Question

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

Answer 1

According to the question, we have:
\[\frac{dp}{dt} = \left( 10 N \right) + \left( 2 N/s \right)t\]
Momentum is zero at time, t = 0
Now, dp = [(10 N) + (2 Ns⁻¹)t]dt

On integrating the above equation, we get:
\[p = \int_0^{10} dp = \int_0^{10} 10 dt + \int_0^{10} \left( 2t dt \right)\]
\[ \Rightarrow p = \left[ 10t + 2\frac{t^2}{2} \right]_0^{10} \]
\[ \Rightarrow p = 10 \times 10 + 100 - 0\]
\[ \therefore p = 100 + 100 = 200 \text{ kg } m/s\]