A 450 g particle has velocity v x = -4.0 m/s at t = - 2 s. Force F x = (4 t 2) N

Question

A 450 g particle has velocity vx = -4.0 m/s at t = - 2 s. Force Fx = (4t2) N, where t is in s, is exerted is exerted on the particle between t = - 2 s and t = 2 s. This force increases from 0 N at t = - 2 s to 4 N at t = 0 s and then back to 0 N at t = 2 s.

What is the particle’s velocity at t = 2 s?

Express your answer with the appropriate units.

Please show the methods used for each step. (conservation of momentum, impulse-momentum theorem, kinematics, combinations of more than one method, etc.)

Explanation / Answer

The impulse is calculated as follows;

J = (4-t^2)dt = 4t - 1/3 t^3 + C

Evaluated from t = -2 to t = 2:
(4*2 - 1/3 (2)^3 + C) - (4*-2 - 1/3 (-2)^3 + C) = (8 - 8/3 + C) - (-8 + 8/3 + C)

= 8 - 8/3 + C + 8 - 8/3 - C

= 16 - 16/3 = 32/3 Ns

Change in momentum is:
dp = mv2 - mv1

= m(v2 - v1)

32/3 Ns = 0.450 kg (v2 - (-4 m/s))

v2 = 19.7 m/s

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