A 0.22 kg hockey puck has a velocity of 2.1 m/s toward the east (the +x directio

Question

A 0.22 kg hockey puck has a velocity of 2.1 m/s toward the east (the +x direction) as it slides over the frictionless surface of an ice hockey rink. What are the (a) magnitude and (b) direction of the constant net force that must act on the puck during a 0.45 s time interval to change the puck's velocity to 3.8 m/s toward the west? What are the (c) magnitude and (d) direction if, instead, the velocity is changed to 3.8 m/s toward the south? Give your directions as positive (counterclockwise) angles measured from the +x direction.

Explanation / Answer

Here a change in momentum is equal to the Impulse applied.
For an object of constant mass, Mathematically that is
(a) mv = Ft
F = mv/t

F = (0.22(2.1 - (-3.8)) / 0.45
F = 2.88 N
(b) As the direction changes 180 degrees, the force is in the West direction or 180°.

(c) Same formula for the second question. This time the east velocity is reduced to zero and the south velocity increases from zero to 3.8 m/s

for the eastbound portion
F = (0.22(2.1 - 0) / 0.45
F = 1.03 N

for the southbound portion
F = (0.22(0 - 3.8) / 0.45
F = -1.86 N

so the total force is
F = (-1.86² + 1.03²)
F = 2.12 N

(d) and the angle is
tan = -1.86/-1.03
as both forces are negative, this is third quadrant
= 241.02

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