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Question Number 190793 by 2kdw last updated on 11/Apr/23
A projectile of mass M explodes at thee highst point of its trajectory when it hase vlocity . The horizontal distance travelede btween launch and explosion is x_0 . Two fragments are produced with initiale velocitis parallel to the ground. They thenfollow their trajectories until they hitt he ground. The fragment of mass m_1 retuns exactly to the launch point of thei orginal projectile (of mass M) while thee othr fragment of mass m_2 hits the grounda t a distance D from this point. Disregardn iteraction with air and assume that massa ws conserved in the explosion (m_1 +m_2 =M) Determine the magnitude of the velocity of fragment 2 just before it hits theground. (a) ((gx_0 )/v) (b)(√((25)/9))v (c) (√(((25)/9)v^2 +(((gx_0 )/5))2)) (d)(√((5/3)x_0 v^2 +(((gx_0 )/v))2))
$$ \\ $$$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{of}\:\mathrm{mass}\:\boldsymbol{{M}}\:\mathrm{explodes}\:\mathrm{at}\:\mathrm{thee} \\ $$$$\mathrm{highst}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{trajectory}\:\mathrm{when}\:\mathrm{it}\:\mathrm{hase} \\ $$$$\mathrm{vlocity}\:.\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{travelede} \\ $$$$\mathrm{btween}\:\mathrm{launch}\:\mathrm{and}\:\mathrm{explosion}\:\mathrm{is}\:\boldsymbol{{x}}_{\mathrm{0}} \:.\:\mathrm{Two} \\ $$$$\mathrm{fragments}\:\mathrm{are}\:\mathrm{produced}\:\mathrm{with}\:\mathrm{initiale} \\ $$$$\mathrm{velocitis}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{They}\: \\ $$$$\mathrm{thenfollow}\:\mathrm{their}\:\mathrm{trajectories}\:\mathrm{until}\:\mathrm{they}\:\mathrm{hitt} \\ $$$$\mathrm{he}\:\mathrm{ground}.\:\mathrm{The}\:\mathrm{fragment}\:\mathrm{of}\:\mathrm{mass}\:\boldsymbol{{m}}_{\mathrm{1}} \:\mathrm{retuns}\:\mathrm{exactly}\:\mathrm{to}\:\mathrm{the}\:\mathrm{launch}\:\mathrm{point}\:\mathrm{of}\:\mathrm{thei} \\ $$$$\mathrm{orginal}\:\mathrm{projectile}\:\left(\mathrm{of}\:\mathrm{mass}\:\mathrm{M}\right)\:\mathrm{while}\:\mathrm{thee} \\ $$$$\mathrm{othr}\:\mathrm{fragment}\:\mathrm{of}\:\mathrm{mass}\:\boldsymbol{{m}}_{\mathrm{2}} \:\mathrm{hits}\:\mathrm{the}\:\mathrm{grounda} \\ $$$$\mathrm{t}\:\mathrm{a}\:\mathrm{distance}\:\boldsymbol{{D}}\:\mathrm{from}\:\mathrm{this}\:\mathrm{point}.\:\mathrm{Disregardn} \\ $$$$\mathrm{iteraction}\:\mathrm{with}\:\mathrm{air}\:\mathrm{and}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{massa} \\ $$$$\mathrm{ws}\:\mathrm{conserved}\:\mathrm{in}\:\mathrm{the}\:\mathrm{explosion}\:\left(\boldsymbol{{m}}_{\mathrm{1}} +\boldsymbol{{m}}_{\mathrm{2}} =\mathrm{M}\right)\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{fragment}\:\mathrm{2}\:\mathrm{just}\:\mathrm{before}\:\mathrm{it}\:\mathrm{hits}\:\mathrm{theground}. \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{g}{x}_{\mathrm{0}} }{{v}} \\ $$$$\left(\mathrm{b}\right)\sqrt{\frac{\mathrm{25}}{\mathrm{9}}}{v} \\ $$$$\left({c}\right)\:\sqrt{\frac{\mathrm{25}}{\mathrm{9}}{v}^{\mathrm{2}} +\left(\frac{{gx}_{\mathrm{0}} }{\mathrm{5}}\right)\mathrm{2}} \\ $$$$\left({d}\right)\sqrt{\frac{\mathrm{5}}{\mathrm{3}}{x}_{\mathrm{0}} {v}^{\mathrm{2}} +\left(\frac{{gx}_{\mathrm{0}} }{{v}}\right)\mathrm{2}} \\ $$
Commented by mr W last updated on 12/Apr/23
terribly many typos! due to the many typos all answers given are wrong! please check your typos and fix them at first! and please make proper line breaks!
$${terribly}\:{many}\:{typos}!\: \\ $$$${due}\:{to}\:{the}\:{many}\:{typos}\:{all}\:{answers}\: \\ $$$${given}\:{are}\:{wrong}! \\ $$$${please}\:{check}\:{your}\:{typos}\:{and}\:{fix}\:{them} \\ $$$${at}\:{first}! \\ $$$${and}\:{please}\:{make}\:{proper}\:{line}\:{breaks}! \\ $$
Answered by mr W last updated on 12/Apr/23
Commented by mr W last updated on 12/Apr/23
v_1 =v t=(x_0 /v)=(D/v_2 ) ⇒v_2 =((Dv)/x_0 ) h=((gt^2 )/2)=(g/2)((x_0 /v))^2 v_3 ^2 =v_2 ^2 +2gh=(((Dv)/x_0 ))^2 +(((gx_0 )/v))^2 ⇒v_3 =(√((((Dv)/x_0 ))^2 +(((gx_0 )/v))^2 ))
$${v}_{\mathrm{1}} ={v} \\ $$$${t}=\frac{{x}_{\mathrm{0}} }{{v}}=\frac{{D}}{{v}_{\mathrm{2}} } \\ $$$$\Rightarrow{v}_{\mathrm{2}} =\frac{{Dv}}{{x}_{\mathrm{0}} } \\ $$$${h}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}=\frac{{g}}{\mathrm{2}}\left(\frac{{x}_{\mathrm{0}} }{{v}}\right)^{\mathrm{2}} \\ $$$${v}_{\mathrm{3}} ^{\mathrm{2}} ={v}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{gh}=\left(\frac{{Dv}}{{x}_{\mathrm{0}} }\right)^{\mathrm{2}} +\left(\frac{{gx}_{\mathrm{0}} }{{v}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{v}_{\mathrm{3}} =\sqrt{\left(\frac{{Dv}}{{x}_{\mathrm{0}} }\right)^{\mathrm{2}} +\left(\frac{{gx}_{\mathrm{0}} }{{v}}\right)^{\mathrm{2}} } \\ $$

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