A women throws an object of mass $500\,g$ with a speed of $25\, ms^{-1}$.
$(a)$ What is the impulse imparted to the object ?
$(b)$ If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object ?
Mass of the object $(m)=500 \mathrm{~g}=0.5 \mathrm{~kg}$
Speed of the object $(v)=25 \mathrm{~m} / \mathrm{s}$
$(a)$ Impulse imparted to the object = change in momentum
$=m v-m u$
$=m(v-u)$
$=0.5(25-0)=12.5 \mathrm{~N} \mathrm{~s}$
$(b)$ Half of initial velocity is,
$\quad=-\frac{25}{2} \mathrm{~m} / \mathrm{s}$
$v^{\prime}=-12.5 \mathrm{~m} / \mathrm{s}$
$\therefore \text { Change in momentum } =m\left(v^{\prime}-v\right)$
$=0.5(-12.5-25)$
$=-18.75 \mathrm{~N} \mathrm{~s}$
Figure shows the position-time graph of a particle of mass $4 \,kg$. What is the
$(a)$ force on the particle for $t\, <\, 0, t \,> \,4\; s, 0 \,<\, t \,< \,4\; s$?
$(b)$ impulse at $t=0$ and $t=4 \;s ?$ (Consider one-dimensional motion only).
A particle is moving in a circle with uniform speed $v$. In moving from a point to another diametrically opposite point
Position-time graph for a particle of mass $100g$ is as shown in figure then impulse acting on the particle at $t = 5$ second is ............ $N-s$
A body of mass $10 kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20,10)$. If $T$ is the time of flight, then its momentum vector, at time $t =\frac{ T }{\sqrt{2}}$, is.
$\left[\right.$ Take $\left.g=10 m / s ^{2}\right]$
The figure shows the position - time $(x-t)$ graph of one-dimensional motion of the body of mass $0.4\; kg$. The magnitude of each impulse is