A 0.19 kg hockey puck has a velocity of 2.1 m/s toward the east (the + x directi

Question

A 0.19 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 0.56 s time interval to change the puck's velocity to 5.7 m/s toward the west? What are the (c) magnitude and(d) direction if, instead, the velocity is changed to 5.7 m/s toward the south? Give your directions as positive (counterclockwise) angles measured from the +x direction.

Explanation / Answer

initial velocity u = 2.1i m/s
final velocity v = -5.7 i m/s
Impulse = change in momnetum (m(v-u)) = Force x time interval
0.19(-5.71i - 2.1i) = F x 0.56
F = - 2.65 i N
a) magnitude = 2.65 N .....Ans
b) direction = towards west ............Ans
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in second part
initial velocity u = 2.1i m/s
final velocity v = -5.7j m/s
Impulse = change in momnetum (m(v-u)) = Force x time interval
0.19(-5.71j - 2.1i) = F x 0.56
F = -0.71i - 1.94 j N
c) magnitude = sqrt(0.71^2 + 1.94^2) = 2.07 N
d) direction = 180+ tan-(1.94 / 0.71) = 249.9 degrees

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