A rigid block of mass m1 = 10 kg having velocity v0 = 2 m/s strikes a stationary block of mass m2 = 30 kg after travelling 1 m along a frictionless horizontal surface as shown in the figure.

The two masses stick together and jointly move by a distance of 0.25 m further along the same frictionless surface, before they touch the mass-less buffer that is connected to the rigid vertical wall by means of a linear spring having a spring constant k = 10⁵ N/m. The maximum deflection of the spring is _________ cm (round off to 2 decimal places).

The correct answer is 1.

Let's break down the physical process step-by-step to understand how to find the maximum deflection of the spring.

Step 1: Understand the system dynamics before and during collision
Initially, we have:
- A block of mass m₁ = 10 kg moving with an initial velocity v₀ = 2 m/s.
- A stationary block of mass m₂ = 30 kg (velocity is zero).
- Since the horizontal surface is frictionless, mass m₁ travels the distance of 1 m with a constant speed of 2 m/s until it strikes m₂.
- During the collision, the two masses stick together. This is a perfectly inelastic collision, meaning they will move jointly with a common final velocity, let's call it v.

Step 2: Apply the Law of Conservation of Linear Momentum
Since no external horizontal forces act on the system of two blocks during the collision, the total linear momentum is conserved:
Initial momentum = Final momentum

$$m_1{v}_0+m_2(0)=(m_1+m_2)v$$

Substituting the given values:
- m₁ = 10 kg
- v₀ = 2 m/s
- m₂ = 30 kg

$$10\times 2=(10+30)\times v$$

$$20=40v$$

Solving for v:

$$v=\frac{20}{40}=0.5\text{m/s}$$

The combined masses move together with a velocity of 0.5 m/s.

Step 3: Analyze the motion after collision up to the spring compression
The combined mass (m₁ + m₂ = 40 kg) moves jointly by a distance of 0.25 m along the frictionless horizontal surface before touching the mass-less buffer. Since the surface is frictionless, no kinetic energy is lost while traversing this 0.25 m, and the blocks reach the spring buffer with the same velocity of 0.5 m/s.

Step 4: Use Conservation of Energy to find maximum deflection
When the combined mass contacts the mass-less buffer and compresses the spring, the spring exerts a restoring force that slows the blocks down. Maximum deflection, x, occurs when the blocks momentarily come to a complete stop. At this point, all of the kinetic energy of the combined mass is converted into elastic potential energy stored in the spring:
Kinetic Energy of combined mass = Potential Energy of spring at maximum deflection

$$\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}k{x}^2$$

Substituting the values:
- Combined mass (m₁ + m₂) = 40 kg
- Velocity v = 0.5 m/s
- Spring constant k = 10⁵ N/m

$$\frac{1}{2}\times 40\times (0.5{)}^2=\frac{1}{2}\times {10}^5\times x^2$$

We can cancel the factor of 1/2 from both sides:

$$40\times 0.25={10}^5\times x^2$$

$$10={10}^5{x}^2$$

$$x^2=\frac{10}{{10}^5}=\frac{1}{{10}^4}$$

Taking the square root on both sides:

$$x=\sqrt{\frac{1}{{10}^4}}=\frac{1}{100}\text{m}$$

Converting the deflection from meters to centimeters:

$$x=0.01\text{m}=1\text{cm}$$

Therefore, the maximum deflection of the spring is 1 cm.