A steel ball of radius $2 \mathrm{~cm}$ is at rest on a frictionless surface. Another steel ball of radius…
A steel ball of radius $2 \mathrm{~cm}$ is at rest on a frictionless surface. Another steel ball of radius $4 \mathrm{~cm}$ moving with velocity of $81 \mathrm{cms}^{-1}$ collides elastically with the ball which is at rest. After collision the ball with radius of $2 \mathrm{~cm}$ moves with speed of
$144 \mathrm{~cm} \mathrm{~s}^{-1}$
$72 \mathrm{~cm} \mathrm{~s}^{-1}$
$216 \mathrm{~cm} \mathrm{~s}^{-1}$
$36 \mathrm{~cm} \mathrm{~s}^{-1}$
Solution
Radius of first steel ball,
$
r_1=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}
$
Radius of second steel ball,
$
r_2=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}
$
$\therefore$ Mass of first steel ball,
$
m_1=V_1 \times \rho=\frac{4}{3} \pi r_1^3 \rho \quad[\rho=\text { density }]
$
Similarly, $\quad m_2=\frac{4}{3} \pi r_2^3 \rho$
$\therefore \quad \frac{m_1}{m_2}=\frac{r_1^3}{r_2^3}=\frac{\left(2 \times 10^{-2}\right)^3}{\left(4 \times 10^{-2}\right)^3}=\frac{1}{8}$
$\Rightarrow \quad m_2=8 m_1$
For elastic collision, velocity of first ball,
$
v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right) u_1+\left(\frac{2 m_2}{m_1+m_2}\right) u_2
$
Here, $u_1=0$ and $u_2=81 \mathrm{cms}^{-1}$
$
\begin{aligned}
\therefore \quad v_1 & =\left(\frac{m_1-m_2}{m_1+m_2}\right) \times 0+\left(\frac{2 m_2}{m_1+m_2}\right) \times 81 \\
& =0+\frac{2 \times 8 m_1}{m_1+8 m_1} \times 81 \quad \text { [from Eq. (i)] } \\
& =\frac{16}{9} \times 81=144 \mathrm{~cm} \mathrm{~s}^{-1} \quad
\end{aligned}
$
[from Eq. (i)]