The $F-t$ graph for an object is shown in the figure. What will be the magnitude of the change in momentum $(\Delta p)$ in the interval of $0.03\, s$ from the start?

$A$ bullet of mass $10\,g$ leaves the barrel of a gun with a velocity of $600\,m/s$. If the barrel of the gun is $50\,cm$ long and the mass of the gun is $3\,kg$,then the value of the impulse supplied to the gun will be $.....\,Ns$.

How can the impulse of a force be determined when a large force acts on an object for a very short interval of time?

$A$ particle of mass $m$, initially at rest, is acted upon by a variable force $F$ for a brief interval of time $T$. It begins to move with a velocity $u$ after the force stops acting. The graph shows $F$ as a function of time, where the curve is a semicircle with peak force $F_0$ at time $T/2$.

$A$ cricket ball of mass $500\, g$ collides with a bat with speed $15\, m/s$ and returns with the same speed within $0.01\, s$. The force acted on the bat is ............ $N$.

$A$ 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: