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`        question 41 of chapter 9 centre of mass and collision`
2 years ago

we will so
101 Points
```							Mass of each block MA and MB = 2kg.Initial velocity of the 1st block, (V) = 1m/sVA = 1 m/s, VB = 0m/sSpring constant of the spring = 100 N/m.The block A strikes the spring with a velocity 1m/s/After the collision, it’s velocity decreases continuously and at a instant the whole system (Block A + thecompound spring + Block B) move together with a common velocity.Let that velocity be V.Using conservation of energy, (1/2) MAVA2 + (1/2)MBVB2 = (1/2)MAv2 + (1/2)MBv2 + (1/2)kx2.(1/2) × 2(1)2 + 0 = (1/2) × 2× v2 + (1/2) × 2 × v2 + (1/2) x2 × 100(Where x = max. compression of spring) 1 = 2v2 + 50x2 …(1)As there is no external force in the horizontal direction, the momentum should be conserved. MAVA + MBVB = (MA + MB)V. 2 × 1 = 4 × v V = (1/2) m/s. …(2)Putting in eq.(1)1 = 2 × (1/4) + 50x+2+ (1/2) = 50x2 x2 = 1/100m2 x = (1/10)m = 0.1m = 10cm
```
2 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions