Question

The friction coefficient between the horizontal surface and each of the blocks shown in figure is 0.2. The collision between the blocks is perfectly elastic. Find the separation between them when they come to rest.

(A) 2 cm

(B) 5 cm

(C) 8 cm

(D) 11 cm

Answer 1

Correct option is (B) 5 cm

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

Given, V₂ = 0

(velocity of 4 kg block)

∴ (1/2) m × u₁² − (1/2)m × (1)² = −m × mg × S

⇒ u₁ =√(1)² − 2 × 0.20 × 10 × 0.16

= 0.6 m/sec

⇒ u₁ = 0.6 m/s

Since it is a perfectly elastic collision.

Let v₁, v₂ → velocity of 2 kg and 4 kg block after collision,

m₁ × u₁ + m₂u₂ = m₁v₁ + m₂v₂

⇒ 2 × 0.6 + 4.0 = 2v₁ + 4v₂

⇒ 2v₁ + 4v₂ = 1.2    ...(i)

Again v₁ − v₂ = + (u₁ − u₂)

= +(0.6 − 0)

= −0.6     ...(ii)

Subtracting (ii) from (i),

3v₂ = 1.2

⇒ v₂ = 0.4 m/s

∴ v₁ = −0.6 + 0.4

= −0.2 m/s

∴ Putting work energy principle for 1st 2kg block when comes to rest

(1/2) × 2 × (0)² + (1/2) × 2 × (0.2)² = −2 × 0.2 × 10 × S₁

⇒ S₁ = 1 cm.

Putting work energy principle for 4 kg block

(1/2) × 4 × (0)² − (1/2) × 4 × (0.4)²= −4 × 0.2 × 10 × S₂

⇒ 2 × 0.4 × 0.4 = 4 × 0.2 × 10 × S₂

⇒ S₂ = 4 cm

Distance between 2 kg and 4 kg block = S₁ + S2

= 1 + 4 

= 5 cm