A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At

ID: 1406968 • Letter: A

Question

A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t = 0 the puck is moving to the right at 3.50 m/s.

(a) Calculate the velocity of the puck (magnitude and direction) after a force of 21.5 N directed to the right has been applied for 0.050 s.
_____m/s

a) +x direction

b)+y direction

c) -x direction

d) -y direction

(b) If instead, a force of 8.8 N directed to the left is applied from t = 0 to t = 0.050 s, what is the final velocity of the puck?
____ m/s

a) +x direction

b) +y direction

c) -x direction

d) -y direction

Here ,

mass of hockey , m = 0.160 Kg

initial speed , u = 3.50 m/s

a) using second law of motion

change in momentum = force * time

0.160 *(v - 3.50) = 21.5 * 0.050

solving

v = 10.21 m/s

the final velocity is 10.21 m/s in a) +x direction

using second law of motion

change in momentum = force * time

0.160 *(v - 3.50) = - 8.8 * 0.050

solving for v

v = 0.75 m/s

the final velocity of puck is 0.75 m/s in the a) +x direction

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