Question Number 23556 by Tinkutara last updated on 01/Nov/17
$$\mathrm{A}\:\mathrm{ball}\:\mathrm{falls}\:\mathrm{vertically}\:\mathrm{on}\:\mathrm{to}\:\mathrm{a}\:\mathrm{floor},\:\mathrm{with} \\ $$$$\mathrm{momentum}\:{p},\:\mathrm{and}\:\mathrm{then}\:\mathrm{bounces} \\ $$$$\mathrm{repeatedly},\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{restitution} \\ $$$$\mathrm{is}\:{e}.\:\mathrm{The}\:\mathrm{total}\:\mathrm{momentum}\:\mathrm{imparted}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{ball}\:\mathrm{to}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 01/Nov/17
![J_1 =m(ev_0 +v_0 ) J_2 =m(e^2 v_0 +ev_0 ) .. .. Σ_(i=1) ^∞ J_i =mv_0 (1+e)[1+e+e^2 +...] =(((1+e)/(1+e)))p .](Q23569.png)
$${J}_{\mathrm{1}} ={m}\left({ev}_{\mathrm{0}} +{v}_{\mathrm{0}} \right) \\ $$$${J}_{\mathrm{2}} ={m}\left({e}^{\mathrm{2}} {v}_{\mathrm{0}} +{ev}_{\mathrm{0}} \right) \\ $$$$.. \\ $$$$.. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{J}_{{i}} \:={mv}_{\mathrm{0}} \left(\mathrm{1}+{e}\right)\left[\mathrm{1}+{e}+{e}^{\mathrm{2}} +...\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\frac{\mathrm{1}+{e}}{\mathrm{1}+{e}}\right){p}\:. \\ $$
Commented by Tinkutara last updated on 01/Nov/17
$$\mathrm{But}\:\mathrm{why}\:\mathrm{not}\:{J}_{\mathrm{1}} ={mv}_{\mathrm{0}} \:\mathrm{only}? \\ $$
Commented by ajfour last updated on 01/Nov/17
![the ground did not just catch the ball, it even threw it back with speed ev_0 . ⇒ change in momentum brought about = m[ev_0 −(−v_0 )].](Q23572.png)
$${the}\:{ground}\:{did}\:{not}\:{just}\:{catch}\:{the} \\ $$$${ball},\:{it}\:{even}\:{threw}\:{it}\:{back}\:{with} \\ $$$${speed}\:{ev}_{\mathrm{0}} . \\ $$$$\Rightarrow\:{change}\:{in}\:{momentum} \\ $$$${brought}\:{about}\:=\:{m}\left[{ev}_{\mathrm{0}} −\left(−{v}_{\mathrm{0}} \right)\right]. \\ $$
Commented by Tinkutara last updated on 01/Nov/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$