Question

A force is applied on a particle of mass 2 kg.
calculate
(1) change in momentum of particle at 0.02 second.
(2) if a particle starts from rest calculate its velocity at 0.02 second.​

Answer 1

Answer:

change in mom.,=2

velocity at. .02sec=5001

Explanation:

f=cp/t

cp=2

vel. at 0.01 =5000

vel. at0.02=5000+50×0.02=5001

Answer 2

(a) Change in momentum = 2 N-m

(b) Velocity, v = 1 m/s

Explanation:

Given that,

Mass of the particle, m = 2 kg

(a) The area under force time graph gives impulse applied. Impulse is also given by the change in momentum as :

$J=100\times 0.02=2\ N-m$

(b) Let v be the velocity of a particle at 0.02 seconds. So,

$J=mv\\\\v=\dfrac{J}{m}\\\\v=\dfrac{2}{2}=1\ m/s$

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