A 0.25 kg hockey puck has a velocity of 1.7 m/s toward the east (the + x directi

Question

A 0.25 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 5.6 m/s toward the west? What are the (c) magnitude and (d) direction if, instead, the velocity is changed to 5.6 m/s toward the south? Give your directions as positive (counterclockwise) angles measured from the +x direction.

Explanation / Answer

(a) in the first, you are going from +1.7 m/s to -5.6 m/s,

a total change of 7.3 m/s.

you do this over a time of .57 s

so the acceleration is 7.3/.57 = 12.8 m /s^2

in the west direction (180 degrees)

F = ma

F = .25*12.8

F = 3.2 N

(b) 180 degre in the west direction

(c) In this the velocity is changed to purely south,

so in the x direction you go from +1.7 m/s to 0 m/s

and in the y direction you go from 0 to - 5.6

so the changes are 2.2 in the x and

5.6 in the y.

the acelerations are then

2.2/.57 and 5.6/0.57

or 3.86 m/s^2 west and 9.82 m/s^2 south.

the magnitude of that acceleration is then

a = sqrt(3.86^2 +9.82^2)

a = 10.55 m/s^2

F = m*a

F = .25*10.55

F = 2.64 N

(d) arctan(9.82/3.86) = 68.54 degrees,

but note that this is in the third quadrant so we have to add 180 degrees.

180+68.54 = 248. 54 degree

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