TS EAMCET · Physics · Center of Mass Momentum and Collision
A \(2 \mathrm{~kg}\) ball moving at \(24 \mathrm{~ms}^{-1}\) undergoes inelastic head-on collision with a \(4 \mathrm{~kg}\) ball moving in the opposite direction at \(48 \mathrm{~ms}^{-1}\). If the coefficient of restitution is \(2 / 3\), their velocities in \(\mathrm{ms}^{-1}\) after impact are
- A \(-56,-8\)
- B \(-28,-4\)
- C \(-14,-2\)
- D \(-7,-1\)
Answer & Solution
Correct Answer
(A) \(-56,-8\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \begin{aligned} & v_1^{\prime}= {\left[\frac{m_1-e m_2}{m_1+m_2}\right] v_1-\left[\frac{(1+e) m_2}{m_1+m_2}\right] v_2 } \\ &=\left[\frac{2-\left(\frac{2}{3}\right) 4}{2+4}\right] 24-\left[\frac{\left(1+\frac{2}{3}\right) 4}{2+4}\right] 48 \\ &=-56 \mathrm{~m}…
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-
Momentum before collision,
\(
p_1=4 m v_1-2 m v_2
\)
Momentum after collision,
\(
p_2=6 m\left(-5 v_2\right)=-30 m v_2
\)
Conservation of momentum gives,
\(
\begin{gathered}
p_1=p_2 \\
\Rightarrow \quad 4 m v_1-2 m v_2=-30 m v_2 \quad \Rightarrow 4 v_1=-32 v_2 \\
\Rightarrow \quad \frac{v_1}{v_2}=\frac{-32}{4}=-8 \text { or }\left|\frac{v_1}{v_2}\right|=8
\end{gathered}
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