JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2025
- Medium
- MCQ
Consider two blocks A and B of masses \(m_1=10 \mathrm{~kg}\) and \(m_2=5 \mathrm{~kg}\) that are placed on a frictionless table. The block A moves with a constant speed \(v=3 \mathrm{~m} / \mathrm{s}\) towards the block B kept at rest. A spring with spring constant \(\mathrm{k}=3000 \mathrm{~N} / \mathrm{m}\) is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is _______. (Neglect the mass of the spring)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(0.1\mathrm{~m}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2=\left(\mathrm{m}_1+\mathrm{m}_2\right) \mathrm{v}_{\mathrm{cm}} \\ & \mathrm{v}_{\mathrm{cm}} \Rightarrow \frac{10 \times 3}{10+5} \Rightarrow \frac{30}{15}=2 \mathrm{~m} / \mathrm{s}\end{aligned}\)
\(\frac{1}{2} \mathrm{kx}^2=\frac{1}{2}(10)(3)^2-\left[\frac{1}{2}(15)(2)^2\right]\)
\(\Rightarrow 90-60=30=3000 x^2\)
\(\begin{aligned} & \mathrm{x}^2 \Rightarrow \frac{30}{3000}=\frac{1}{100} \\ & \mathrm{x} \Rightarrow \frac{1}{10} \mathrm{~m}.\end{aligned}\)
\(\frac{1}{2} \mathrm{kx}^2=\frac{1}{2}(10)(3)^2-\left[\frac{1}{2}(15)(2)^2\right]\)
\(\Rightarrow 90-60=30=3000 x^2\)
\(\begin{aligned} & \mathrm{x}^2 \Rightarrow \frac{30}{3000}=\frac{1}{100} \\ & \mathrm{x} \Rightarrow \frac{1}{10} \mathrm{~m}.\end{aligned}\)
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