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The velocity of a 2 kg particle varies with time as v = (2t i + 4 j) m/s, where t is in seconds. Find the impulse delivered to the particle between t = 0 and t = 2 s.

8 i N-s
10 i N-s
12 i N-s
16 i N-s

Correct answer: 8 i N-s

Solution

v(0) = 4 j, v(2) = 4 i + 4 j; change in velocity = 4 i, so impulse = m*delta v = 2*4 i = 8 i N-s.

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