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Question Number 24178 by NECx last updated on 13/Nov/17
A block is released from rest at the  top of a frictionless incline  plane 16.00m long.It reaches the  bottom 4.0s later.A second block  is projected up the plane from the  bottom at the instant the first one  is released in such a way that it  returns to the bottom simultaneously  with the first block.Find:  a)the acceleration of each block  on the incline plane  b)the initial velocity of the first  block.  c)how far up the incline plane  did the second block travel.  d)What angle does the plane makes  with the horizontal.  (g=10m/s^2 )
$${A}\:{block}\:{is}\:{released}\:{from}\:{rest}\:{at}\:{the} \\ $$$${top}\:{of}\:{a}\:{frictionless}\:{incline} \\ $$$${plane}\:\mathrm{16}.\mathrm{00}{m}\:{long}.{It}\:{reaches}\:{the} \\ $$$${bottom}\:\mathrm{4}.\mathrm{0}{s}\:{later}.{A}\:{second}\:{block} \\ $$$${is}\:{projected}\:{up}\:{the}\:{plane}\:{from}\:{the} \\ $$$${bottom}\:{at}\:{the}\:{instant}\:{the}\:{first}\:{one} \\ $$$${is}\:{released}\:{in}\:{such}\:{a}\:{way}\:{that}\:{it} \\ $$$${returns}\:{to}\:{the}\:{bottom}\:{simultaneously} \\ $$$${with}\:{the}\:{first}\:{block}.{Find}: \\ $$$$\left.{a}\right){the}\:{acceleration}\:{of}\:{each}\:{block} \\ $$$${on}\:{the}\:{incline}\:{plane} \\ $$$$\left.{b}\right){the}\:{initial}\:{velocity}\:{of}\:{the}\:{first} \\ $$$${block}. \\ $$$$\left.{c}\right){how}\:{far}\:{up}\:{the}\:{incline}\:{plane} \\ $$$${did}\:{the}\:{second}\:{block}\:{travel}. \\ $$$$\left.{d}\right){What}\:{angle}\:{does}\:{the}\:{plane}\:{makes} \\ $$$${with}\:{the}\:{horizontal}. \\ $$$$\left({g}=\mathrm{10}{m}/{s}^{\mathrm{2}} \right) \\ $$
Commented by NECx last updated on 14/Nov/17
please I ve issues with this question
$${please}\:{I}\:{ve}\:{issues}\:{with}\:{this}\:{question} \\ $$
Commented by ajfour last updated on 14/Nov/17
Commented by ajfour last updated on 14/Nov/17
Let block B was projected with  initial velocity u.  s_A =l  and s_B =0  , in a time of 4s.  ⇒ (1/2)(gsin θ)t^2  = l   and   −ut+(1/2)(gsin θ)t^2  = 0  or    ⇒  8gsin θ =16     sin θ =(1/5)  or  θ =sin^(−1) ((1/5))      ⇒  4u =16              u = 4m/s .  0=u^2 +2(gsin θ)s_(up the incline)   s_(up the incline) =(u^2 /(2gsin θ)) =((16×5)/(20×1)) =4m .  Acceleration of both blocks down  the incline is mgsin θ/m = gsin θ  As already found above sin θ=(1/5)  so angle of incline is θ=sin^(−1) ((1/5)).
$${Let}\:{block}\:{B}\:{was}\:{projected}\:{with} \\ $$$${initial}\:{velocity}\:\boldsymbol{{u}}. \\ $$$$\boldsymbol{{s}}_{\boldsymbol{{A}}} ={l}\:\:{and}\:\boldsymbol{{s}}_{\boldsymbol{{B}}} =\mathrm{0}\:\:,\:{in}\:{a}\:{time}\:{of}\:\mathrm{4}{s}. \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\left({g}\mathrm{sin}\:\theta\right){t}^{\mathrm{2}} \:=\:{l}\:\:\:{and} \\ $$$$\:−{ut}+\frac{\mathrm{1}}{\mathrm{2}}\left({g}\mathrm{sin}\:\theta\right){t}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$${or}\:\:\:\:\Rightarrow\:\:\mathrm{8}{g}\mathrm{sin}\:\theta\:=\mathrm{16} \\ $$$$\:\:\:\mathrm{sin}\:\theta\:=\frac{\mathrm{1}}{\mathrm{5}}\:\:{or}\:\:\theta\:=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\:\:\:\:\Rightarrow\:\:\mathrm{4}{u}\:=\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{u}}\:=\:\mathrm{4}{m}/{s}\:. \\ $$$$\mathrm{0}={u}^{\mathrm{2}} +\mathrm{2}\left({g}\mathrm{sin}\:\theta\right){s}_{{up}\:{the}\:{incline}} \\ $$$${s}_{{up}\:{the}\:{incline}} =\frac{{u}^{\mathrm{2}} }{\mathrm{2}{g}\mathrm{sin}\:\theta}\:=\frac{\mathrm{16}×\mathrm{5}}{\mathrm{20}×\mathrm{1}}\:=\mathrm{4}{m}\:. \\ $$$${Acceleration}\:{of}\:{both}\:{blocks}\:{down} \\ $$$${the}\:{incline}\:{is}\:{mg}\mathrm{sin}\:\theta/{m}\:=\:{g}\mathrm{sin}\:\theta \\ $$$${As}\:{already}\:{found}\:{above}\:\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${so}\:{angle}\:{of}\:{incline}\:{is}\:\theta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right). \\ $$
Commented by NECx last updated on 14/Nov/17
wow.... Thanks
$${wow}….\:{Thanks} \\ $$$$ \\ $$
Commented by NECx last updated on 14/Nov/17
oh..... i dont think you′ve completed  it. It remains (a) (c) n (d)
$${oh}…..\:{i}\:{dont}\:{think}\:{you}'{ve}\:{completed} \\ $$$${it}.\:{It}\:{remains}\:\left({a}\right)\:\left({c}\right)\:{n}\:\left({d}\right) \\ $$
Commented by NECx last updated on 15/Nov/17
thanks boss.
$${thanks}\:{boss}. \\ $$$$ \\ $$

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