Physics · Laws Of Motion · NEET
Yes, as long as the string is inextensible (does not stretch) and stays taut, both bodies move the same distance in the same time, so they share the SAME magnitude of acceleration. Directions differ (one goes up, one goes down over a pulley), but the size of a is identical. This single fact lets you treat the two masses as one system to find a.
First find the common acceleration using the whole system: a = (net force pulling the system) / (total mass). Then pick ONE body, draw its free body diagram, and apply F = ma to that body alone. The tension T appears in that single-body equation. Example: for a hanging mass m going down, mg - T = ma, so T = m(g - a).
Yes, if the string is massless and the pulley is smooth (frictionless), the tension has the same value all along the string, including both sides of the pulley. If the pulley had mass or friction, the two sides would differ, but NEET's standard assumption is a light string over a smooth pulley, so T is one single value.
The heavier block's weight wins. Over a pulley, the heavier mass m1 pulls down harder than the lighter mass m2, so the net force drags the heavier one down and the lighter one up. The common acceleration is a = (m1 - m2)g / (m1 + m2). The system accelerates in the direction of the heavier body.
The moment the string is cut, its tension instantly becomes zero, but a spring (if present) keeps its force for that instant because a spring cannot change length suddenly. So each body's acceleration jumps to whatever the remaining forces give. This is exactly the 2017 NEET trap covered in the PYQ below.
Two blocks A and B of masses 3m and m respectively are connected by a massless inextensible string. The whole system is suspended from a massless spring as shown. The magnitudes of the accelerations of A and B immediately after the string is cut are, respectively:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For masses m1 (heavier) and m2 hanging on either side of a smooth pulley: acceleration a = (m1 - m2)g / (m1 + m2), and tension T = 2 m1 m2 g / (m1 + m2). The system moves in the direction of the heavier mass.
No. NEET assumes a light (massless) inextensible string, so the string has no weight and the tension is the same throughout. This is why you can transfer the same T value across the pulley.
Treat both bodies as one system: a = (hanging weight - friction) / (total mass). Then apply F = ma to one body to get T. In NCERT Example 4.9, a 3 kg hanging block pulls a 20 kg trolley: 30 - T = 3a and T - f = 20a solved together.
Tension is zero if the string goes slack (both sides free-fall together, or the string is cut). If a hanging block is in free fall with nothing opposing, T = 0. This is the key idea in the string-cut PYQ.
Because internal forces like tension cancel when you consider the whole system, so a = (external net force)/(total mass) is fast and avoids errors. Then you only need ONE single-body equation to pull out the tension.
