A block of mass (10 ) g , where is a constant is moving with velocity 3 ~m / s to the right collides inelastically with the block on the right with mass 10 ~g and sticks to it. The right block is connected to three springs as shown in the figure. The spring constant of each spring is 2 ~N / m . If the amplitude of the resulting simple harmonic motion is 1 2 2 m , then the value of is

Step-by-step solution

Equivalent spring constant of system of springs = 3 k 2 =3 ~N / m . Momentum is conserved in collision. Let blocks moves with velocity v just after collision, then gathered m₁ v₁= (m₁+m₂ ) v 10 10⁻³ 3=(10 +10) 10⁻³ v v= 30 10( +1) = 3 +1 ~ms ⁻¹ gathered Displacement of blocks is 1 2 2 ~m . So, energy conservation gives, aligned & 1 2 (m₁+m₂ ) v^2= 1 2 k_ eq x^2 [friction is absent] & (10 +10) 10⁻³ 9 ^2 ( +1)^2 =3 1 8 & 9 ^2 100( +1) = 3 8 & 24 ^2=100 +100 =5 aligned