# is common ratio and #|r|<1#

Hence we get total time taken by the ball to come to rest

#T=t_1xx 1/(1-(1-x/100)^(1/2))#
#T=sqrt((2h)/g)xx1/(1-(1-x/100)^(1/2))#

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