The-system-is-initially-at-origin-and-is-moving-with-a-velocity-3-m-s-k-A-force-120t-newton-i-acts-on-mass-m-2-where-t-is-time-in-seconds-The-man-throws-a-light-ball-at-the-instant-when

Question Number 22574 by Tinkutara last updated on 20/Oct/17

$$\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{origin}\:\mathrm{and}\:\mathrm{is} \\ $$$$\mathrm{moving}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\left(\mathrm{3}\:\mathrm{m}/\mathrm{s}\right)\:\overset{\wedge} {{k}}.\:\mathrm{A} \\ $$$$\mathrm{force}\:\left(\mathrm{120}{t}\:\mathrm{newton}\right)\:\overset{\wedge} {{i}}\:\mathrm{acts}\:\mathrm{on}\:\mathrm{mass}\:{m}_{\mathrm{2}} , \\ $$$$\mathrm{where}\:{t}\:\mathrm{is}\:\mathrm{time}\:\mathrm{in}\:\mathrm{seconds}. \\ $$$$\mathrm{The}\:\mathrm{man}\:\mathrm{throws}\:\mathrm{a}\:\mathrm{light}\:\mathrm{ball},\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{instant}\:\mathrm{when}\:{m}_{\mathrm{1}} \:\mathrm{starts}\:\mathrm{slipping}\:\mathrm{on}\:{m}_{\mathrm{2}} , \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{5}\:\mathrm{m}/\mathrm{s}\:\mathrm{vertically}\:\mathrm{up}\:\mathrm{w}.\mathrm{r}.\mathrm{t} \\ $$$$\mathrm{himself}.\:\mathrm{Taking}\:\mathrm{the}\:\mathrm{masses}\:\mathrm{of}\:\mathrm{blocks} \\ $$$$\mathrm{and}\:\mathrm{man}\:\mathrm{as}\:\mathrm{60}\:\mathrm{kg}\:\mathrm{each}\:\mathrm{and}\:\mathrm{assuming} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{man}\:\mathrm{never}\:\mathrm{slips}\:\mathrm{on}\:{m}_{\mathrm{2}} ,\:\mathrm{find} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{The}\:\mathrm{time}\:\mathrm{at}\:\mathrm{which}\:\mathrm{man}\:\mathrm{throws}\:\mathrm{the} \\ $$$$\mathrm{ball}\:\mathrm{and} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{The}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where} \\ $$$$\mathrm{ball}\:\mathrm{lands}.\:\mathrm{Neglect}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{system}.\:\left({g}\:=\:\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right) \\ $$

Commented by Tinkutara last updated on 20/Oct/17

Answered by Sahib singh last updated on 21/Oct/17

if m_1 = m_2 = m_(man) v_(initial) = 3 ^ k ^� m/s μ = 0.2 ⇒ a_( maximum of m_2 ) = (F_(friction) /m_2 ) = (((60×0.2×10)i ^� )/(60)) m/s^2 =2 i ^� m/s^2 ⇒ when F = 360 i ^� N as total mass is 180 kg ⇒ at t = 3 s along x axis F = 120 t ⇒ (dp/dt) = 120 t ⇒ dp = 120 t dt ⇒ ∫_0 ^( v) dp = ∫_(0 ) ^( 3) 120 t dt ⇒ mΔv = 60 × 9 ⇒ Δv = ((60×9)/(180)) = 3 i ^� m/s ⇒ v_(man) = 3 i ^� & (dx/dt) = (t^2 /3) ⇒ x_(at t = 3) = 3 & v_z = 3 ⇒ z_(at t = 3) = 9 ⇒ r_(ball initially) ^→ = 3i ^� +9k ^� ⇒v_(horizonral of ball ) =( 3i ^� +3k ^� )m/s ⇒∣v_(horizontal) ∣ = 3(√(2 )) ⇒ Range = ((2×∣v_(vertical) ∣×∣v_(horizontal) ∣)/g) ⇒ R = ((2×5×3(√2))/(10)) ⇒ R(in meters) = 3(√2) ⇒ as the horizontal velocity had direction (((3i ^� + 3k ^� )/(3(√2)))) ⇒ R^(→ ) will also have same direction ⇒ change in position vector of the ball r_(final) ^→ − r_(initial) ^→ =(3 i ^� + 3 k ^� ) m ⇒r_(final) ^→ = (6 i ^� + 9 k ^� )m

