Question and Answers Forum

All Questions   Topic List

MechanicsQuestion and Answers: Page 7

Question Number 194193    Answers: 0   Comments: 0

Question Number 192540    Answers: 0   Comments: 1

Question Number 192429    Answers: 0   Comments: 0

Question Number 192428    Answers: 0   Comments: 0

Question Number 192044    Answers: 0   Comments: 0

Question Number 191868    Answers: 2   Comments: 0

A particle of mass m moves under the central repulsive force ((mb)/r^3 ) and is initially moving at a distance ′a′ from the origin of a force with velocity ′v′ at right angle to ′a′. show that rcos pθ=a where p =(b/(a^2 v^2 ))+1.

$${A}\:{particle}\:{of}\:{mass}\:{m}\:{moves}\:{under}\:{the}\:{central} \\ $$$${repulsive}\:{force}\:\frac{{mb}}{{r}^{\mathrm{3}} }\:\:{and}\:{is}\:{initially}\:{moving} \\ $$$${at}\:{a}\:{distance}\:'{a}'\:\:{from}\:{the}\:{origin}\:{of}\:\:{a}\:{force} \\ $$$${with}\:{velocity}\:\:'{v}'\:{at}\:{right}\:{angle}\:{to}\:\:'{a}'. \\ $$$${show}\:{that}\:\:\: \\ $$$$\:\:\:\:\:{r}\mathrm{cos}\:{p}\theta={a}\:\:{where}\:{p}\:=\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }+\mathrm{1}. \\ $$$$ \\ $$

Question Number 190962    Answers: 0   Comments: 1

A massless spring with a spring constant K=10Nm^(−1) is suspended from rigid support carries a mass m=100g at it lower end the system is subjected to resistive force −pv.it observed that the system oscillates and its energy to one third of it initial value in one and half minutes calculates the value of p

$$\mathrm{A}\:\mathrm{massless}\:\mathrm{spring}\:\mathrm{with}\:\mathrm{a}\:\mathrm{spring}\:\mathrm{constant} \\ $$$$\mathrm{K}=\mathrm{10Nm}^{−\mathrm{1}} \:\mathrm{is}\:\mathrm{suspended}\:\mathrm{from}\:\mathrm{rigid}\:\mathrm{support} \\ $$$$\mathrm{carries}\:\mathrm{a}\:\mathrm{mass}\:\mathrm{m}=\mathrm{100g}\:\mathrm{at}\:\mathrm{it}\:\mathrm{lower}\:\mathrm{end} \\ $$$$\mathrm{the}\:\mathrm{system}\:\mathrm{is}\:\mathrm{subjected}\:\mathrm{to}\:\mathrm{resistive}\:\mathrm{force} \\ $$$$−\mathrm{pv}.\mathrm{it}\:\mathrm{observed}\:\mathrm{that}\:\mathrm{the}\:\mathrm{system}\:\mathrm{oscillates} \\ $$$$\mathrm{and}\:\mathrm{its}\:\mathrm{energy}\:\mathrm{to}\:\mathrm{one}\:\mathrm{third}\:\mathrm{of}\:\mathrm{it}\:\mathrm{initial}\:\mathrm{value} \\ $$$$\mathrm{in}\:\mathrm{one}\:\mathrm{and}\:\mathrm{half}\:\mathrm{minutes}\:\mathrm{calculates}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\mathrm{p} \\ $$

Question Number 190903    Answers: 1   Comments: 0

Question Number 190902    Answers: 1   Comments: 0

Question Number 190813    Answers: 1   Comments: 0

Question Number 190793    Answers: 1   Comments: 1

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))

