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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A block of mass 4 kg is sliding on a frictionless horizontal...

A block of mass 4 kg4\text{ kg} is sliding on a frictionless horizontal surface with a kinetic energy of 18 J18\text{ J}. It strikes a rigid wall perpendicularly and rebounds in the opposite direction with a kinetic energy of 8 J8\text{ J}. The magnitude of the impulse imparted to the block by the wall is :

Options

  1. A4 kg m s−14\text{ kg m s}^{-1}
  2. B10 kg m s−110\text{ kg m s}^{-1}
  3. C12 kg m s−112\text{ kg m s}^{-1}
  4. D20 kg m s−120\text{ kg m s}^{-1}

Correct answer

D. 20 kg m s−120\text{ kg m s}^{-1}

Step-by-step solution

Initial kinetic energy Ki=18 JK_i = 18\text{ J}

Magnitude of initial momentum pi=2mKi=2×4×18=12 kg m s−1p_i = \sqrt{2mK_i} = \sqrt{2 \times 4 \times 18} = 12\text{ kg m s}^{-1}

Final kinetic energy Kf=8 JK_f = 8\text{ J}

Magnitude of final momentum pf=2mKf=2×4×8=8 kg m s−1p_f = \sqrt{2mK_f} = \sqrt{2 \times 4 \times 8} = 8\text{ kg m s}^{-1}

Since the block rebounds in the opposite direction, the initial and final momenta have opposite signs.

Magnitude of impulse J=∣Δp⃗∣=∣pf−(−pi)∣=pf+piJ = |\Delta \vec{p}| = |p_f - (-p_i)| = p_f + p_i

J=8+12=20 kg m s−1J = 8 + 12 = 20\text{ kg m s}^{-1}

Answer: 20 kg m s−120\text{ kg m s}^{-1}
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