A 0.18 kg hockey puck has a velocity of 1.7 m/s toward the east (the +x directio

Question

A 0.18 kg hockey puck has a velocity of 1.7 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 0.57 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. Number units Number units Number units Number units

Explanation / Answer

Impuse = F*t = change in momentum mv = [1.7 -(-3.8)]*0.18 = 0.99 kg*m/s

F = 0.99/0.57 = 1.7368 N west

The change in momentum will be the negative of that in the x so the horizontal momentum is zero and the change in the vertical momentum is 0.18*3.8 south.
mv = 0.18*1.7 = 0.306 kgm/s west and 0.18*3.8 = 0.684 kgm/s south

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