$$ \\ $$$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{of}\:\mathrm{mass}\:\boldsymbol{{M}}\:\mathrm{explodes}\:\mathrm{at}\:\mathrm{thee} \\ $$$$\mathrm{highst}\:\mathrm{point}\:\mathrm{of}\:\mathrm{its}\:\mathrm{trajectory}\:\mathrm{when}\:\mathrm{it}\:\mathrm{hase} \\ $$$$\mathrm{vlocity}\:.\:\mathrm{The}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{travelede} \\ $$$$\mathrm{btween}\:\mathrm{launch}\:\mathrm{and}\:\mathrm{explosion}\:\mathrm{is}\:\boldsymbol{{x}}_{\mathrm{0}} \:.\:\mathrm{Two} \\ $$$$\mathrm{fragments}\:\mathrm{are}\:\mathrm{produced}\:\mathrm{with}\:\mathrm{initiale} \\ $$$$\mathrm{velocitis}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{They}\: \\ $$$$\mathrm{thenfollow}\:\mathrm{their}\:\mathrm{trajectories}\:\mathrm{until}\:\mathrm{they}\:\mathrm{hitt} \\ $$$$\mathrm{he}\:\mathrm{ground}.\:\mathrm{The}\:\mathrm{fragment}\:\mathrm{of}\:\mathrm{mass}\:\boldsymbol{{m}}_{\mathrm{1}} \:\mathrm{retuns}\:\mathrm{exactly}\:\mathrm{to}\:\mathrm{the}\:\mathrm{launch}\:\mathrm{point}\:\mathrm{of}\:\mathrm{thei} \\ $$$$\mathrm{orginal}\:\mathrm{projectile}\:\left(\mathrm{of}\:\mathrm{mass}\:\mathrm{M}\right)\:\mathrm{while}\:\mathrm{thee} \\ $$$$\mathrm{othr}\:\mathrm{fragment}\:\mathrm{of}\:\mathrm{mass}\:\boldsymbol{{m}}_{\mathrm{2}} \:\mathrm{hits}\:\mathrm{the}\:\mathrm{grounda} \\ $$$$\mathrm{t}\:\mathrm{a}\:\mathrm{distance}\:\boldsymbol{{D}}\:\mathrm{from}\:\mathrm{this}\:\mathrm{point}.\:\mathrm{Disregardn} \\ $$$$\mathrm{iteraction}\:\mathrm{with}\:\mathrm{air}\:\mathrm{and}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{massa} \\ $$$$\mathrm{ws}\:\mathrm{conserved}\:\mathrm{in}\:\mathrm{the}\:\mathrm{explosion}\:\left(\boldsymbol{{m}}_{\mathrm{1}} +\boldsymbol{{m}}_{\mathrm{2}} =\mathrm{M}\right)\:\mathrm{Determine}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{fragment}\:\mathrm{2}\:\mathrm{just}\:\mathrm{before}\:\mathrm{it}\:\mathrm{hits}\:\mathrm{theground}. \\ $$$$\left(\mathrm{a}\right)\:\frac{\mathrm{g}{x}_{\mathrm{0}} }{{v}} \\ $$$$\left(\mathrm{b}\right)\sqrt{\frac{\mathrm{25}}{\mathrm{9}}}{v} \\ $$$$\left({c}\right)\:\sqrt{\frac{\mathrm{25}}{\mathrm{9}}{v}^{\mathrm{2}} +\left(\frac{{gx}_{\mathrm{0}} }{\mathrm{5}}\right)\mathrm{2}} \\ $$$$\left({d}\right)\sqrt{\frac{\mathrm{5}}{\mathrm{3}}{x}_{\mathrm{0}} {v}^{\mathrm{2}} +\left(\frac{{gx}_{\mathrm{0}} }{{v}}\right)\mathrm{2}} \\ $$

Question Number 190629    Answers: 0   Comments: 1

Question Number 189205    Answers: 0   Comments: 6

Question Number 188092    Answers: 0   Comments: 0

Question Number 188062    Answers: 1   Comments: 8

Question Number 187789    Answers: 1   Comments: 0

Question Number 187675    Answers: 1   Comments: 1

Question Number 187640    Answers: 2   Comments: 1

Question Number 187513    Answers: 0   Comments: 1

Question Number 187364    Answers: 2   Comments: 3

Question Number 187355    Answers: 1   Comments: 2

Question Number 187216    Answers: 1   Comments: 1

Question Number 186692    Answers: 1   Comments: 1

Question Number 186645    Answers: 2   Comments: 1

Question Number 186439    Answers: 1   Comments: 1

Question Number 186416    Answers: 1   Comments: 1

  Pg 2      Pg 3      Pg 4      Pg 5      Pg 6      Pg 7      Pg 8      Pg 9      Pg 10      Pg 11   

Terms of Service

Privacy Policy

Contact: [email protected]