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Difficulty: MediumNewton's Laws of Motion and Linear Momentum

A body of mass 4.0 kg4.0\text{ kg} is initially moving due east across a frictionless horizontal surface at a constant velocity of 15 m s−115\text{ m s}^{-1}. A constant horizontal force of 20 N20\text{ N} directed due west is then applied to the body for 5.0 s5.0\text{ s}. What is the final velocity of the body?

  1. 10 m s−110\text{ m s}^{-1} due westAnswer
  2. B
    10 m s−110\text{ m s}^{-1} due east
  3. C
    40 m s−140\text{ m s}^{-1} due east
  4. D
    25 m s−125\text{ m s}^{-1} due west

Answer

10 m s−110\text{ m s}^{-1} due west
According to the impulse-momentum theorem (FΔt=mv−muF \Delta t = m v - m u), setting East as the positive direction gives u=+15 m s−1u = +15\text{ m s}^{-1}, F=−20 NF = -20\text{ N}, and t=5.0 st = 5.0\text{ s}. Substituting these values yields −100=4.0(v−15)-100 = 4.0(v - 15), leading to v−15=−25v - 15 = -25, so v=−10 m s−1v = -10\text{ m s}^{-1}. The negative sign confirms the final velocity is 10 m s−110\text{ m s}^{-1} due west.

Step-by-Step Solution

1
Assign direction signs to vectors and determine initial momentum.
Taking East as positive (++) and West as negative (−-), initial velocity u=+15 m s−1u = +15\text{ m s}^{-1}, force F=−20 NF = -20\text{ N}, and mass m=4.0 kgm = 4.0\text{ kg}. Initial momentum pi=mu=4.0×(+15)=+60 kg m s−1p_i = m u = 4.0 \times (+15) = +60\text{ kg m s}^{-1}.
Linear momentum is a vector quantity, so direction must be tracked consistently.
2
Calculate the impulse exerted by the retarding force.
\text{Impulse } I = F \Delta t = (-20) \times 5.0 = -100\text{ N s} \text{ (or kg m s}^{-1}\text{)}.
The impulse equals the change in linear momentum according to Newton's Second Law.
3
Determine the final momentum and final velocity.
Final momentum pf=pi+I=+60+(−100)=−40 kg m s−1p_f = p_i + I = +60 + (-100) = -40\text{ kg m s}^{-1}. Final velocity v=pfm=−404.0=−10 m s−1v = \frac{p_f}{m} = \frac{-40}{4.0} = -10\text{ m s}^{-1}.
Dividing the final momentum by mass gives the final velocity, where the negative sign indicates motion due west.

Key Concept

Impulse-Momentum Theorem and Directional Vector Conventions

Alternative Method

Using Newton's Second Law directly: Acceleration a=Fm=−20 N4.0 kg=−5.0 m s−2a = \frac{F}{m} = \frac{-20\text{ N}}{4.0\text{ kg}} = -5.0\text{ m s}^{-2}. Using the kinematic formula v=u+at=15+(−5.0×5.0)=15−25=−10 m s−1v = u + a t = 15 + (-5.0 \times 5.0) = 15 - 25 = -10\text{ m s}^{-1}. Thus, the final velocity is 10 m s−110\text{ m s}^{-1} due west.
Estimated Time:1m 15s
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Topics

Newton's Laws of Motion and Linear Momentum

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