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

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 2.90 m/s.

(a) Calculate the velocity of the puck (magnitude and direction) after a force of 32.0 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 12.5 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

Explanation / Answer

here,
taking Right Direction as +X and left as -X

m = 0.160 kg
U = 2.90 m/s towards Right
F = 32.0 N towards Right

as Force = mass* Acceleration
32 = 0.160 * a
a = 200 m/s^2

from First Eqn of moiton we have,
V = U + a*t
V = 2.90 + 200 * .05s

V = 12.9 m/s towards right i.e towards +x

Part B:
Now F = 12.5 N towards Left
so
-12.5 = 0.160*a

a = -78.125 m/s^2

Therefore,from First Eqn of moiton we have,
V = U + a*t
V = 2.90 - 78.125*.05

V = -1.006 m/s towards left or -x

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