QJEEAdv18Ph9fbc: JEE Advanced 2018 Physics - Class 11 JEEAdv18Ph9fbc Solution: A spring-block system is resting on a frictionless floor as…

After collision the 2.0 kg block will perform simple harmonic oscillation with time period

$T = 2\pi \sqrt {{m \over k}}$

Given m = 2.0 kg and k = 2.0 N m ^{$-$ 1}, we have

T = 2 $\pi$

Thus, the block returns to its original position in time

$t = {T \over 2}$ = $\pi$ s

That is, t = 3.14 s

Now, if v₁ and v₂ are velocities of 1.0 kg block and 2.0 kg block, respectively, before collision; v'₁ and v'₂ are velocities of 1.0 kg block and 2.0 block, respectively, after collision. So, by conservation of momentum

m₁v₁ + m₂v₂ = m₁v'₁ + m₂v'₂

Here, m₁ = 1.0 kg, v₁ is initial speed of 1.0 kg block = 2.0 m s ^{$-$ 1}, m₂ = 2.0 kg, v₂ is initial speed of 2.0 kg block = 0.0 m s ^{$-$ 1}, v'₁ is final speed of 1.0 kg block after collision and v'₂ is final speed of 2.0 kg block after collision. Then, 1.0 kg $\times$ 2.0 m/s + 2.0 kg $\times$ 0 m/s = 1.0 v'₁ + 2.0 v'₂

v'₁ + 2v'₂ = 2 ..... (1)

Also, using definition of coefficient of restitution

v'₂ $-$ v'₁ = $\varepsilon$ (v₁ $-$ v₂)

Since collision is elastic, So $\varepsilon$ = 1

$\Rightarrow$ v'₂ $-$ v'₁ = v₁ $-$ v₂

$\Rightarrow$ v'₂ $-$ v'₁ = 2 $-$ 0

$\Rightarrow$ v'₂ $-$ v'₁ = 2 ...... (2)

From Eqs. (1) and (2), we get

$v{'_2} = {4 \over 3}$ m s ^{$-$ 1} and $v{'_1} = {-2 \over 3}$ m s ^{$-$ 1}

Therefore, distance between the blocks is given as

$s = v{'_1} \times t = {{ - 2} \over 3} \times 3.14 = 2.09$ m