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The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x axis. The net force F is acting on the particle along the x axis. F is a constant force. Part A The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0. Express your answer in terms of any or all of m, F, and t.

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Answer:

[tex]p=F\times t[/tex]

Explanation:

We are given:

  • mass of the particle moving along x-axis = m
  • force acting on paricle along x-axis = F
  • time = t

momentum, p = ?

From the Newton's second law of motion:

[tex]F=\frac{dp}{dt}[/tex]

where:

dp & dt are the change in momentum and change in time respectively.

Using the above eq. of the Newton's second law of motion:

[tex]dp=F\times dt[/tex]

according to given:

[tex]p=F\times t[/tex]

We also have for impulse:

[tex]i=F\times t[/tex]

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