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If a body of mass ‘m’ elastically collide with another body of mass ‘nm’ at rest $(n > 1)$ . Then
List 1 List 2
I) Fractional loss of K.E is 8/9, if n is equal to P) n=3
II) If after the collision, the first body moves with 3/4^th of its previous speed, then n is equal to Q) n=2
III) Maximum transfer of K.E of first body when n equal to R) n=1
IV) Fraction loss of K.E of first body is 3/4, then n is equal to S) n=7
T) n=5

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$(I) \to (Q); (II) \to (S); (III) \to (R); (IV) \to (P)$

$(I) \to (R); (II) \to (Q); (III) \to (P); (IV) \to (T)$

$(I) \to (S); (II) \to (Q); (III) \to (R); (IV) \to (P)$

$(I) \to (Q); (II) \to (P); (III) \to (T); (IV) \to (R)$

answer is A.

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Detailed Solution

A) fractional loss in $K E = 1 - {(\frac{v_{1}}{u_{1}})}^{2} = \frac{8}{9} o n$ solving n=2

B) if

$\begin{array}{l} V_{1} = \frac{u_{2}}{2} = (\frac{m - n m}{m + n m}) u_{1} \Rightarrow \frac{1 - n}{1 + n} = \frac{1}{2} \\ \Rightarrow 2 - 2 n = 1 + n \Rightarrow 1 = 3 n \Rightarrow n = 1 / 3 \end{array}$

C) If $m_{1} = m_{2}$ maximum K.E transfers

D) $\begin{array}{l} 1 - {(\frac{v_{1}}{u_{1}})}^{2} = \frac{3}{4} \Rightarrow n = 3 \\ v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1}, V_{2} = \frac{2 m_{1} u_{1}}{(m_{1} + m_{2})} \end{array}$

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