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NEET PHYSICSEasy

A rigid ball of mass MM strikes a rigid wall at 60∘60^{\circ} and gets reflected without loss of speed, as shown in the figure. The value of the impulse imparted by the wall on the ball will be:

A

MvMv

B

2Mv2Mv

C

Mv2\frac{Mv}{2}

D

Mv3\frac{Mv}{3}

Step-by-Step Solution

  1. Concept: Impulse is defined as the change in momentum (J⃗=Δp⃗=p⃗f−p⃗i\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i). The force exerted by the wall acts only in the direction perpendicular to the wall (normal direction) [NCERT Class 11, Physics Part I, Sec 5.7].
  2. Resolve Components: Let the angle of incidence with the normal be θ=60∘\theta = 60^{\circ}.
  • Initial Momentum (p⃗i\vec{p}_i): pix=Mvcos⁡60∘p_{ix} = Mv \cos 60^{\circ} (towards the wall) piy=Mvsin⁡60∘p_{iy} = Mv \sin 60^{\circ} (parallel to the wall)
  • Final Momentum (p⃗f\vec{p}_f): pfx=−Mvcos⁡60∘p_{fx} = -Mv \cos 60^{\circ} (away from the wall) pfy=Mvsin⁡60∘p_{fy} = Mv \sin 60^{\circ} (parallel to the wall, unchanged)
  1. Calculate Impulse:
  • Along the wall (y-axis): Δpy=pfy−piy=0\Delta p_y = p_{fy} - p_{iy} = 0.
  • Perpendicular to the wall (x-axis): Δpx=pfx−pix=(−Mvcos⁡60∘)−(Mvcos⁡60∘)=−2Mvcos⁡60∘\Delta p_x = p_{fx} - p_{ix} = (-Mv \cos 60^{\circ}) - (Mv \cos 60^{\circ}) = -2Mv \cos 60^{\circ}.
  1. Magnitude: ∣J⃗∣=∣−2Mvcos⁡60∘∣=2Mv(12)=Mv|\vec{J}| = |-2Mv \cos 60^{\circ}| = 2Mv \left(\frac{1}{2}\right) = Mv (Reference: NCERT Class 11, Physics Part I, Chapter 5, Laws of Motion, Example 4.5).
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