# A 2 kg object is allowed to fall freely at t = 0 s. Callculate its momentum at (a) t = 0, (b) t = 1 s and (c) t = 2 s during its free-fall.

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A 2 kg object is allowed to fall freely at t = 0 s. Callculate its momentum at (a) t = 0, (b) t = 1 s and (c) t = 2 s during its free-fall.

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(a) As velocity of the object at t = 0 s is zero, the initial momentum of the object will also be zero.

(b) At t = 1s, the velocity of the object will be 9.8 ms–1 [use v = v0 + at] pointing downward. So the momentum of the object will be

p1 = (2 kg) × (9.8 ms–1) = 19.6 kgms–1 pointing downward.

(c) At t = 2 s, the velocity of the object will be 19.6 m s–1 pointing downward. So the momentum of the object will now be

p2 = (2 kg) × (19.6 ms–1) = 39.2 kgms–1 pointing downward.

Thus, we see that the momentum of a freely-falling body increases continuously in magnitude and points in the same direction. Now think what causes the momentum of a freely-falling body to change in magnitude?