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Question-29400

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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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