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NEET PHYSICSLAWS OF MOTIONEasy

Question

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).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from LAWS OF MOTION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSLAWS OF MOTIONstrikesreflectedwithoutfigureimpulse

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