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(B) Let the masses be $m_A$ and $m_B$,initial velocities be $u_A$ and $u_B$,and the retarding force be $F$. Since the force is the same,the retardatio

Vedclass · Accepted Answer

$6: 1$

Vedclass · Answer

(B) Let the masses be $m_A$ and $m_B$,initial velocities be $u_A$ and $u_B$,and the retarding force be $F$. Since the force is the same,the retardation $a = F/m$ is different for both. For $A$: $a_A = F/m_A$,$v_A = 0$,$t_A = 2t_B$,$s_A = \frac{2}{3} s_B$. Using $v = u + at$,we get $0 = u_A - (F/m_A)(2t_B) \implies u_A = (2Ft_B)/m_A$. Using $s = ut + \frac{1}{2}at^2$,we get $s_A = u_A(2t_B) - \frac{1}{2}(F/m_A)(2t_B)^2 = (4Ft_B^2)/m_A - (2Ft_B^2)/m_A = (2Ft_B^2)/m_A$. For $B$: $a_B = F/m_B$,$v_B = 0$,$t_B$,$s_B$. Similarly,$u_B = (Ft_B)/m_B$ and $s_B = (Ft_B^2)/(2m_B)$. Given $s_A = \frac{2}{3} s_B$,we have $(2Ft_B^2)/m_A = \frac{2}{3} 	imes (Ft_B^2)/(2m_B)$. Simplifying,$2/m_A = 1/(3m_B) \implies m_A/m_B = 6/1$. Thus,the ratio of masses $m_A:m_B$ is $6: 1$.

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