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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 level ground, the magnitude of change in its momentum will be:

A

2mv2mv

B

mv/2mv/\sqrt{2}

C

mv2mv\sqrt{2}

D

zero

Step-by-Step Solution

  1. Identify Velocity Components:
  • Initial velocity vector: v⃗i=vcos⁡θi^+vsin⁡θj^\vec{v}_i = v\cos\theta \hat{i} + v\sin\theta \hat{j}.
  • Final velocity vector (on landing): Due to gravity acting vertically downwards, the horizontal component remains constant, while the vertical component reverses direction (magnitude remains same on level ground).
  • v⃗f=vcos⁡θi^−vsin⁡θj^\vec{v}_f = v\cos\theta \hat{i} - v\sin\theta \hat{j} .
  1. Calculate Change in Momentum:
  • Change in momentum Δp⃗=mv⃗f−mv⃗i\Delta\vec{p} = m\vec{v}_f - m\vec{v}_i.
  • Δp⃗=m(vcos⁡θi^−vsin⁡θj^)−m(vcos⁡θi^+vsin⁡θj^)\Delta\vec{p} = m(v\cos\theta \hat{i} - v\sin\theta \hat{j}) - m(v\cos\theta \hat{i} + v\sin\theta \hat{j}).
  • Δp⃗=−2mvsin⁡θj^\Delta\vec{p} = -2mv\sin\theta \hat{j}.
  1. Magnitude of Change:
  • ∣Δp⃗∣=2mvsin⁡θ|\Delta\vec{p}| = 2mv\sin\theta.
  1. Substitute Given Values:
  • 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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