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Question-56585

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Question Number 56585 by Tinkutara last updated on 18/Mar/19
Commented by Tinkutara last updated on 18/Mar/19
Commented by mr W last updated on 19/Mar/19
v_0 =speed of rocket at t=0. speed of ball 1=v_1 =v_0 +0.3 speed of ball 2=v_2 =v_0 −0.2 speed of rocket at t: v_R =v_0 +2t let v_R =v_1 v_0 +2t=v_0 +0.3 ⇒t=0.15 s i.e. after 0.15 seconds the wall of chamber and ball 1 are together. in this time the ball 1 has moved 0.15(v_0 +0.3) and the ball 2 has moved 0.15(v_0 −0.2) ⇒0.15(v_0 +0.3)−0.15(v_0 −0.2)=0.075m i.e. at t=0.15, the distance between ball 1 and ball 2 is 4−0.075=3.925m. after t=0.15 s the ball 1 moves with the rocket, let′s say both balls collide after time t, then (v_0 +0.3)t+((2t^2 )/2)−(v_0 −0.2)t=3.925 0.5t+t^2 −3.925=0 ⇒t=((−0.5+(√(0.5^2 +4×3.925)))/2)≈1.75 s 0.15+1.75=1.90 s i.e. both balls collide in 1.90 s.
$${v}_{\mathrm{0}} ={speed}\:{of}\:{rocket}\:{at}\:{t}=\mathrm{0}. \\ $$$${speed}\:{of}\:{ball}\:\mathrm{1}={v}_{\mathrm{1}} ={v}_{\mathrm{0}} +\mathrm{0}.\mathrm{3} \\ $$$${speed}\:{of}\:{ball}\:\mathrm{2}={v}_{\mathrm{2}} ={v}_{\mathrm{0}} −\mathrm{0}.\mathrm{2} \\ $$$${speed}\:{of}\:{rocket}\:{at}\:{t}: \\ $$$${v}_{{R}} ={v}_{\mathrm{0}} +\mathrm{2}{t} \\ $$$${let}\:{v}_{{R}} ={v}_{\mathrm{1}} \\ $$$${v}_{\mathrm{0}} +\mathrm{2}{t}={v}_{\mathrm{0}} +\mathrm{0}.\mathrm{3} \\ $$$$\Rightarrow{t}=\mathrm{0}.\mathrm{15}\:{s} \\ $$$${i}.{e}.\:{after}\:\mathrm{0}.\mathrm{15}\:{seconds}\:{the}\:{wall}\:{of}\:{chamber} \\ $$$${and}\:{ball}\:\mathrm{1}\:{are}\:{together}.\:{in}\:{this}\:{time} \\ $$$${the}\:{ball}\:\mathrm{1}\:{has}\:{moved}\:\mathrm{0}.\mathrm{15}\left({v}_{\mathrm{0}} +\mathrm{0}.\mathrm{3}\right)\:{and} \\ $$$${the}\:{ball}\:\mathrm{2}\:{has}\:{moved}\:\mathrm{0}.\mathrm{15}\left({v}_{\mathrm{0}} −\mathrm{0}.\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{0}.\mathrm{15}\left({v}_{\mathrm{0}} +\mathrm{0}.\mathrm{3}\right)−\mathrm{0}.\mathrm{15}\left({v}_{\mathrm{0}} −\mathrm{0}.\mathrm{2}\right)=\mathrm{0}.\mathrm{075}{m} \\ $$$${i}.{e}.\:{at}\:{t}=\mathrm{0}.\mathrm{15},\:{the}\:{distance}\:{between} \\ $$$${ball}\:\mathrm{1}\:{and}\:{ball}\:\mathrm{2}\:{is}\:\mathrm{4}−\mathrm{0}.\mathrm{075}=\mathrm{3}.\mathrm{925}{m}. \\ $$$${after}\:{t}=\mathrm{0}.\mathrm{15}\:{s}\:{the}\:{ball}\:\mathrm{1}\:{moves}\:{with} \\ $$$${the}\:{rocket},\:{let}'{s}\:{say}\:{both}\:{balls}\:{collide} \\ $$$${after}\:{time}\:{t},\:{then} \\ $$$$\left({v}_{\mathrm{0}} +\mathrm{0}.\mathrm{3}\right){t}+\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{2}}−\left({v}_{\mathrm{0}} −\mathrm{0}.\mathrm{2}\right){t}=\mathrm{3}.\mathrm{925} \\ $$$$\mathrm{0}.\mathrm{5}{t}+{t}^{\mathrm{2}} −\mathrm{3}.\mathrm{925}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{−\mathrm{0}.\mathrm{5}+\sqrt{\mathrm{0}.\mathrm{5}^{\mathrm{2}} +\mathrm{4}×\mathrm{3}.\mathrm{925}}}{\mathrm{2}}\approx\mathrm{1}.\mathrm{75}\:{s} \\ $$$$\mathrm{0}.\mathrm{15}+\mathrm{1}.\mathrm{75}=\mathrm{1}.\mathrm{90}\:{s} \\ $$$${i}.{e}.\:{both}\:{balls}\:{collide}\:{in}\:\mathrm{1}.\mathrm{90}\:{s}. \\ $$
Commented by Tinkutara last updated on 19/Mar/19
Thank you so much Sir! Awesome explanation!

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