A particle is acted upon by a force for 1 second whose X component remains constant at F_x= 30 N but y and z components vary with time as
shown in the graph. What angle does the change in momentum $(Δ \vec{P})$
P make with X axis at t = 1 s ?

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$\cos^{- 1} \frac{3}{\sqrt{29}}$

$\cos^{- 1} \frac{3}{\sqrt{39}}$

$\cos^{- 1} \frac{4}{\sqrt{29}}$

$\cos^{- 1} \frac{2}{\sqrt{39}}$

answer is A.

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Detailed Solution

$Δ \vec{P} = \int \vec{F} dt = \hat{i} \int F_{x} dt + \hat{j} \int F_{y} dt + \hat{k} \int F_{z} dt = \hat{i} (30 \times 1) + \hat{j} (\frac{1}{2} \times 40 \times 1) + \hat{k} (\frac{1}{2} \times 80 \times 1) = 30 \hat{i} + 20 \hat{j} + 40 \hat{k} [Area under F_{y} - t graph gives \int F_{y} dt] ∴ | Δp | = \sqrt{30^{2} + 20^{2} + 40^{2}} = 10 \sqrt{29} N - s If angle made by Δ \vec{P} with x axis is θ Then, \cos θ = \frac{30}{10 \sqrt{29}} = \frac{3}{\sqrt{29}}$