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MHT CET 2026 (14 May, Shift 1) · Impulse & Momentum Conservation

Impulse & Momentum Conservation question from MHT CET 2026 (14 May, Shift 1)

Q4MHT CET 2026 · 14 MAY · SHIFT 1Medium

Impulse & Momentum Conservation

A bullet is fired from a gun. The force on the bullet is F=(600−2×105t)F = (600 - 2 \times 10^{5} t) N where tt is in second. The force on the bullet becomes zero as soon as it leaves the barrel. The average impulse imparted to the bullet is

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Answer: (c)

  1. Force becomes zero when 600−2×105t=0⇒t=3×10−3600 - 2 \times 10^{5} t = 0 \Rightarrow t = 3 \times 10^{-3} s.
  2. Impulse J=∫0tF dt=600t−105t2J = \int_0^{t} F\,dt = 600t - 10^{5}t^{2}
  3. J=600(3×10−3)−105(9×10−6)=1.8−0.9=0.9J = 600(3 \times 10^{-3}) - 10^{5}(9 \times 10^{-6}) = 1.8 - 0.9 = 0.9 N s
  4. So the answer is (c) 0.90.9 N-s.
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