JEE Physics 2025 Question with Solution

Given: $m_1 = 10 \, \text{kg}$, $m_2 = 5 \, \text{kg}$, initial velocity of block $A$ is $v_1 = 3 \, \text{m/s}$, initial velocity of block $B$ is $v_2 = 0 \, \text{m/s}$, and spring constant $k = 3000 \, \text{N/m}$.

Find: The compression $x$ in the spring when both blocks move together with the spring in compressed state.

Using conservation of linear momentum:

$$m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_{cm}$$ $$(10 \, \text{kg})(3 \, \text{m/s}) + (5 \, \text{kg})(0 \, \text{m/s}) = (10 \, \text{kg} + 5 \, \text{kg})v_{cm}$$ $$30 \, \text{kg m/s} = 15 \, \text{kg} \cdot v_{cm}$$ $$v_{cm} = \frac{30}{15} = 2 \, \text{m/s}$$

The kinetic energy lost during the inelastic collision is stored as spring potential energy.

Initial kinetic energy:

$$KE_i = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(10 \, \text{kg})(3 \, \text{m/s})^2 = 45 \, \text{J}$$

Final kinetic energy:

$$KE_f = \frac{1}{2}(m_1 + m_2)v_{cm}^2 = \frac{1}{2}(15 \, \text{kg})(2 \, \text{m/s})^2 = 30 \, \text{J}$$

So, energy stored in the spring is:

$$PE_{\text{spring}} = KE_i - KE_f = 45 \, \text{J} - 30 \, \text{J} = 15 \, \text{J}$$

Now,

$$PE_{\text{spring}} = \frac{1}{2}kx^2$$ $$15 = \frac{1}{2}(3000)x^2$$ $$15 = 1500x^2$$ $$x^2 = \frac{15}{1500} = \frac{1}{100}$$ $$x = \sqrt{\frac{1}{100}} = \frac{1}{10} \, \text{m} = 0.1 \, \text{m}$$

Therefore, the compression in the spring is $0.1 \, \text{m}$. The correct option is C.

Given: Block $A$ of mass $10 \, \text{kg}$ moves with speed $3 \, \text{m/s}$ toward block $B$ of mass $5 \, \text{kg}$ at rest. The spring constant is $3000 \, \text{N/m}$.

Find: Spring compression $x$.

First compute the common velocity after collision from momentum conservation:

$$p_{\text{initial}} = m_1 \cdot v = 10 \, \text{kg} \cdot 3 \, \text{m/s} = 30 \, \text{kg m/s}$$ $$p_{\text{final}} = (m_1 + m_2)v_f = (10 + 5) \, \text{kg} \cdot v_f = 15 \, \text{kg} \cdot v_f$$ $$30 = 15v_f \Rightarrow v_f = 2 \, \text{m/s}$$

Now compare kinetic energies.

Initial kinetic energy:

$$KE_{\text{initial}} = \frac{1}{2}m_1 v^2 = \frac{1}{2} \cdot 10 \cdot 3^2 = 45 \, \text{J}$$

Final kinetic energy:

$$KE_{\text{final}} = \frac{1}{2}(m_1 + m_2)v_f^2 = \frac{1}{2} \cdot 15 \cdot 2^2 = 30 \, \text{J}$$

Hence the energy stored in the spring is:

$$\text{PE}_{\text{spring}} = KE_{\text{initial}} - KE_{\text{final}} = 45 - 30 = 15 \, \text{J}$$

For the spring,

$$\frac{1}{2}kx^2 = 15$$ $$\frac{1}{2} \cdot 3000 \cdot x^2 = 15$$ $$1500x^2 = 15 \Rightarrow x^2 = 0.01$$ $$x = \sqrt{0.01} = 0.1 \, \text{m}$$

Therefore, the spring compression is $0.1 \, \text{m}$.