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⚛️ Physics  ·  Laws of Motion  ·  NEET & JEE

A 0.1 kg ball moving at 20 m s<sup>-1</sup> is struck and rebounds along the same line at 30 m s<sup>-1</sup>. If the contact time is 0.02 s, the average force on the ball is:

Answer: 250 N.

  • A 250 N
  • B 200 N
  • C 150 N
  • D 300 N

Correct answer: A. 250 N

Explanation: Change in momentum = 0.1 × (30 - (-20)) = 0.1 × 50 = 5 N s. Force = 5 / 0.02 = 250 N.

incline surfaceθmgNfmg sinθmg cosθ

Free-body diagram of a block on an incline: weight (mg) resolves into mg sinθ along the slope and mg cosθ into the slope, balanced by the normal force N and friction f.

Concept context

Newton's three laws, friction, circular motion, and free body diagrams.

Read the full Laws of Motion notes →