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Question Number 18265 by Tinkutara last updated on 17/Jul/17

A particle is projected at an angle 60° with speed 10(√3) m/s from the point A as shown in the figure. At the same time the wedge is made to move with speed 10(√3) m/s toward right as shown in figure. Find the time after which particle will strike the wedge.

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{60}° \\ $$$$\mathrm{with}\:\mathrm{speed}\:\mathrm{10}\sqrt{\mathrm{3}}\:\mathrm{m}/\mathrm{s}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:{A} \\ $$$$\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\:\mathrm{At}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{time}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{is}\:\mathrm{made}\:\mathrm{to}\:\mathrm{move}\:\mathrm{with} \\ $$$$\mathrm{speed}\:\mathrm{10}\sqrt{\mathrm{3}}\:\mathrm{m}/\mathrm{s}\:\mathrm{toward}\:\mathrm{right}\:\mathrm{as}\:\mathrm{shown} \\ $$$$\mathrm{in}\:\mathrm{figure}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{time}\:\mathrm{after}\:\mathrm{which} \\ $$$$\mathrm{particle}\:\mathrm{will}\:\mathrm{strike}\:\mathrm{the}\:\mathrm{wedge}. \\ $$

Commented by Tinkutara last updated on 17/Jul/17

Commented by ajfour last updated on 18/Jul/17

In order to collide at time t 10(√3)t=ycot 30°+(10(√3))[cos 60°]t 10(√3)t=y(√3)+5(√3)t y=5t Also y=(10(√3))(((√3)/2))t−((gt^2 )/2) ⇒ 5t=15t−5t^2 t^2 −2t=0 t=2s .

$$\mathrm{In}\:\mathrm{order}\:\mathrm{to}\:\mathrm{collide}\:\mathrm{at}\:\mathrm{time}\:\mathrm{t} \\ $$$$\mathrm{10}\sqrt{\mathrm{3}}\mathrm{t}=\mathrm{ycot}\:\mathrm{30}°+\left(\mathrm{10}\sqrt{\mathrm{3}}\right)\left[\mathrm{cos}\:\mathrm{60}°\right]\mathrm{t} \\ $$$$\mathrm{10}\sqrt{\mathrm{3}}\mathrm{t}=\mathrm{y}\sqrt{\mathrm{3}}+\mathrm{5}\sqrt{\mathrm{3}}\mathrm{t} \\ $$$$\mathrm{y}=\mathrm{5t} \\ $$$$\mathrm{Also}\:\:\:\:\:\mathrm{y}=\left(\mathrm{10}\sqrt{\mathrm{3}}\right)\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{t}−\frac{\mathrm{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{5t}=\mathrm{15t}−\mathrm{5t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}^{\mathrm{2}} −\mathrm{2t}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{t}=\mathrm{2s}\:. \\ $$

Commented by ajfour last updated on 18/Jul/17

and just now.

$$\mathrm{and}\:\mathrm{just}\:\mathrm{now}. \\ $$

Commented by Tinkutara last updated on 18/Jul/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Commented by Tinkutara last updated on 07/Mar/18

Sir I have a doubt now. Sorry for too late! But why ycot30° is used?

Commented by ajfour last updated on 08/Mar/18

the wedge has to travel this extra horizontal distance to collide with the particle if they collide at level y=y .

$${the}\:{wedge}\:{has}\:{to}\:{travel}\:{this}\:{extra} \\ $$$${horizontal}\:{distance}\:{to}\:{collide}\:{with} \\ $$$${the}\:{particle}\:{if}\:{they}\:{collide}\:{at}\:{level} \\ $$$${y}={y}\:. \\ $$

Commented by Tinkutara last updated on 08/Mar/18

Oh yes thanks! ☺������

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