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Question-29400




Question Number 29400 by ajfour last updated on 08/Feb/18
Commented by ajfour last updated on 08/Feb/18
Find b in terms of a,u,𝛂, and e .  (e being the coefficient of  restitution)
$${Find}\:\boldsymbol{{b}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}},\boldsymbol{{u}},\boldsymbol{\alpha},\:{and}\:\boldsymbol{{e}}\:. \\ $$$$\left(\boldsymbol{{e}}\:{being}\:{the}\:{coefficient}\:{of}\right. \\ $$$$\left.{restitution}\right)\: \\ $$
Answered by 33 last updated on 08/Feb/18
a = (u cos α ) t_1   b = (eu cos α ) t_2   t_(1 ) + t_(2  ) = t  (a/(u cos α )) + (b/(eu cos α )) = ((2u sin α)/g)  ⇒ b = (  ((e u^2  sin 2α)/g) − ae )
$${a}\:=\:\left({u}\:{cos}\:\alpha\:\right)\:{t}_{\mathrm{1}} \\ $$$${b}\:=\:\left({eu}\:{cos}\:\alpha\:\right)\:{t}_{\mathrm{2}} \\ $$$${t}_{\mathrm{1}\:} +\:{t}_{\mathrm{2}\:\:} =\:{t} \\ $$$$\frac{{a}}{{u}\:{cos}\:\alpha\:}\:+\:\frac{{b}}{{eu}\:{cos}\:\alpha\:}\:=\:\frac{\mathrm{2}{u}\:{sin}\:\alpha}{{g}} \\ $$$$\Rightarrow\:{b}\:=\:\left(\:\:\frac{{e}\:{u}^{\mathrm{2}} \:{sin}\:\mathrm{2}\alpha}{{g}}\:−\:{ae}\:\right) \\ $$
Commented by ajfour last updated on 08/Feb/18
Thanks.
$${Thanks}. \\ $$
Answered by mrW2 last updated on 08/Feb/18
Commented by mrW2 last updated on 08/Feb/18
L=((u^2  sin 2α)/g)  b′=L−a  b=eb′=e(L−a)=e(((u^2  sin 2α)/g)−a)  the wall acts like a mirror.
$${L}=\frac{{u}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\alpha}{{g}} \\ $$$${b}'={L}−{a} \\ $$$${b}={eb}'={e}\left({L}−{a}\right)={e}\left(\frac{{u}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\alpha}{{g}}−{a}\right) \\ $$$${the}\:{wall}\:{acts}\:{like}\:{a}\:{mirror}. \\ $$
Commented by 33 last updated on 09/Feb/18
but sir how can we use the eq of  projectile motion when we know  that the path from A to B is not  at all parabolic.  Sir, i do not think it can be done  in this way tho you got the same  snswer.( no offence)
$${but}\:{sir}\:{how}\:{can}\:{we}\:{use}\:{the}\:{eq}\:{of} \\ $$$${projectile}\:{motion}\:{when}\:{we}\:{know} \\ $$$${that}\:{the}\:{path}\:{from}\:{A}\:{to}\:{B}\:{is}\:{not} \\ $$$${at}\:{all}\:{parabolic}. \\ $$$${Sir},\:{i}\:{do}\:{not}\:{think}\:{it}\:{can}\:{be}\:{done} \\ $$$${in}\:{this}\:{way}\:{tho}\:{you}\:{got}\:{the}\:{same} \\ $$$${snswer}.\left(\:{no}\:{offence}\right) \\ $$
Commented by mrW2 last updated on 09/Feb/18
To get the result that b=eb′, we even  don′t need to know any equation.    Assume the ball hits the wall at  point D.  Let′s image that the ball becomes two  balls after the hit. One ball continues the  original motion to the right as if the  wall doesn′t exist. The other ball is  reflected by the wall and goes to the  left. Both balls have the same   vertical speed at point D, but different   horizontal speed. Since they have  the same vertical start speed, they  need the same time to hit the ground.  And the horizontal distance each ball   covers in this time is proportional  to its horizontal speed at point D.  The ball to the right follows the path  DB with horizontal distance b′ and  the ball to the left follows the path  DC with horizontal distance b. Since  the horizontal speed of the ball to the  left is e times of the horizontal speed  of the ball to the right, therefore  b=eb′.  So I think my method is correct  and its result is correct not just by  accident.  Please think about it again sir.
