Physics Q29 – JEE-MAIN 2025

Home Physics Work, Energy & PowerQuestion #656

Consider two blocks A and B of masses

m_{1} = 10

kg and

m_{2} = 5

kg that are placed on a frictionless table. The block A moves with a constant speed

v = 3

m/s towards the block B kept at rest. A spring with spring constant

k = 3000

N/m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is. (Neglect the mass of the spring)

Clue 1 of 3: Collision Dynamics

In a collision involving a spring, the maximum compression occurs when the relative velocity between the colliding objects becomes zero, meaning they move with a common velocity.

🥷

Ninja Strategy•Energy Upper Bound

Calculate the maximum possible compression if all initial kinetic energy of the moving block were converted into spring potential energy; any option exceeding this value can be eliminated.

🧠 Full Solution Path

Step 1: Calculate the common velocity of the blocks✦ Active

Apply the conservation of linear momentum for the system of blocks A and B. The initial momentum of block A is $m_{1} v_{1}$ , and block B is at rest. At maximum compression, both blocks move together with a common velocity $V_{c o m m o n}$ .

m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) V_{c o m m o n}

10 kg \times 3 m/s + 5 kg \times 0 m/s = (10 kg + 5 kg) V_{c o m m o n}

30 = 15 V_{c o m m o n}

V_{c o m m o n} = 2 m/s

Step 2: Apply conservation of mechanical energy to find maximum compression○ Expand

The total mechanical energy of the system (blocks A, B, and spring) is conserved from the initial state (block A moving, block B at rest, spring uncompressed) to the state of maximum compression (both blocks moving with common velocity $V_{c o m m o n}$ , spring compressed by $x_{m a x}$ ). The initial potential energy of the spring is zero.

\frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} (m_{1} + m_{2}) V_{c o m m o n}^{2} + \frac{1}{2} k x_{m a x}^{2}

\frac{1}{2} (10) (3)^{2} + \frac{1}{2} (5) (0)^{2} = \frac{1}{2} (10 + 5) (2)^{2} + \frac{1}{2} (3000) x_{m a x}^{2}

45 = 30 + 1500 x_{m a x}^{2}

💡 Teacher's Secret Hint

Remember to include both kinetic energy of the combined mass and potential energy of the spring in the final energy state.

Step 3: Solve for the compression in the spring○ Expand

From the energy conservation equation, isolate and solve for $x_{m a x}$ .

15 = 1500 x_{m a x}^{2}

x_{m a x}^{2} = \frac{15}{1500} = \frac{1}{100}

x_{m a x} = \sqrt{\frac{1}{100}} = 0.1 m

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Physics Question 29 – JEE-MAIN 2025

Clue 1 of 3: Collision Dynamics

🧠 Full Solution Path