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NEET PHYSICSEasy

A particle of mass mm is projected with velocity vv making an angle of 45∘45^\circ with the horizontal. When the particle lands on the level ground, the magnitude of the change in its momentum will be:

A

2mv2mv

B

mv/2mv/\sqrt{2}

C

mv2mv\sqrt{2}

D

zero

Step-by-Step Solution

  1. Concept: In projectile motion, the horizontal component of velocity remains constant (ignoring air resistance), while the vertical component reverses its direction (but retains the same magnitude) when the particle returns to the same level [NCERT Class 11, Physics Part I, Sec 4.9 Projectile Motion].
  2. Velocity Components:
  • Initial Velocity: v⃗i=vcos⁡θi^+vsin⁡θj^\vec{v}_i = v\cos\theta \hat{i} + v\sin\theta \hat{j}
  • Final Velocity (on landing): v⃗f=vcos⁡θi^−vsin⁡θj^\vec{v}_f = v\cos\theta \hat{i} - v\sin\theta \hat{j}
  1. Change in Momentum (Δp⃗\Delta \vec{p}): Δp⃗=m(v⃗f−v⃗i)\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) Δp⃗=m[(vcos⁡θi^−vsin⁡θj^)−(vcos⁡θi^+vsin⁡θj^)]\Delta \vec{p} = m[(v\cos\theta \hat{i} - v\sin\theta \hat{j}) - (v\cos\theta \hat{i} + v\sin\theta \hat{j})] Δp⃗=−2mvsin⁡θj^\Delta \vec{p} = -2mv\sin\theta \hat{j}
  2. Magnitude: ∣Δp⃗∣=2mvsin⁡θ|\Delta \vec{p}| = 2mv\sin\theta Given θ=45∘\theta = 45^\circ: ∣Δp⃗∣=2mvsin⁡(45∘)=2mv(12)=2mv|\Delta \vec{p}| = 2mv \sin(45^\circ) = 2mv \left(\frac{1}{\sqrt{2}}\right) = \sqrt{2}mv
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