A 0.22 kg hockey puck has a velocity of 2.3 m/s toward the east (the + x directi
ID: 1332049 • Letter: A
Question
A 0.22 kg hockey puck has a velocity of 2.3 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 0.35 s time interval to change the puck's velocity to 6.7 m/s toward the west? What are the (c) magnitude and(d) direction if, instead, the velocity is changed to 6.7 m/s toward the south? Give your directions as positive (counterclockwise) angles measured from the +x direction.
Explanation / Answer
a) v1 = 2.3 m/s
v2 = -6.7 m/s
a = (v2 - v1)/t
= (-6.7 - 2.3)/0.35
= -25.7 m/s^2
F = m*|a|
= 0.22*25.7
= 5.66 N
b) direction : West
c) v1x = 2.3 m/s
v1y = 0
v2x = 0
v2y = -6.7 m/s
ax = (v2x - v1x)/t
= (0 - 2.3)/0.35
= -6.57 m/s^2
ay = (v2y - v1y)/t
= (-6.7 - 0)/0.35
= -19.14 m/s^2
a = sqrt(ax^2 + ay^2)
= sqrt(6.57^2 + 19.14^2)
= 20.24 m/s^2
F = m*|a|
= 0.22*20.24
= 4.45 N
d)
direction : theta = tan^-1(ay/ax)
= tan^-1(-19.14/-6.57)
= 71 degrees below -x axis
= 109 degrees with +x axis counterclockwise
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