$${To}\:{get}\:{the}\:{result}\:{that}\:{b}={eb}',\:{we}\:{even} \\ $$$${don}'{t}\:{need}\:{to}\:{know}\:{any}\:{equation}. \\ $$$$ \\ $$$${Assume}\:{the}\:{ball}\:{hits}\:{the}\:{wall}\:{at} \\ $$$${point}\:{D}. \\ $$$${Let}'{s}\:{image}\:{that}\:{the}\:{ball}\:{becomes}\:{two} \\ $$$${balls}\:{after}\:{the}\:{hit}.\:{One}\:{ball}\:{continues}\:{the} \\ $$$${original}\:{motion}\:{to}\:{the}\:{right}\:{as}\:{if}\:{the} \\ $$$${wall}\:{doesn}'{t}\:{exist}.\:{The}\:{other}\:{ball}\:{is} \\ $$$${reflected}\:{by}\:{the}\:{wall}\:{and}\:{goes}\:{to}\:{the} \\ $$$${left}.\:{Both}\:{balls}\:{have}\:{the}\:{same}\: \\ $$$${vertical}\:{speed}\:{at}\:{point}\:{D},\:{but}\:{different}\: \\ $$$${horizontal}\:{speed}.\:{Since}\:{they}\:{have} \\ $$$${the}\:{same}\:{vertical}\:{start}\:{speed},\:{they} \\ $$$${need}\:{the}\:{same}\:{time}\:{to}\:{hit}\:{the}\:{ground}. \\ $$$${And}\:{the}\:{horizontal}\:{distance}\:{each}\:{ball}\: \\ $$$${covers}\:{in}\:{this}\:{time}\:{is}\:{proportional} \\ $$$${to}\:{its}\:{horizontal}\:{speed}\:{at}\:{point}\:{D}. \\ $$$${The}\:{ball}\:{to}\:{the}\:{right}\:{follows}\:{the}\:{path} \\ $$$${DB}\:{with}\:{horizontal}\:{distance}\:{b}'\:{and} \\ $$$${the}\:{ball}\:{to}\:{the}\:{left}\:{follows}\:{the}\:{path} \\ $$$${DC}\:{with}\:{horizontal}\:{distance}\:{b}.\:{Since} \\ $$$${the}\:{horizontal}\:{speed}\:{of}\:{the}\:{ball}\:{to}\:{the} \\ $$$${left}\:{is}\:{e}\:{times}\:{of}\:{the}\:{horizontal}\:{speed} \\ $$$${of}\:{the}\:{ball}\:{to}\:{the}\:{right},\:{therefore} \\ $$$${b}={eb}'. \\ $$$${So}\:{I}\:{think}\:{my}\:{method}\:{is}\:{correct} \\ $$$${and}\:{its}\:{result}\:{is}\:{correct}\:{not}\:{just}\:{by} \\ $$$${accident}. \\ $$$${Please}\:{think}\:{about}\:{it}\:{again}\:{sir}. \\ $$
Commented by 33 last updated on 10/Feb/18
yes sir you are right. it be done  that way. rather its better method  and encourages higher order  thinking. Excellent !
$${yes}\:{sir}\:{you}\:{are}\:{right}.\:{it}\:{be}\:{done} \\ $$$${that}\:{way}.\:{rather}\:{its}\:{better}\:{method} \\ $$$${and}\:{encourages}\:{higher}\:{order} \\ $$$${thinking}.\:{Excellent}\:! \\ $$
Commented by mrW2 last updated on 10/Feb/18
Thank you for checking!
$${Thank}\:{you}\:{for}\:{checking}! \\ $$
Commented by 33 last updated on 10/Feb/18
no problem sir
$${no}\:{problem}\:{sir} \\ $$

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