Question:
The momentum $\rho$ of a particle changes with time $t$ according to the relation $\frac{d \rho}{d t}=(10 N)+(2 N / s) t$. If the momentum is zero at $\mathrm{t}=0$, what will be the momentum be at $\mathrm{t}=10 \mathrm{~s}$ ?
Solution:
It is given that $\frac{d p}{d t}=(10 N)+\left(\frac{2 N}{s}\right) t$
Now, $p=0$ at $t=0 .$ So,
$p=\int_{0}^{10}(10 d t+2 t d t)=10(10-0)+(100-0)=200 \mathrm{~N} / \mathrm{s}=200 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$