Question

A force in the negative direction (on the x axis) is applied for 27 ms at 0.40 kg ball initially moving at a speed of 14 m / s in the positive direction of x. The force varies in magnitude and the momentum has a magnitude of 32.4 N.s. What is a) speed b) the direction of the ball after the force is applied? c) What is the magnitude of the average force applied to the ball?

Answer 1

a)

Impulse = ΔP

              = mv - mu    .................... (1)

therefore ,

        mv - mu = (-32.4 N.s)   

      (0.4 kg)v- (0.40 kg)(14 m/s) = (-32.4 N.s)

      velocity v = -67 m/s

it is moving negative x- axis with speed -67 m/s

b)

time t = 27 ms = 0.027 s

from kinematic equations,

     v = u + at

therefore ,

      acceleration  a = (v-u)/t

                            = (-67 m/s -14 m/s)/(0.027 s)

                            = 3000 m/s^2

   the average force applied to the ball is

            F = ma

               = (0.4 kg)(3000 m/s^2)

               = 1200 N

  

Answer 2

a) Impulse = mv - mu

      -32.4 = 0.40v- 0.40* 14

             v = -67 m/s

b) a = (v-u)/t

       = (-67 -14)/0.027

       = -3000 m/s^2

   F = ma

      = 0.40 * 3000

      = 1200 N (to 3 sig fig)