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1) A merry go-round rotates with a constant angular speed \(\omega \). Relative to the rotating frame of the merry go-round, an outside object(say a man) stationary on the ground appears to revolve with the same angular speed \(\omega \) in the opposite sense. Explain the motion of the object in terms of pseudo-forces acting on it in the rotating frame of reference.

Do the complete Vector analysis

2) Show that is a collision is to take place between a fast moving electron and an electron at rest is to lead to production of an electron-positron pair, the K.E of the fast electron should be atleast \(.6m{ c }^{ 2 }\), where m is the mass of electron and c is the speed of light.

Note by Tushar Gopalka
2 years, 9 months ago

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Yep, @Tushar Gopalka solved, actually question meant that due to the collision, the energy released should be converted to positron and electron , the preexisting electrons are still present, hence charge conserved, heres the solution.,

Let the initial spped be u, corresponding gamma factor is \(\gamma \) , and mass of electron 'm', we wish that it should create a pair,, so on RHS, we have 3 electrons and one posistron, each has mass 'm',

for min energy, we want maximum inelasticity in colision so that most energy is lost by the 1st electron and this is used for pair creation, hence we want all 4 particles to move in same line after collision,,

Let them move with velocity 'v' and have gamma factor \(\alpha =\frac { 1 }{ \sqrt { 1-{ v }^{ 2 } } } \)

hence, conserving momentum and energy (note however that in relativity, momentum is given as \(\gamma mv\) and energy as \(\gamma m{ c }^{ 2 }\) )

*also i take c=1 , for convenient calulation *

now here we go,

\(\gamma m+m=4\alpha m\)


\(\gamma mv\quad =4mv\alpha \)

dividing both

\(v=\frac { \gamma }{ \gamma +1 } u\)

substituting this in 2nd equation and using \(\varepsilon =\frac { 1 }{ \sqrt { 1-{ v }^{ 2 } } } \)

we have

\(16=\frac { 1-{ u }^{ 2 } }{ 1-{ u }^{ 2 } } +2\gamma +1=2+2\gamma \quad or\quad \gamma =7\\ \\ hence\quad KE\quad =\quad (\gamma m-m){ c }^{ 2 }=6m{ c }^{ 2 }\quad \)

hence proved

Mvs Saketh - 2 years, 9 months ago

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Yeah ,, see using concept of pseudoforces,, the forces on the man are centripetal and coriolis forces,, now the centripetal force acts radially outward (since the corresponding acceleration is radially inward) and it has magnitude mw(^2)R

, now the coriolis force is is given by -2m(w x v ) (where x means cross product) , you can check that it acts inward and it has magnitude 2mw(^2)R , hence net force is inward which is mw(^2)R and hence the man rotates in a circle ,,,

It is in opposite sense which is fairly obvious due to relative velocity

i shall solve the second one and report back , but you are sure that is the exact wording of the problem? because charge must be conserved,

also can you please provide me the source to those INPHO papers,, i only got recent ones from net

Mvs Saketh - 2 years, 9 months ago

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Yes the wording is correct.

Tushar Gopalka - 2 years, 9 months ago

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You have to buy it from hbcse website. But it takes two months to get the delivery

Tushar Gopalka - 2 years, 9 months ago

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got it, on it, if i succeed, i shall tell

Mvs Saketh - 2 years, 9 months ago

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Comment deleted Jan 21, 2015

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Anywhere on the ground what difference does it make

Rohit Shah - 2 years, 9 months ago

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