Question

Two particles \(A\) and \(B\) move from rest along a straight line with constant accelerations \(f\) and f' respectively. If \(A\) takes \(\mathrm{m}\) sec. more than that of \(\mathrm{B}\) and describes \(\mathrm{n}\) units more than that of \(\mathrm{B}\) in acquiring the same velocity, then
(A) \(\left(f+f^{\prime}\right) m^{2}=f f^{\prime} n\)
(B) \(\left(f-f f^{\prime}\right) m^{2}=f f^{\prime} n\)
(C) \(\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^{2}\)
(D) \(\frac{1}{2}\left(f+f^{\prime}\right) m=f f^{\prime} n^{2}\)

Answer 1

Ans: (C)
Hint: \(A: u=0\)
B: \(\quad u=0\)
\(a_{1}=f \quad t_{1}=t+m\)
\(s=n+s\)
\(\mathrm{a}_{2}=\mathrm{f}^{\prime} \quad \mathrm{t}_{2}=\mathrm{t} \quad \mathrm{s}=\mathrm{s}\)
\(s+n=\frac{1}{2} f .(t+m)^{2} \cdots\) (i) and \(s=\frac{1}{2} f^{\prime}(t)^{2} \cdots\) (ii) \(\therefore f^{\prime}(t)^{2}+n=\frac{1}{2} f(t+m)^{2} \cdots\) (iii)
\(v_{1}=u_{1}+a_{1} t_{1}=0+f \cdot(t+m)\)
\(v_{2}=u_{2}+a_{2} t_{2}=0+f^{\prime} . t\)
\(\therefore f(t+m)=f^{\prime} t\)
\(t=\frac{f m}{f^{\prime}-f}\)
from (iii)
\(\frac{1}{2} f^{\prime} \cdot\left(\frac{f m}{f^{\prime}-f}\right)^{2}+n=\frac{1}{2} f\left(\frac{f m}{f^{\prime}-f}+m\right)^{2}\)
\(\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^{2}\)

Answer 2

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