Question

A 0.21 kg hockey puck has a velocity of 2.2 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 0.43 s time interval to change the puck's velocity to 3.7 m/s toward the west?

and what are the:

(c) magnitude and (d) direction if, instead, the velocity is changed to 3.7 m/s toward the south?

Give your directions as positive (counterclockwise) angles measured from the +x direction.

Answer 1

Given

the hockey puck has mass m = 0.21 kg

velocity towards east is u = 2.2 m/s

the constant net force should act on it during 0.43 s to change the puck's velocity to 3.7 m/s towards west F = ?

we know that the impulse = change in momentum

F*dT = m *dV

F = m*dV/dT

F = 0.21*(-3.7-2.2)/(0.43) N

F = -2.8814 N

the direction is to west (-ve sign indicates )

b) if the velocity is changed to 3.7 s towards south from East then the force is  

here first the horizontal force that is along east and vertical force along south we can calculate later by vector form we get the magnitude and direction

force along +x direction (EAST) Fx = m(dv/dt)

Fx = (0.21)(0-2.2)/(0.43) N = -1.07442 N

along south

Fy = (0.21)(-3.7-0)/(0.43) N = -1.81 N

the magnitude of the force is

F = sqrt(Fx^2+Fy^2)

F = sqrt((-1.07442)^2+(-1.81)^2) N

F = 2.11 N

the direction is  

theta = arc tan (1.81/1.07442) = 59.30 degrees

that is 180+59.30 = 239.30 degrees from the +x axis (counter clock wise dierection)