TS EAMCET · Physics · Center of Mass Momentum and Collision

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,
\(
\begin{gathered}
p_1=p_2 \\
\Rightarrow \quad 4 m v_1-2 m v_2=-30 m v_2 \quad \Rightarrow 4 v_1=-32 v_2 \\
\Rightarrow \quad \frac{v_1}{v_2}=\frac{-32}{4}=-8 \text { or }\left|\frac{v_1}{v_2}\right|=8
\end{gathered}
\)
- A 5
- B 2.5
- C 7
- D 10
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
Equivalent spring constant of system of springs \(=\frac{3 k}{2}=3 \mathrm{~N} / \mathrm{m}\). Momentum is conserved in collision. Let blocks moves with velocity \(v\) just after collision, then…
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