A small particle of mass m is projected at an angle θ with the x-axis with an initial velocity v₀ in the x-y plane as shown in the figure. At a time t where {i}, {j} and {k} are unit vectors along x, y and z-axis respectively. [AIEEE 2010]

Correct Answer is : −12mgv0t2cos⁡θk^-1/2 m g v₀ t² θ {k}−21mgv0t2cosθk^

Correct Answer is : −12mgv0t2cos⁡θk^-1/2 m g v₀ t² θ {k}−21mgv0t2cosθk^

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AIEEE 2010 Solved Paper

Section: Physics

Question : 59 of 89

Marks: +1, -0

m

\theta

x

v_0

x-y

t < \frac{v_0 \sin\theta}{g}

\hat{i}, \hat{j}

\hat{k}

x, y

z

Solution:

Position vector of the particle from the original at any time

t

\vec{r}=v_0 t \cos\theta \hat{i}+ \left(v_0 t \sin\theta-\frac{1}{2} g t^2\right) \hat{j}

As Velocity vector,

\vec{v}=\frac{\vec{dr}}{dt}

\therefore \vec{v}=\frac{d}{dt}\left(v_0 t \cos\theta \hat{i}+ \left(v_0 t \sin\theta-\frac{1}{2} g t^2\right) \hat{j}\right)

=v_0 \cos\theta \hat{i}+ (v_0 \sin\theta-g t) \hat{j}

Angular momentum of the particle about the origin is

\vec{L}=m(\vec{r}\times \vec{v})

\;\vec{L}=m \hat{\left[ v_0 \cos\theta t {i}+ \left(v_0 \sin\theta t-\frac{1}{2} g t^2\right) \hat{j}\right]}

\times \left[ v_0 \cos\theta \hat{i}+ (v_0 \sin\theta-g t) \hat{j}\right]

=m v_0 \cos\theta t \left[-\frac{1}{2} g t\right] \hat{k}

=-\frac{1}{2} m g v_0 t^2 \cos\theta \hat{k}

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