$A$ body of mass $0.05 \ kg$ is observed to fall with an acceleration of $9.5 \ ms^{-2}$. The opposing force of air on the body is $(g = 9.8 \ ms^{-2})$.

The variation of momentum $(p)$ with time $(t)$ for one of the bodies in a two-body collision is shown in the figure. The instantaneous force is maximum corresponding to which point?

The engine of a jet aircraft applies a thrust force of $10^5\, N$ during take-off and causes the plane to attain a velocity of $1\, km/s$ in $10\, s$. The mass of the plane is:

$A$ force $\vec{F}=(40 \hat{i}+10 \hat{j}) \, N$ acts on a body of mass $5 \, kg$. If the body starts from rest,its position vector $\vec{r}$ at time $t=10 \, s$ will be -

At any instant,the velocity of a particle of mass $500 \, g$ is $(2t \hat{i} + 3t^2 \hat{j}) \, ms^{-1}$. If the force acting on the particle at $t = 1 \, s$ is $(\hat{i} + x \hat{j}) \, N$,then the value of $x$ will be:

$A$ body of mass $4 \ kg$ experiences two forces $\overrightarrow{F}_1 = 5 \hat{i} + 8 \hat{j} + 7 \hat{k}$ and $\overrightarrow{F}_2 = 3 \hat{i} - 4 \hat{j} - 3 \hat{k}$. The acceleration acting on the body is :