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NEET PHYSICSEasy

A body of mass MM hits normally a rigid wall with velocity vv and bounces back with the same velocity. The impulse experienced by the body is:

A

1.5Mv1.5Mv

B

2Mv2Mv

C

zero

D

MvMv

Step-by-Step Solution

  1. Concept: Impulse is defined as the change in momentum of a body (Δp⃗=p⃗f−p⃗i\Delta \vec{p} = \vec{p}_f - \vec{p}_i). This is a direct application of Newton's Second Law involving time duration [NCERT Physics Class 11, Laws of Motion, Section 4.5/5.7].
  2. Setup:
  • Let the initial velocity be directed towards the wall. Taking this direction as positive, v⃗i=+v\vec{v}_i = +v.
  • The body bounces back normally (perpendicularly) with the same speed. The direction is reversed, so v⃗f=−v\vec{v}_f = -v.
  1. Calculation:
  • Initial Momentum (pip_i) = MvMv
  • Final Momentum (pfp_f) = M(−v)=−MvM(-v) = -Mv
  • Impulse (JJ) = Change in Momentum = pf−pip_f - p_i
  • J=−Mv−Mv=−2MvJ = -Mv - Mv = -2Mv
  1. Magnitude: The magnitude of the impulse imparted by the wall on the body is ∣−2Mv∣=2Mv|-2Mv| = 2Mv. (Reference: NCERT Physics Class 11, Laws of Motion, Example 4.5 deals with a similar case of billiards balls striking a wall).
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