Question

The momentum $p$ of a particle changes with time $t$ according to the relation $\frac{dp}{dt}=\left(10N\right)+\left(2N/s\right)t$. If the momentum is zero at $t=0$, what will the momentum be at $t=10s$?

Answer 1

The momentum of the system s given as:

$\frac{dp}{dt}=10N+\left(2N/s\right)t$

According to the question, the momentum at $t=0$ is zero.

$P\left(0\right)$ at $t=0$

$\frac{dp}{dt}=10+2t$

Integrate the above expression for the momentum from $0to10sec$.

${\int }_o^{p}dp={\int }_{t=0}^{10}(10+2t)dt$

$p-0=10+2t$

$=[10t+t^2{]}_0^{10}$

$=200Ns$