A reference frame attached to the earth:

A particle of mass $m$ moving with velocity $v$ towards east strikes another particle of same mass moving with the same velocity $v$ towards north. After striking, two particles fuse together. With what velocity this new particle of mass $2m$ will move in north-east direction?

A body of mass $\left(4m\right)$ is lying in $X-Y$ plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass $\left(m\right)$ move perpendicular to each other with equal speeds $\left(v\right)$. The kinetic energy generated due to the explosion is:

Two balls, having linear momenta ${\overrightarrow{p}}_1=p\hat{i}$ and ${\overrightarrow{p}}_2=-p\hat{i}$, undergo a collision in free space. There is no external force acting on the balls. Let ${p_1}^{\text{'}}$ and ${p_2}^{\text{'}}$ be their final momenta. The following option (s) is (are) not allowed for any non-zero value of $p, a_1, a_2, b_1, b_2, c_1$ and $c_2$:

A shell is fired from a cannon with a velocity of $v {\mathrm{ms}}^{-1}$at an angle $\theta$ with the horizontal. It explodes into two pieces of equal masses at the highest point of its path. One of the pieces retraces its path to cannon. The speed $\left(\mathrm{in} {\mathrm{ms}}^{-1}\right)$ of the other piece immediately after the explosion is:

An isolated particle of mass $m$ is moving in a horizontal plane $\left(X-Y\right)$ along the $X-\mathrm{axis}$, at a certain height above the ground. It suddenly explodes into two fragments of masses $\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ and $\raisebox{1ex}{$3m$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$. An instant later, the smaller fragment is at $y=+15 \mathrm{cm}$. The larger fragment at this instant is at:

A balloon of mass $M$ is descending down with an acceleration $a\left(\mathrm{where} a<g\right)$. How much mass should be removed from it so that it starts moving up with an acceleration $a$?

The force $F$ acting on a particle of mass $m$ is indicated by the force-time graph shown adjoining. The change in momentum of the particle over the time interval from $0 \mathrm{s} \mathrm{to} 8 \mathrm{s}$is:

A particle moves in the x-y plane under the action of a force $\overrightarrow{F}$ such that the value of its linear momentum $\overrightarrow{P}$ at any time $t$ is $p_x=2 \cos t, p_y=2\sin t$. The angle $\theta$ between $\overrightarrow{F}$ and $\overrightarrow{P}$ at a given time $t$will be: