Velocity of a particle of mass 2 kg varies with time 't' according to the equation v→=(2ti^−4j^)m/s m/s. Here 'r' is in seconds. The impulse imparted to the particle in the time interval 0 ≤ t ≤ 2s is:
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a
(4i^)kg−m/s
b
(8i^−4j^)kg−m/s
c
(8i^+4j^)kg−m/s
d
(8i^)kg−m/s
answer is D.
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Detailed Solution
Impulse = Change in momentum=2[(2×2i^−4j^)−(2×0i^−4j^)]=(8i^)kg−m/s