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Question Number 29118 by Tinkutara last updated on 04/Feb/18

Commented by Tinkutara last updated on 04/Feb/18

Use the values: e=3/5 m=7 kg M=15 kg u=30 m/s Theta=30°

Commented by Tinkutara last updated on 04/Feb/18

Thank you very much Sir! I got the answer.

Commented by ajfour last updated on 04/Feb/18

Commented by ajfour last updated on 04/Feb/18

v+Vsin θ=eusin θ    (along normal to incline of wedge)  MV+mucos^2 θ−mvsin θ=mu     (momentum conservation         along horizontal)  ⇒ MV−m(eusin θ−Vsin θ)sin θ                           =musin^2 θ     V  =((mu(1+e)sin^2 θ)/(M+msin^2 θ)) .

$${v}+{V}\mathrm{sin}\:\theta={eu}\mathrm{sin}\:\theta \\ $$$$\:\:\left({along}\:{normal}\:{to}\:{incline}\:{of}\:{wedge}\right) \\ $$$${MV}+{mu}\mathrm{cos}\:^{\mathrm{2}} \theta−{mv}\mathrm{sin}\:\theta={mu} \\ $$$$\:\:\:\left({momentum}\:{conservation}\right. \\ $$$$\left.\:\:\:\:\:\:\:{along}\:{horizontal}\right) \\ $$$$\Rightarrow\:{MV}−{m}\left({eu}\mathrm{sin}\:\theta−{V}\mathrm{sin}\:\theta\right)\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={mu}\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\boldsymbol{{V}}\:\:=\frac{{mu}\left(\mathrm{1}+{e}\right)\mathrm{sin}\:^{\mathrm{2}} \theta}{{M}+{m}\mathrm{sin}\:^{\mathrm{2}} \theta}\:. \\ $$

Commented by mrW2 last updated on 04/Feb/18

e_⊥ =e=((V sin θ+v)/(u sin θ))  ⇒V sin θ+v=eu sin θ    e_(//) =1=((v_(//) −V cos θ)/(u cos θ))   (∗)  ⇒v_(//) =(u+V) cos θ≠u cos θ    assume parallel restitution coef. =1

$${e}_{\bot} ={e}=\frac{{V}\:\mathrm{sin}\:\theta+{v}}{{u}\:\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{V}\:\mathrm{sin}\:\theta+{v}={eu}\:\mathrm{sin}\:\theta \\ $$$$ \\ $$$${e}_{//} =\mathrm{1}=\frac{{v}_{//} −{V}\:\mathrm{cos}\:\theta}{{u}\:\mathrm{cos}\:\theta}\:\:\:\left(\ast\right) \\ $$$$\Rightarrow{v}_{//} =\left({u}+{V}\right)\:\mathrm{cos}\:\theta\neq{u}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$${assume}\:{parallel}\:{restitution}\:{coef}.\:=\mathrm{1} \\ $$

Commented by ajfour last updated on 04/Feb/18

ball did not receive any impulse  parallel to incline , the wedge  did receieve an impulsive   component from ground,   parallel to its incline, so  v_(//)   of ball = ucos θ  and not  v_(//) −Vcos θ = ucos θ ,  Sir!

$${ball}\:{did}\:{not}\:{receive}\:{any}\:{impulse} \\ $$$${parallel}\:{to}\:{incline}\:,\:{the}\:{wedge} \\ $$$${did}\:{receieve}\:{an}\:{impulsive}\: \\ $$$${component}\:{from}\:{ground},\: \\ $$$${parallel}\:{to}\:{its}\:{incline},\:{so} \\ $$$${v}_{//} \:\:{of}\:{ball}\:=\:{u}\mathrm{cos}\:\theta \\ $$$${and}\:{not}\:\:{v}_{//} −{V}\mathrm{cos}\:\theta\:=\:{u}\mathrm{cos}\:\theta\:, \\ $$$${Sir}! \\ $$

Answered by ajfour last updated on 04/Feb/18

V_M = 5.04 m/s .

$${V}_{{M}} =\:\mathrm{5}.\mathrm{04}\:{m}/{s}\:. \\ $$

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