Question

A particle of mass m is projected with velocity v₀ in X-Y plane at an angle θ with X-axis as shown in figure. For time \(t<\frac{v_0\sin\theta}{g}\), angular momentum of the particle is, where \(\hat{i},\ \hat{j}\ and\ \hat{k}\) are unit vectors along X, Y and Z axes respectively:

\((a) -mgv_0t^2\cos\theta\hat{i}\)

\((b)\ mgv_0t\cos\theta\hat{K}\)

\((C)\ -\frac{1}{2}\ mgv_0t^2\cos\theta\ \hat {k}\)

\((d)\ \frac{1}{2}mgv_0{}^2t^2\cos\theta\hat{i}\)

Answer 1

Correct option is : \((C)\ -\frac{1}{2}\ mgv_0t^2\cos\theta\ \hat {k}\)

Angular momentum,

\(\because\ \vec L = m(\vec r\times\vec v)\)

If (x, y, t) are the co-ordinates of point P, then

x = v₀.cosθ x t

and y = v₀ sinθ x t - \(\frac{1}{2}gt^2\)

∴ Position vector of P,

\(\vec r = x\hat{i} + y\hat{j} = [v_0\cos\theta.t\hat{i} + \{v_0\sin\theta\times t - \frac{1}{2}gt^2\}\hat{j}]\)

AT point P,

v_x = v₀ cosθ

and v_y = v₀ sinθ - gt

\(\therefore\ \vec v = v_x\hat{i} + v_g\hat{j}\) 

\(= v_0\cos\theta.\hat{i} + (v_0\sin\theta - gt)\hat{j}\) 

Therefore angular momentum,

Therefore, option (c) is correct.