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"The string is cut. Find the acceleration of each block just after." This one line turns an easy equilibrium problem into one that many JEE and NEET students get wrong, usually by letting the spring force vanish along with the string. This post gives the one rule that settles these questions and applies it to four set-ups, including a pair that look alike but have different answers.
Is this in the syllabus?
Yes, as an application of Newton's second law. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list Newton's laws of motion and "equilibrium of concurrent forces" in Unit 3. Section 4.9 of the NCERT Class 11 Physics chapter Laws of Motion (Reprint 2026-27) introduces the two forces involved, tension and the spring force.
Why does a spring force not change instantly?
NCERT writes the spring force as F = −kx, where x is the displacement from the unstretched state. So the force depends only on how much the spring is stretched. Stretch changes only when the blocks move, and a block cannot move a finite distance in zero time. So, just after the cut, every spring has the same stretch, and the same force, as just before.
A string is different. NCERT notes that an inextensible string has a very high force constant. A tiny change in its length is enough to change its tension by a lot, so string tension can jump in an instant. Once a string is cut, its tension is zero.
Take g = 10 m/s² throughout.
Worked example 1: cut the string below a spring
Block A (2 kg) hangs from the ceiling by a spring. Block B (3 kg) hangs from A by a string.
Before the cut, the string holds B: T = 3 × 10 = 30 N. The spring holds both: S = 5 × 10 = 50 N.
Cut the string.
B: only its weight acts, so a_B = 10 m/s² downward.
A: the spring still pulls up with 50 N, and its weight is 20 N down. Net force 30 N up, so a_A = 30 ÷ 2 = 15 m/s² upward.
The spring's force was set for 5 kg, and it now acts on 2 kg, so A shoots up.
Worked example 2: cut the spring instead
Same arrangement, but now the spring is cut. Its force becomes zero, because the spring itself is gone. A and B are joined by a string with nothing else holding them. Both fall with a = 10 m/s², and the string goes slack: T = 0.
Worked example 3: string on top, spring below
Now the order is reversed. A (2 kg) hangs from the ceiling by a string, and B (3 kg) hangs from A by a spring. Before the cut, the spring force is 30 N and the string tension is 50 N.
Cut the string.
B: the spring still pulls up with 30 N, equal to its weight. Net force zero, so a_B = 0.
A: its weight, 20 N, plus the spring pulling down with 30 N. Net 50 N down, so a_A = 50 ÷ 2 = 25 m/s² downward.
A falls faster than g, which surprises many students. It is the spring dragging it down.
Worked example 4: spring or string, held sideways
A 0.5 kg ball hangs from a spring that makes 37° with the vertical, and a horizontal string holds it in place. Use sin 37° ≈ 0.6 and cos 37° ≈ 0.8.
Before the cut, vertical balance gives S × 0.8 = 5 N, so S = 6.25 N. Horizontal balance gives T = S × 0.6 = 3.75 N.
Cut the horizontal string. The spring keeps its 6.25 N, so the net force is what the string used to cancel: 3.75 N horizontal, away from where the string was. a = 3.75 ÷ 0.5 = 7.5 m/s².
Now swap the spring for a string at the same angle. At the instant of cutting, the ball is at rest, so it starts to swing on a circle. Its acceleration must be along the tangent. Tension adjusts at once to cancel the weight's component along the string: T = mg cos 37° = 4 N. The net force is mg sin 37° = 3 N, perpendicular to the string, so a = 6.0 m/s².
Same picture, different answer: 7.5 m/s² with a spring, 6.0 m/s² with a string.
Which mistakes cost marks?
Letting the spring force vanish. Only a cut spring loses its force.
Keeping the old tension in an uncut string. Tension re-adjusts instantly; work it out from the constraint.
Adding centripetal terms. Just after release the speed is zero, so v²/r = 0.
Using the whole system's mass for a single block. Draw a separate free-body diagram for each.
Thinking no body can fall faster than g. Example 3 shows a spring pulling one down.
Our guide to free-body diagrams and Newton's second law covers drawing each diagram, and the spring-block energy post covers what happens after the first instant.
A practice routine
For examples 1 and 3, change the masses to 4 kg and 1 kg and redo both before checking.
Draw before-and-after free-body diagrams for every cut problem, side by side.
Write the one-line rule from the callout on the first page of your mechanics notebook.
Solve three past-paper cut-the-string questions, marking which forces stayed and which changed.
When you cannot see which force changes, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Try this now: in example 3, what are the accelerations if the spring is cut instead of the string?
