| \[{{S}_{1}}:\] None zero work has to be done on a moving particle to change its momentum. |
| \[{{S}_{2}}:\] To change momentum of a particle a non-zero net force should act on it. |
| \[{{S}_{3}}:\] Two particles undergo rectilinear motion along different straight lines. Then the centre of mass of system of given two particles also always moves along a straight line. |
| \[{{S}_{4}}:\] If direction of net momentum of a system of particles (having nonzero net momentum) is fixed, the centre of mass of given system moves along a straight line. |
A) TTFT
B) FTFT
C) FTTT
D) FFTF
Correct Answer: B
Solution :
\[{{S}_{1}}\] : No work is done by net force, it only changes direction of momentum of particle. Hence \[{{S}_{1}}\] is false. \[{{S}_{2}}\]: True by definition. \[{{S}_{3}}\]: Nothing is said about acceleration of both particles. Hence angle between velocity and acceleration of centre of mass may not be zero. Consequently centre of mass may not move along a straight line. Hence \[{{S}_{3}}\] is false. \[{{S}_{4}}:{{\vec{V}}_{cm}}=\frac{{{m}_{1}}{{{\vec{v}}}_{1}}+{{m}_{2}}{{{\vec{v}}}_{2}}+....+{{m}_{n}}{{{\vec{v}}}_{n}}}{{{m}_{1}}+{{m}_{2}}+.....+{{m}_{n}}}=\frac{{{{\vec{F}}}_{net}}}{({{m}_{1}}+{{m}_{2}}+.....+{{m}_{B}})}\] Direction at \[{{\vec{P}}_{net}}\]. is fixed so \[{{\vec{V}}_{cm}}\] is also constant in the direction. So path of CM will be straight line.
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