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प्रश्न
A railroad car of mass M is at rest on frictionless rails when a man of mass m starts moving on the car towards the engine. If the car recoils with a speed v backward on the rails, with what velocity is the man approaching the engine?
उत्तर
Given:
The mass of the railroad car is M.
The mass of the man is m.
The car recoils with a speed v, backwards on the rails.
Let the man of mass m approaches towards the engine with a velocity v' w.r.t the engine.
∴ The velocity of man w.r.t earth is v' − v, towards right.
\[V_{centre of mass} = 0 (\text{Initially at rest })\]
\[ \therefore 0 = - Mv + m(v' - v)\]
\[ \Rightarrow Mv = m(v' - v)\]
\[ \Rightarrow mv' = Mv + mv\]
\[ \Rightarrow v' = \left( \frac{M + m}{m} \right)v\]
\[ \Rightarrow v' = \left( 1 + \frac{M}{m} \right)v\]
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-
Show K = K′ + 1/2MV2
where K is the total kinetic energy of the system of particles, K′ is the total kinetic energy of the
system when the particle velocities are taken with respect to the centre of mass and MV2/2 is the
kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the
system). The result has been used in Sec. 7.14. - Show where `"L""'" = sum"r""'"_"t" xx "p""'"_"t"` is the angular momentum of the system about the centre of mass with
velocities taken relative to the centre of mass. Remember `"r"_"t" = "r"_"t" - "R"`; rest of the notation is the standard notation used in the chapter. Note L′ and MR × V can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles. - Show `"dL"^"'"/"dt" = ∑"r"_"i"^"'" xx "dP"^"'"/"dt"`
Further show that `"dL"^'/"dt" = τ_"ext"^"'"`
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(Hint: Use the definition of centre of mass and third law of motion. Assume the internal forces between any two particles act along the line joining the particles.)
