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question 41 of chapter 9 centre of mass and collision

question 41 of chapter 9 centre of mass and collision

Grade:11

1 Answers

we will so
101 Points
6 years ago
Mass of each block MA and MB = 2kg.
Initial velocity of the 1st block, (V) = 1m/s
VA = 1 m/s, VB = 0m/s
Spring 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 + the
compound spring + Block B) move together with a common velocity.
Let that velocity be V.
Using conservation of energy, (1/2) MAVA
2 + (1/2)MBVB
2 = (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

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