$A$ particle of mass $m$ is moving with a uniform velocity $v_1$. It is given an impulse such that its velocity becomes $v_2$. The impulse is equal to

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

$Assertion$ : $A$ quick collision between two bodies is more violent than a slow collision,even when the initial and final velocities are identical.
$Reason$ : The rate of change of momentum is greater in the first case.

In projectile motion,two particles of masses $m_1$ and $m_2$ have velocities $\vec{V}_1$ and $\vec{V}_2$ respectively at time $t=0$. Their velocities become $\vec{V}_1^{\prime}$ and $\vec{V}_2^{\prime}$ at time $2t$ while still moving in air. The value of $[(m_1 \vec{V}_1^{\prime} + m_2 \vec{V}_2^{\prime}) - (m_1 \vec{V}_1 + m_2 \vec{V}_2)]$ is ($g =$ acceleration due to gravity).

$A$ body of mass $2 \,kg$ collides with a wall with a speed of $100 \,ms^{-1}$ and rebounds with the same speed. If the time of contact is $1/50 \,s$,the force exerted on the wall is

$A$ stream of glass beads, each with a mass of $15 \ g$, comes out of a horizontal tube at a rate of $100 \ \text{beads per second}$. The beads fall a distance of $5 \ m$ to a balance pan and bounce back to their original height. How much mass (in $kg$) must be placed in the other pan of the balance to keep the pointer at zero?