Question

The momentum p of a particle changes with time t according to the relation dpdt=(10 N)+(2 N/s)t. If the momentum is zero at t = 0, what will the momentum be at t = 10 s?

Answer 1

Momentum at, t = 10 s is 200 Kg m/s.

Explanation:    

According to the question, we have :  

The momentum p of a particle changes with time t, according to,

$\frac{d p}{d t}=(10 N)+(2 N / s) t$

Where,                      

If the momentum is zero at time, t = 0 :

$\text { Now, } \mathrm{dp}=\left[(10 \mathrm{N})+\left(2 \mathrm{Ns}^{-1}\right) \mathrm{t}\right] \mathrm{d} \mathrm{t}$            

If the momentum at time, t = 10 s :

Integrate both sides of the above equation, we get  

$p=\int_{0}^{10} d p=\int_{0}^{10} 10 d t+\int_{0}^{10}(2 t d t)$          

Substitute limit of the integrals,                

$\ p=\left[10 t+2 \frac{t^{2}}{2}\right]_{0}^{10}$                                          

$\ p=10 \times 10+100-0$

Therefore,  $p=100+100=200 \mathrm{kg} m / s$    

Thus, the momentum is 200 Kg m/s at t = 10 s.