A steel sphere of radius 1.2 cm collides wit | vedclass

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(D) Let the mass of the first sphere be $m_1$ and the second be $m_2$. Since both are steel spheres,their density $\rho$ is the same. Thus,$m = \rho V

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

(D) Let the mass of the first sphere be $m_1$ and the second be $m_2$. Since both are steel spheres,their density $\rho$ is the same. Thus,$m = \rho V = \rho (\frac{4}{3} \pi r^3)$.So,$m_1 \propto r_1^3$ and $m_2 \propto r_2^3$.For an elastic collision in one dimension where the second body is at rest,the final velocity of the first body $v_1'$ is given by $v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1$.Given $v_1' = \frac{7}{9} v_1$,we have $\frac{m_1 - m_2}{m_1 + m_2} = \frac{7}{9}$.Cross-multiplying: $9m_1 - 9m_2 = 7m_1 + 7m_2$.$2m_1 = 16m_2 \implies m_1 = 8m_2$.Since $m \propto r^3$,we have $r_1^3 = 8r_2^3$.Taking the cube root: $r_1 = 2r_2$.Given $r_1 = 1.2 \ cm$,we get $1.2 = 2r_2 \implies r_2 = 0.6 \ cm$.

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