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.
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 × 10−2 = u21−1 ⇒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, V1−V2 = −(u1−u2) = -(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.