m1v1+m2v2=(m1+m2)vcmvcm⇒10+510×3⇒1530=2 m/s
21kx2=21(10)(3)2−[21(15)(2)2]
⇒90−60=30=3000x2
x2⇒300030=1001x⇒101 m.
JEE Main 2025 Physics, Center of Mass Momentum and Collision — question figure Consider two blocks A and B of masses m_1=10 ~kg and m_2=5 ~kg that…
JEE Main 2025 — Physics Mechanics
2025mcqmedium
Consider two blocks A and B of masses m1=10 kg and m2=5 kg that are placed on a frictionless table. The block A moves with a constant speed v=3 m/s towards the block B kept at rest. A spring with spring constant k=3000 N/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)
Official previous-year question
Held on 3 Apr 2025 · Verified 6 Jul 2026.
Options
- A
0.2 m
- B
0.4 m
- C
0.1 m
- D
0.3 m
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