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Laws of Motion question

2019 · 10 Apr · Shift 2 · Q63
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Laws of Motion question

2019 · 10 Apr · Shift 2 · Q63

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A bullet of mass 20 g has an initial speed of 1 ms–1 , just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5 × 10–2 N, the speed of the bullet after emerging from the other side of the wall is close to :
  1. A
    0.3 ms-1
  2. B
    0.1 ms-1
  3. C
    0.7 ms-1
  4. D
    0.4 ms-1
View written solutionFree

Correct answer: C

  1. Given data

    • Mass of bullet: m=20 g=0.02 kgm = 20\,\text{g} = 0.02\,\text{kg}m=20g=0.02kg
    • Initial speed: u=1 m s−1u = 1\,\text{m s}^{-1}u=1m s−1
    • Thickness of wall: s=20 cm=0.2 ms = 20\,\text{cm} = 0.2\,\text{m}s=20cm=0.2m
    • Mean resisting force: F=2.5×10−2 NF = 2.5 \times 10^{-2}\,\text{N}F=2.5×10−2N

  2. Find retardation produced by the wall

    Since the resisting force opposes motion, a=−Fm=−2.5×10−20.02=−1.25 m s−2a = -\frac{F}{m} = -\frac{2.5 \times 10^{-2}}{0.02} = -1.25\,\text{m s}^{-2}a=−mF​=−0.022.5×10−2​=−1.25m s−2

  3. Use equation of motion

    We use v2=u2+2asv^2 = u^2 + 2asv2=u2+2as

    Substituting values: v2=(1)2+2(−1.25)(0.2)v^2 = (1)^2 + 2(-1.25)(0.2)v2=(1)2+2(−1.25)(0.2) v2=1−0.5=0.5v^2 = 1 - 0.5 = 0.5v2=1−0.5=0.5

  4. Calculate final speed

    v=0.5≈0.707 m s−1v = \sqrt{0.5} \approx 0.707\,\text{m s}^{-1}v=0.5​≈0.707m s−1

    So the speed after emerging is close to 0.7 m s−10.7\,\text{m s}^{-1}0.7m s−1

  5. Check options

    • A: 0.3 m s−10.3\,\text{m s}^{-1}0.3m s−1 ❌
    • B: 0.1 m s−10.1\,\text{m s}^{-1}0.1m s−1 ❌
    • C: 0.7 m s−10.7\,\text{m s}^{-1}0.7m s−1 ✅
    • D: 0.4 m s−10.4\,\text{m s}^{-1}0.4m s−1 ❌

Therefore, the correct option is C.

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