Spring-Mass System PYQ | JEE Main Physics 2025 | Acceleration Trick

 

❓ Question

Two masses $m_1$ and $m_2$ are connected through two springs as shown. Initially, the system is in equilibrium.

If the lower spring $k_2$ is suddenly cut, find the initial acceleration of mass $m_1$.

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✍️ Solution

Initially, the system is in equilibrium.

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Equilibrium Condition for $m_2$

For mass $m_2$,

$$T_2=m_2g.$$

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Equilibrium Condition for $m_1$

For mass $m_1$,

$$T_1=m_1g+T_2.$$

Substituting $T_2=m_2g$,

$$T_1=(m_1+m_2)g.$$

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After Cutting Spring $k_2$

Immediately after the spring is cut,

• The extension of the upper spring does not change instantaneously.
• Hence the force exerted by the upper spring remains

$$T_1=(m_1+m_2)g.$$

Also,

$$T_2=0.$$

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Applying Newton's Second Law to $m_1$

The forces on $m_1$ are:

• Upward: $T_1$
• Downward: $m_1g$
  

Therefore,

$$T_1-m_1g=m_1a.$$

Substitute $T_1=(m_1+m_2)g$

$$(m_1+m_2)g-m_1g=m_1a.$$
$$m_2g=m_1a.$$

Hence,

$$\boxed{a=\frac{m_2g}{m_1}}.$$

The acceleration is upward.

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✅ Final Answer

$$\boxed{a=\frac{m_2g}{m_1}\ \text{(upward)}}$$