For a particle, the momentum p varies with time t such that (dp)/(dt) = (10 N) + (2 N/s)t . Assuming the momentum is zero at t = 0 , determine the momentum at t = 10 s .

Correct answer: C. 200 kg m/s Given (dp)/(dt) = 10 + 2t dp = (10 + 2t) dt Integrating both sides from t = 0 to t = 10 s : ∫_ 0 ^ p dp = ∫_ 0 ^ 10 (10 + 2t) dt p = [ 10t + t^2 ]_ 0 ^ 10 p = 10(10) + (10)^2 p = 100 + 100 = 200 kg m/s

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Physics and Mathematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Physics and Mathematics

For a particle, the momentum p varies with time t such that...

For a particle, the momentum

p

varies with time

t

such that

\dfrac{dp}{dt} = (10 \text{ N}) + (2 \text{ N/s})t

. Assuming the momentum is zero at

t = 0

, determine the momentum at

t = 10 \text{ s}

Options

A $120 \text{ kg m/s}$
B $300 \text{ kg m/s}$
C $200 \text{ kg m/s}$
D $100 \text{ kg m/s}$

Correct answer

C. $200 \text{ kg m/s}$

Step-by-step solution

Given

\dfrac{dp}{dt} = 10 + 2t

dp = (10 + 2t) dt

Integrating both sides from

t = 0

t = 10 \text{ s}

\int_{0}^{p} dp = \int_{0}^{10} (10 + 2t) dt

p = \left[ 10t + t^2 \right]_{0}^{10}

p = 10(10) + (10)^2

p = 100 + 100 = 200 \text{ kg m/s}

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