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Grade: 11
        
 
question 37 of chapter 9 centre of mass and collision
7 months ago

Answers : (1)

we will so
101 Points
							
Final K.E. = 0.2J
Initial K.E. = ½ mV1
2 + 0 = ½ × 0.1 u2 = 0.05 u2
mv1 = mv2 = mu
Where 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 = 
2
u
2
u
u  
v1 =
2
u
(1 – ℓ)
Given (1/2)mv1
2 +(1/2)mv2
2 = 0.2
v1
2 + v2
2 = 4
(1 ) 4
4
u
(1 )
4
u 2
2
2
2
       (1 )
2
u 2
2
  = 4  u2 = 1 2
8
 
For maximum value of u, denominator should be minimum,
 ℓ = 0.
 u2 = 8  u = 2 2 m/s
For minimum value of u, denominator should be maximum,
 ℓ = 1
u2 = 4  u = 2 m/s
7 months ago
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