$$ \\ $$$$ \\ $$$${if}\:\:\:\:\:\:\:\:{m}_{\mathrm{1}} =\:{m}_{\mathrm{2}} =\:{m}_{{man}} \\ $$$${v}_{{initial}} \:=\:\mathrm{3}\overset{\:} {\:}{k}\hat {\:}\:\:{m}/{s}\: \\ $$$$\mu\:=\:\mathrm{0}.\mathrm{2}\: \\ $$$$\Rightarrow\:{a}_{\:{maximum}\:{of}\:{m}_{\mathrm{2}} } \:=\:\frac{{F}_{{friction}} }{{m}_{\mathrm{2}} } \\ $$$$=\:\frac{\left(\mathrm{60}×\mathrm{0}.\mathrm{2}×\mathrm{10}\right){i}\hat {\:}}{\mathrm{60}}\:{m}/{s}^{\mathrm{2}} \\ $$$$=\mathrm{2}\:{i}\hat {\:}\:{m}/{s}^{\mathrm{2}} \\ $$$$\Rightarrow\:{when}\:{F}\:=\:\mathrm{360}\:{i}\hat {\:}\:{N} \\ $$$${as}\:{total}\:{mass}\:{is}\:\mathrm{180}\:{kg} \\ $$$$\Rightarrow\:{at}\:\:\:{t}\:=\:\mathrm{3}\:{s} \\ $$$${along}\:{x}\:{axis} \\ $$$$\:{F}\:=\:\mathrm{120}\:{t} \\ $$$$\Rightarrow\:\frac{{dp}}{{dt}}\:=\:\mathrm{120}\:{t} \\ $$$$\Rightarrow\:{dp}\:=\:\mathrm{120}\:{t}\:{dt} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:{v}} \:\:{dp}\:=\:\int_{\mathrm{0}\:} ^{\:\mathrm{3}} \:\mathrm{120}\:{t}\:{dt} \\ $$$$\Rightarrow\:{m}\Delta{v}\:=\:\mathrm{60}\:×\:\mathrm{9} \\ $$$$\Rightarrow\:\Delta{v}\:=\:\frac{\mathrm{60}×\mathrm{9}}{\mathrm{180}}\:=\:\mathrm{3}\:{i}\hat {\:}{m}/{s} \\ $$$$\Rightarrow\:{v}_{{man}} =\:\mathrm{3}\:{i}\hat {\:} \\ $$$$\& \\ $$$$\:\frac{{dx}}{{dt}}\:=\:\frac{{t}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\:{x}_{{at}\:\:{t}\:=\:\mathrm{3}} \:=\:\mathrm{3}\: \\ $$$$\&\: \\ $$$${v}_{{z}} \:=\:\mathrm{3}\: \\ $$$$\Rightarrow\:{z}_{{at}\:\:{t}\:=\:\mathrm{3}} \:=\:\mathrm{9} \\ $$$$\Rightarrow\:\overset{\rightarrow} {{r}}_{{ball}\:{initially}} =\:\mathrm{3}{i}\hat {\:}+\mathrm{9}{k}\hat {\:} \\ $$$$ \\ $$$$\Rightarrow{v}_{{horizonral}\:{of}\:{ball}\:} =\left(\:\mathrm{3}{i}\hat {\:}+\mathrm{3}{k}\hat {\:}\right){m}/{s} \\ $$$$\Rightarrow\mid{v}_{{horizontal}} \mid\:=\:\mathrm{3}\sqrt{\mathrm{2}\:} \\ $$$$\Rightarrow\:{Range}\:=\:\frac{\mathrm{2}×\mid{v}_{{vertical}} \mid×\mid{v}_{{horizontal}} \mid}{{g}} \\ $$$$\Rightarrow\:{R}\:=\:\frac{\mathrm{2}×\mathrm{5}×\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{10}} \\ $$$$\Rightarrow\:{R}\left({in}\:{meters}\right)\:=\:\mathrm{3}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{as}\:{the}\:{horizontal}\:{velocity} \\ $$$${had}\:{direction}\:\left(\frac{\mathrm{3}{i}\hat {\:}+\:\mathrm{3}{k}\hat {\:}}{\mathrm{3}\sqrt{\mathrm{2}}}\right) \\ $$$$\Rightarrow\:\overset{\rightarrow\:} {{R}}\:{will}\:{also}\:{have}\:{same} \\ $$$${direction} \\ $$$$\Rightarrow\:{change}\:{in}\:{position}\:{vector}\:{of}\:{the}\: \\ $$$$\:{ball} \\ $$$$\overset{\rightarrow} {{r}}_{{final}} −\:\overset{\rightarrow} {{r}}_{{initial}} \:=\left(\mathrm{3}\:{i}\hat {\:}+\:\mathrm{3}\:{k}\hat {\:}\right)\:{m}\: \\ $$$$\Rightarrow\overset{\rightarrow} {{r}}_{{final}} =\:\left(\mathrm{6}\:{i}\hat {\:}\:+\:\mathrm{9}\:{k}\hat {\:}\right){m} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$