Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

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Question 4 of 130


Consider two blocks A and B of masses m1=10 kgm_1=10 \mathrm{~kg} and m2=5 kgm_2=5 \mathrm{~kg} that are placed on a frictionless table. The block A moves with a constant speed v=3 m/sv=3 \mathrm{~m} / \mathrm{s} towards the block B kept at rest. A spring with spring constant k=3000 N/m\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)

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Answer: (C) 0.1 m0.1\mathrm{~m}

m1v1+m2v2=(m1+m2)vcmvcm⇒10×310+5⇒3015=2 m/s\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}
12kx2=12(10)(3)2−[12(15)(2)2]\frac{1}{2} \mathrm{kx}^2=\frac{1}{2}(10)(3)^2-\left[\frac{1}{2}(15)(2)^2\right]
⇒90−60=30=3000x2\Rightarrow 90-60=30=3000 x^2
x2⇒303000=1100x⇒110 m.\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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