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`         question 37 of chapter 9 centre of mass and collision`
2 years ago

we will so
101 Points
```							Final K.E. = 0.2JInitial K.E. = ½ mV12 + 0 = ½ × 0.1 u2 = 0.05 u2mv1 = mv2 = muWhere v1 and v2 are final velocities of 1st and 2nd block respectively. v1 + v2 = u …(1)(v1 – v2) + ℓ (a1 – u2) = 0  ℓa = v2 – v1 ..(2)u2 = 0, u1= u.Adding Eq.(1) and Eq.(2)2v2 = (1 + ℓ)u  v2 = (u/2)(1 + ℓ) v1 = 2u2uu  v1 =2u(1 – ℓ)Given (1/2)mv12 +(1/2)mv22 = 0.2v12 + v22 = 4(1 ) 44u(1 )4u 2222       (1 )2u 22  = 4  u2 = 1 28 For maximum value of u, denominator should be minimum, ℓ = 0. u2 = 8  u = 2 2 m/sFor minimum value of u, denominator should be maximum, ℓ = 1u2 = 4  u = 2 m/s
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions