Newton's Laws of Motion: PYQ Walkthrough (4)

easy • 3 min read
Tags Newtons Laws Of Motion

Question

A block of mass m=4m = 4 kg rests on a smooth horizontal surface. A horizontal force F=20F = 20 N is applied. A second block of mass M=6M = 6 kg is placed in front of the first so the two move together. Find the contact force between the blocks. (NEET 2023, modified)

Solution — Step by Step

Both blocks share the same acceleration because they stay in contact. Total mass is m+M=10m + M = 10 kg, so a=F/(m+M)=20/10=2a = F / (m+M) = 20/10 = 2 m/s2^2. We start with the system because finding aa first is the cleanest route.

Draw a free body diagram for M=6M = 6 kg alone. The only horizontal force on it is the contact force NN from the first block. So N=M⋅aN = M \cdot a.

 N=6×2=12 NN = 6 \times 2 = 12 \text{ N}

For block mm: F−N=m⋅a⇒20−12=4×2=8F - N = m \cdot a \Rightarrow 20 - 12 = 4 \times 2 = 8. Both sides match, so the answer is consistent.

Final answer: contact force =12= 12 N.

Why This Works

Newton’s second law applies to any chosen body or group of bodies. The trick toppers use is to first apply it to the full system — that gives acceleration without involving internal forces. Once aa is known, isolating any single block reveals the internal contact force.

Internal forces between the two blocks cancel when we treat them as one system (third law pair). That is why the system equation is so clean — only the external FF shows up.

Alternative Method

Apply Newton’s second law to each block separately from the start:

  • Block mm: F−N=maF - N = m a
  • Block MM: N=MaN = M a

Add the equations: F=(m+M)aF = (m+M)a, giving a=2a = 2 m/s2^2. Then N=6×2=12N = 6 \times 2 = 12 N. Same answer, slightly more algebra.

A very common slip — students write N=F=20N = F = 20 N, treating the contact force as equal to the applied force. The contact force only equals FF when block MM is the only one accelerating; here block mm also has to be accelerated, so NN must be smaller than FF.

Common Mistake

Forgetting that the direction of the contact force on block MM is the same as FF, while the contact force on block mm is opposite. Students mix these up and get a sign error in the FBD. Always draw the FBD with the surface in mind — the front block is pushed forward by the back block.

PYQ shortcut: when two bodies move together under a force applied to one of them, the contact force on the body in front is MfrontMtotal×F\frac{M_{\text{front}}}{M_{\text{total}}} \times F. Here 610×20=12\frac{6}{10} \times 20 = 12 N. Saves 30 seconds in the exam.

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