In the figure shown a semicircular area is removed from a uniform square plate of side $l$ and mass (before removing) $m.$ The $x-$coordinate of centre of mass of remaining portion is (The origin is at the centre of square)

In the figure shown the system is at rest initially. Two persons $\text{'}A\text{'}$ and $\text{'}B\text{'}$ of masses $40 \mathrm{kg}$ each move with speeds $v_1$ and $v_2$ respectively towards each other on a plank lying on a smooth horizontal surface as shown in the figure. Plank travels a distance of $20 \mathrm{m}$ towards right direction in $5 \sec .$ (Here $v_1$ and $v_2$ are given with respect to the plank). Then:

(i) $v_1=10 \mathrm{m} {\mathrm{s}}^{-1}, {\mathrm{v}}_2=0 \mathrm{m} {\mathrm{s}}^{-1}$
(ii)$v_1=15 \mathrm{m} {\mathrm{s}}^{-1}, {\mathrm{v}}_2=5 \mathrm{m} {\mathrm{s}}^{-1}$
(iii)$v_1=20 \mathrm{m} {\mathrm{s}}^{-1}, {\mathrm{v}}_2=10 \mathrm{m} {\mathrm{s}}^{-1}$

When a block is placed on a wedge as shown in the figure, the block starts sliding down and the wedge also starts sliding on the ground. All surfaces are rough. The centre of mass of (wedge + block) system will move

A small block of mass $\text{'}m\text{'}$ is placed on bigger block of mass $M\mathit{,}$ which is placed on a frictionless horizontal surface. The two blocks are given equal speed $u,$ but opposite directions, as shown in the figure. After some time, it is observed that both the blocks are moving in the direction of motion of the lower block, with a
speed greater than $\frac{u}{2}.$ It can be concluded that -

A particle of constant mass $m$ moves in straight line under a time varying force. The change in momentum of the particle in a time interval $t_1$ to $t_2$ can be given as
$\mathrm{\Delta }p={\int }_{t_1}^{t_2}F\mathrm{d}t$.
The force $F$ acting on the particle as a function of time $t$ is shown in the graph. If velocity of the particle is zero at $t=0$, then