You should have the velocity as a function of time either given explicitly or implicitly (a graph)
v = ds/dt (differentiating the position vector)
integrating the acceleration.
you can use impulse or work and energy principle and also newton law of motion to find acceleration then velocity
NOT SURE IF THAT WHAT YOU WANT.
The answer is 5kg m/s with the second momentum being 10 kg m/s. i took the test and it was right. hope this helps yalls
Answer:

Explanation:
Given that
Radius =r
Electric filed =E
Q=Charge on the ring
The electric filed at distance x given as

For maximum condition



For maximum condition




At
the electric field will be maximum.