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Question Number 17835 by Tinkutara last updated on 11/Jul/17

A rocket is moving in a gravity free space with a constant acceleration of 2 m/s^2 along + x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 m/s relative to the rocket. At the same time, another ball is thrown in −x direction with a speed of 0.2 m/s from its right end relative to the rocket. The time in seconds when the two balls hit each other is

$$\mathrm{A}\:\mathrm{rocket}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{a}\:\mathrm{gravity}\:\mathrm{free} \\ $$$$\mathrm{space}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of} \\ $$$$\mathrm{2}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{along}\:+\:\mathrm{x}\:\mathrm{direction}\:\left(\mathrm{see}\:\mathrm{figure}\right). \\ $$$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{a}\:\mathrm{chamber}\:\mathrm{inside}\:\mathrm{the} \\ $$$$\mathrm{rocket}\:\mathrm{is}\:\mathrm{4}\:\mathrm{m}.\:\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{left}\:\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{chamber}\:\mathrm{in}\:+{x}\:\mathrm{direction} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{0}.\mathrm{3}\:\mathrm{m}/\mathrm{s}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{rocket}.\:\mathrm{At}\:\mathrm{the}\:\mathrm{same}\:\mathrm{time},\:\mathrm{another}\:\mathrm{ball} \\ $$$$\mathrm{is}\:\mathrm{thrown}\:\mathrm{in}\:−{x}\:\mathrm{direction}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed} \\ $$$$\mathrm{of}\:\mathrm{0}.\mathrm{2}\:\mathrm{m}/\mathrm{s}\:\mathrm{from}\:\mathrm{its}\:\mathrm{right}\:\mathrm{end}\:\mathrm{relative}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{rocket}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{in}\:\mathrm{seconds}\:\mathrm{when} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{hit}\:\mathrm{each}\:\mathrm{other}\:\mathrm{is} \\ $$

Commented by Tinkutara last updated on 11/Jul/17

Answered by ajfour last updated on 11/Jul/17

let rocket′s initial speed=u (u+0.3)t=4+(u−0.2)t ⇒ t=(4/(0.5)) =8 s .

$$\mathrm{let}\:\mathrm{rocket}'\mathrm{s}\:\mathrm{initial}\:\mathrm{speed}=\mathrm{u} \\ $$$$\:\left(\mathrm{u}+\mathrm{0}.\mathrm{3}\right)\mathrm{t}=\mathrm{4}+\left(\mathrm{u}−\mathrm{0}.\mathrm{2}\right)\mathrm{t} \\ $$$$\Rightarrow\:\:\:\:\mathrm{t}=\frac{\mathrm{4}}{\mathrm{0}.\mathrm{5}}\:=\mathrm{8}\:\mathrm{s}\:. \\ $$$$ \\ $$

Answered by mrW1 last updated on 11/Jul/17

v=velocity of rocket as the balls are thrownn v_1 =v+0.3 v_2 =v−0.2 (v+0.3)t=4+(v−0.2)t ⇒0.5t=4 ⇒t=(4/(0.5))=8 s

$$\mathrm{v}=\mathrm{velocity}\:\mathrm{of}\:\mathrm{rocket}\:\mathrm{as}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{thrownn} \\ $$$$\mathrm{v}_{\mathrm{1}} =\mathrm{v}+\mathrm{0}.\mathrm{3} \\ $$$$\mathrm{v}_{\mathrm{2}} =\mathrm{v}−\mathrm{0}.\mathrm{2} \\ $$$$\left(\mathrm{v}+\mathrm{0}.\mathrm{3}\right)\mathrm{t}=\mathrm{4}+\left(\mathrm{v}−\mathrm{0}.\mathrm{2}\right)\mathrm{t} \\ $$$$\Rightarrow\mathrm{0}.\mathrm{5t}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{t}=\frac{\mathrm{4}}{\mathrm{0}.\mathrm{5}}=\mathrm{8}\:\mathrm{s} \\ $$

Commented by ajfour last updated on 11/Jul/17

Rocket acquires a velocity of 0.3m/s in a time t_0 =0.3/2=0.15s after this time rocket moves faster than ball 1 and ball 1 hits the point from where it was thrown in a time equal to 2t_0 =0.3s ; the second ball is still on its way and reaches the left chamber wall of rocket in a time given by: (0.2)t+(1/2)(2)t^2 =4 or t^2 +(t/5)−4=0 or 5t^2 +t−20=0 ⇒ t=((−1+(√(1+400)))/(10)) ≈ 1.9 s .

$$\mathrm{Rocket}\:\mathrm{acquires}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{0}.\mathrm{3m}/\mathrm{s} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{time}\:\mathrm{t}_{\mathrm{0}} =\mathrm{0}.\mathrm{3}/\mathrm{2}=\mathrm{0}.\mathrm{15s} \\ $$$$\mathrm{after}\:\mathrm{this}\:\mathrm{time}\:\mathrm{rocket}\:\mathrm{moves}\:\mathrm{faster} \\ $$$$\mathrm{than}\:\mathrm{ball}\:\mathrm{1}\:\mathrm{and}\:\mathrm{ball}\:\mathrm{1}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{from}\:\mathrm{where}\:\mathrm{it}\:\mathrm{was}\:\mathrm{thrown}\:\mathrm{in}\:\mathrm{a}\:\mathrm{time} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{2t}_{\mathrm{0}} =\mathrm{0}.\mathrm{3s}\:;\:\mathrm{the}\:\mathrm{second}\:\mathrm{ball} \\ $$$$\mathrm{is}\:\mathrm{still}\:\mathrm{on}\:\mathrm{its}\:\mathrm{way}\:\mathrm{and}\:\mathrm{reaches}\:\mathrm{the} \\ $$$$\mathrm{left}\:\mathrm{chamber}\:\mathrm{wall}\:\mathrm{of}\:\mathrm{rocket}\:\mathrm{in}\:\mathrm{a}\:\mathrm{time} \\ $$$$\mathrm{given}\:\mathrm{by}: \\ $$$$\:\:\:\:\:\:\left(\mathrm{0}.\mathrm{2}\right)\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\mathrm{t}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{or}\:\:\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{t}}{\mathrm{5}}−\mathrm{4}=\mathrm{0}\:\:\:\mathrm{or}\:\:\:\mathrm{5t}^{\mathrm{2}} +\mathrm{t}−\mathrm{20}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\mathrm{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{400}}}{\mathrm{10}}\:\approx\:\mathrm{1}.\mathrm{9}\:\mathrm{s}\:. \\ $$$$\: \\ $$

Commented by Tinkutara last updated on 11/Jul/17

Thanks Sir!

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

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