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Laws of Motion question

2021 · 18 Mar · Shift 1 · Q70
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Laws of Motion question

2021 · 18 Mar · Shift 1 · Q70

JEE MainPhysicsLaws of MotionNumerical+4 / −1
A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10 m/s and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is 'x' N. The value of 'x' to the nearest integer is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 10

  1. Given data

    • Mass of bullet: m=0.1 kgm = 0.1\,\text{kg}m=0.1kg
    • Initial velocity: u=10 m/su = 10\,\text{m/s}u=10m/s
    • Final velocity: v=0v = 0v=0 (bullet stops)
    • Distance travelled inside wood: s=50 cm=0.5 ms = 50\,\text{cm} = 0.5\,\text{m}s=50cm=0.5m

  2. Use the equation of motion Since deceleration is uniform, v2=u2+2asv^2 = u^2 + 2asv2=u2+2as Substituting the values, 0=(10)2+2a(0.5)0 = (10)^2 + 2a(0.5)0=(10)2+2a(0.5) 0=100+a0 = 100 + a0=100+a a=−100 m/s2a = -100\,\text{m/s}^2a=−100m/s2

  3. Find the retarding force Using Newton's second law, F=maF = maF=ma F=0.1×(−100)=−10 NF = 0.1 \times (-100) = -10\,\text{N}F=0.1×(−100)=−10N

  4. Magnitude of retarding force ∣F∣=10 N|F| = 10\,\text{N}∣F∣=10N

  5. Nearest integer x=10x = 10x=10

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