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Question

The friction coefficient between the horizontal surface and each of the blocks shown in figure is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take g = 10 m/s2.


A

4 cm

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B

3 cm

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C

5 cm

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D

6 cm

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Solution

The correct option is C

5 cm


Let velocity of 2kg block on reaching the 4kg block before collision = u1.

Given, V2 = 0 (velocity of 4kg block).

From work energy principle,

(12)m ×u21(12)m × 12 = m × ug × s

64 × 102 = u211 u1 = 0.6 m/s

Since it is a perfectly elastic collision.

Let V1, V2 velocity of 2kg & 4kg block after collision.

m1V1+m2V2 = m1v1+m2v2

2×0.6+4×0 = 2v1+4v2 v1+2v2 = 0.6 ..........(1)

Again, V1V2 = (u1u2) = -(0.6 - 0) = -0.6 .........(2)

Subtracting (2) from (1)

3v2 = 1.2 v2 = 0.4 m/s.

v1 = -0.6 + 0.4 = -0.2 m/s

Putting work energy principle for 2kg block after collision

(12) × 2 × 02 (12) × 2 × (0.2)2 = 2 × 0.2 × 10 × s

(12) × 2 × 0.2 × 0.2 = 2 × 0.2 × 10 × s S1 = 1cm.

Putting work energy principle for 4kg block.

(12) × 4 × 02 (12) × 4 × (0.4)2 = 4 × 0.2 × 10 × s

(2 × 0.4 × 0.4 = 4 × 0.2 × 10 × s S2 = 4 cm.

Distance between 2kg & 4kg block = S1+S2 = 1 + 4 = 5 cm.


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