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A block is released on a smooth wedge that sits on a smooth floor. As the block slides down, it pushes the wedge backwards. Most wrong answers to this classic JEE problem come from one assumption: that the block still moves along the incline and that the normal force is still mg cos θ. Neither is true here. This post shows the constraint that replaces them, works one example fully and checks it three ways.
Where does this problem come from?
The JEE (Main) 2026 syllabus lists Newton's laws and conservation of linear momentum under Unit 3, and the JEE (Advanced) 2026 brochure lists Newton's laws and "Inertial and uniformly accelerated frames of reference" under mechanics. The NCERT Class 11 chapter on laws of motion has inclined planes but no movable wedge, so this is a JEE extension of school material. Check the syllabus for your exam year.
What is the constraint?
The block stays in contact with the wedge's surface. So relative to the wedge, the block can only slide along the incline. Written in the ground frame:
block's acceleration = wedge's acceleration + acceleration along the incline (relative to the wedge).
A second way to say the same thing: perpendicular to the incline, the block and the wedge must have the same component of acceleration, or they would separate or overlap.
This is the same idea as in pulley constraint relations: geometry fixes how the accelerations are linked before any forces are written.
Worked example: a block on a free wedge
A 2 kg block slides on a smooth 4 kg wedge of angle 37°, which rests on a smooth floor. Take g = 10 m/s², sin 37° = 0.6 and cos 37° = 0.8. Find the wedge's acceleration A, the block's acceleration relative to the wedge a_r, and the normal force N. (Sample masses for practice.)
Let the incline slope down to the right, so the block slides right and the wedge is pushed left.
Step 1: the wedge. The only horizontal force on the wedge is the horizontal part of the block's push, N sin θ. So N sin θ = MA, that is 0.6N = 4A.
Step 2: the block, perpendicular to the incline. The block's acceleration has a component perpendicular to the incline coming only from the wedge's motion: A sin θ, directed into the wedge, because the surface moves away from under the block. Taking that direction as positive, mg cos θ − N = mA sin θ, that is 16 − N = 1.2A.
Step 3: solve. From step 1, N = 4A/0.6 ≈ 6.667A. Then 16 = 6.667A + 1.2A = 7.867A, so A ≈ 2.03 m/s² and N ≈ 13.6 N.
Step 4: the block, along the incline. mg sin θ = m(a_r − A cos θ), so a_r = g sin θ + A cos θ = 6 + 2.03 × 0.8 ≈ 7.63 m/s².
The general results are A = mg sin θ cos θ/(M + m sin² θ) and N = Mmg cos θ/(M + m sin² θ). With these numbers: 9.6/4.72 ≈ 2.03 m/s² and 64/4.72 ≈ 13.6 N.
How do you check the answer?
Check 1: momentum. No horizontal external force acts on the block-plus-wedge system, so horizontal momentum stays zero. The block's horizontal acceleration is a_r cos θ − A = 6.10 − 2.03 = 4.07 m/s². Then m × 4.07 ≈ 8.14 and M × A ≈ 8.14. They match.
Check 2: a limit. Let the wedge become very heavy (M much larger than m). Then A goes to 0, N goes to mg cos θ and a_r goes to g sin θ: the fixed-incline results, as expected.
Check 3: the path. In the ground frame, the block's acceleration has components 4.07 m/s² across and 7.63 × 0.6 ≈ 4.58 m/s² down. Its path is at about 48° below the horizontal, steeper than the 37° incline. That is the visual proof that the block does not move along the incline as seen from the ground.
How far does the wedge move?
Suppose the block slides 1.5 m along the incline, starting from rest. Because horizontal momentum stays zero, the centre of mass does not move sideways. The block moves 1.5 cos θ = 1.2 m to the right relative to the wedge. If the wedge moves X to the left, then m(1.2 − X) = MX, so 2(1.2 − X) = 4X and X = 0.4 m.
Check with the accelerations: both start from rest and accelerate for the same time, so X/1.5 = A/a_r = 2.03/7.63 ≈ 0.267, giving X ≈ 0.4 m.
Could you solve it in the wedge's frame?
Yes. The wedge accelerates, so in its frame you add a pseudo force mA on the block, pointing opposite to the wedge's acceleration. Our post on pseudo forces shows the method for a wedge pushed at a set acceleration. Here A is unknown, so you still need the wedge's own equation. Pick one frame and stay in it.
Common mistakes and how to avoid them
Taking N = mg cos θ. Only true if the wedge is fixed.
Using g sin θ as the block's ground acceleration. That is its relative acceleration only for a fixed wedge.
Sign slips in the constraint. Draw the wedge's acceleration arrow and resolve it along and perpendicular to the incline on the diagram.
A practice routine
Day 1: Redo the example with M = 2 kg. You should get A = 9.6/2.72 ≈ 3.53 m/s².
Day 2: Solve the same problem in the wedge's frame and confirm the answers.
Day 3: Two problems where the wedge is pushed by a known force F, so the constraint stays but the wedge equation gains a term.
Day 4: One problem with friction between block and wedge. Check first whether the block slides at all.
When you are unsure whether your constraint equation is right, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare its constraint line with yours.
Tonight, compute N for M = 2 kg and check it is about 11.8 N. Then explain in one sentence why a lighter wedge pushes less.
