Question:
A steel sphere of radius $1.2~\text{cm}$ collides with another steel sphere at rest.
If the collision is elastic and the first sphere moves with $\dfrac{7}{9}$ of its initial velocity after collision,
then the radius of the second sphere is
A steel sphere of radius $1.2~\text{cm}$ collides with another steel sphere at rest.
If the collision is elastic and the first sphere moves with $\dfrac{7}{9}$ of its initial velocity after collision,
then the radius of the second sphere is
If the collision is elastic and the first sphere moves with $\dfrac{7}{9}$ of its initial velocity after collision,
then the radius of the second sphere is
Show Hint
Relate mass ratio to cube of radius when density is constant. Use elastic collision formula.
Updated On: Jun 4, 2025
- $1.8~\text{cm}$
- $2.4~\text{cm}$
- $1.2~\text{cm}$
- $0.6~\text{cm}$
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The Correct Option is D
Solution and Explanation
Let masses be $m_1$ and $m_2$
From elastic collision: $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2} u_1 = \dfrac{7}{9}u_1$
Solving: $\dfrac{m_1 - m_2}{m_1 + m_2} = \dfrac{7}{9}$
Cross-multiplying: $9m_1 - 9m_2 = 7m_1 + 7m_2 \Rightarrow 2m_1 = 16m_2 \Rightarrow \dfrac{m_1}{m_2} = 8$
Mass $\propto$ Volume $\propto r^3$
$\Rightarrow \left( \dfrac{r_1}{r_2} \right)^3 = 8 \Rightarrow \dfrac{r_1}{r_2} = 2 \Rightarrow r_2 = \dfrac{1.2}{2} = 0.6~\text{cm}$
From elastic collision: $v_1 = \dfrac{m_1 - m_2}{m_1 + m_2} u_1 = \dfrac{7}{9}u_1$
Solving: $\dfrac{m_1 - m_2}{m_1 + m_2} = \dfrac{7}{9}$
Cross-multiplying: $9m_1 - 9m_2 = 7m_1 + 7m_2 \Rightarrow 2m_1 = 16m_2 \Rightarrow \dfrac{m_1}{m_2} = 8$
Mass $\propto$ Volume $\propto r^3$
$\Rightarrow \left( \dfrac{r_1}{r_2} \right)^3 = 8 \Rightarrow \dfrac{r_1}{r_2} = 2 \Rightarrow r_2 = \dfrac{1.2}{2} = 0.6~\text{cm}$
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