Question

please solve it.......​

Answer 1

Answer:

ATQ<

KE of man = 1/2mv^2

KE of boy = 1/2[ 1/2 * m/2 * v1^2]

on equating ,

v1^2  = 4 v^2

v1= 2v.......................(1)

on speeding up by 1 m/s,

KE of man = 1/2 * m * ( v+1)^2

KE of boy = 1/2 * m./2 * v1^2

on equating ,

2(v+1)^2 = v1^2

$\sqrt{2}$ (v+1) = v1.....................(2)

solve n get ur desired ans

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Original\:speed\:of\:man=(\sqrt{2}+1)\:m/s}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{Given : } \\ \implies K.E \: of \: man = \frac{1}{2} K.E \: of \: boy \\ \\ \: \underline \bold{To \: Find : } \\ \implies Original \: speed \: of \: man = ?$

• According to given question :

$\bold{Let \: Speed \:of \: man = v_{1}} \\ \: \: \: \: \: \: \: \bold{Speed \: of \: boy = v_{2}} \\ \\ \bold{Mass \: of \: man = m \: kg} \\ \bold{Mass \: of \: boy = 0.5m \: kg} \\ \\ \bold{At \: original \: speed \: of \: man} \\ \implies K.E \: of \: man = \frac{1}{2} K.E \: of \: boy \\ \\ \implies \frac{1}{2}{m v_{1} }^{2} = \frac{1}{2} \times \frac{1}{2} {mv_{2}}^{2} \\ \\ \implies \frac{1}{2} {m v_{1} }^{2} = \frac{1}{4} {m v_{2} }^{2} \\ \\ \implies v_{1} = \frac{ v_{2}}{2} - - - - - (1) \\ \\ \bold{At \: man \: speed \: increase \: by \: 1 \: ms} \\ \implies K.E \: of \: man = \frac{1}{2} K.E\: of \: boy \\ \\ \implies \frac{1}{2} {m( v_{1} + 1) }^{2} = \frac{1}{2} \frac{m (v_{2})^{2} }{2} \\ \\ \implies \frac{1}{ \cancel2} { \cancel{m}( v_{1} + 1)}^{2} \times \frac{ \cancel{2} \times 2}{ \cancel{m}} = (v_{2})^{2} \\ \\ \implies ( v_{1} + 1)^{2} \times 2 =({ 2 v_{1} })^{2} \\ \\ \implies ({ v_{1} + 1 })^{2} = 2 {( v_{1}})^{2} \\ \\ \implies v_{1} + 1 = \sqrt{2 {( v_{1} })^{2} } \\ \\ \implies v_{1} + 1 = \sqrt{2} v_{1} \\ \\ \implies v_{1} ( \sqrt{2 } - 1 ) = 1 \\ \\ \implies v_{1} = \frac{1}{ \sqrt{2} - 1} \\ \\ \bold{\implies v_{1} = ( \sqrt{2} + 1) \: m/s}$