A particle in three-dimensional space moves along the path:
$\vec { r } ( t ) = \left\langle t ^ { 2 } , \sin ( t ) , e ^ { t } \right\rangle$.
Find the time when the particle changes direction.

A particle is moving with velocity $\vec { v } = K ( y \hat { i } + x \hat { j } )$, where $K$ is a constant. The general equation for its path is
(1) $y = x ^ { 2 } +$ constant
(2) $y ^ { 2 } = x +$ constant
(3) $x y =$ constant
(4) $y ^ { 2 } = x ^ { 2 } +$ constant

A small particle of mass $m$ is projected at an angle $\theta$ with the x-axis with an initial velocity $\mathrm { v } _ { 0 }$ in the $\mathrm { x } - \mathrm { y }$ plane as shown in the figure. At a time $t < \frac { v _ { 0 } \sin \theta } { g }$, the angular momentum of the particle is where $\hat { \mathrm { i } } , \hat { \mathrm { j } }$ and $\hat { \mathrm { k } }$ are unit vectors along $\mathrm { x } , \mathrm { y }$ and z-axis respectively.
(1) $- \mathrm { mgv } _ { 0 } \mathrm { t } ^ { 2 } \cos \theta \hat { \mathrm { j } }$
(2) $\mathrm { mgv } _ { 0 } t \cos \theta \hat { \mathrm { k } }$
(3) $- \frac { 1 } { 2 } m g v _ { 0 } t ^ { 2 } \cos \theta \hat { k }$
(4) $\frac { 1 } { 2 } m g v _ { 0 } t ^ { 2 } \cos \theta \hat { i }$

The time dependence of the position of a particle of mass $m = 2$ is given by $\vec { r } t = 2 t \hat { i } - 3 t ^ { 2 } \hat { j }$. Its angular momentum, with respect to the origin, at time $\mathrm { t } = 2$ is:
(1) 36 k
(2) $48 \hat { i } + \hat { j }$
(3) $- 48 \hat{k}$
(4) $- 34 \mathrm { k } - \hat { \mathrm { i } }$

A particle is moving along the $x$-axis with its coordinate with time $t$ given by $x ( t ) = 10 + 8 t - 3 t ^ { 2 }$. Another particle is moving along the $y$-axis with its coordinate as a function of time given by $y ( t ) = 5 - 8 t ^ { 3 }$. At $t = 1 \mathrm {~s}$, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt { v }$. Then $v$ (in $\mathrm { m s } ^ { - 1 }$) is $\_\_\_\_$.

A particle of mass $m$ is moving in time $t$ on a trajectory given by, $$\vec{r} = 10\alpha t^{2}\hat{\mathrm{i}} + 5\beta(t - 5)\hat{\mathrm{j}}$$ where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for $t = 0$ at time $t =$ \_\_\_\_ seconds.

A disc is rolling without slipping on a surface. The radius of the disc is $R$. At $t = 0$, the top most point on the disc is $A$ as shown in figure. When the disc completes half of its rotation, the displacement of point $A$ from its initial position is
(1) $2R$
(2) $R\sqrt{\left(\pi^2 + 4\right)}$
(3) $R\sqrt{\left(\pi^2 + 1\right)}$
(4) $2R\sqrt{\left(1 + 4\pi^2\right)}$