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A 0.16 kg hockey puck is moving on an icy, frictionless, horizontal surface at a speed of 3 m/s to the right.

calculate the velocity of the puck (magnitude and direction) after a force of 25 n directed to the right has been applied for 0.05 s.
_____ m/s

a. +x direction.
b. +y direction.
c. -x direction.
d. -y direction.

User Cybergen
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1 Answer

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Final answer:

The velocity of the puck after the force has been applied is 10.8125 m/s in the +x direction.

Step-by-step explanation:

To find the velocity of the puck after the force has been applied, we can use the equation F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, we have F = 25 N and m = 0.16 kg. As the force is applied for 0.05 s, we can also use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. The initial velocity u is given as 3 m/s to the right.

Rearranging the equation, we get v = u + at, where v = ?, u = 3 m/s, a = F/m, and t = 0.05 s. Plugging in the values, we have v = 3 + (25/0.16)(0.05) = 3 + 7.8125 = 10.8125 m/s. So the velocity of the puck after the force has been applied is 10.8125 m/s to the right. Therefore, the correct answer is a. +x direction.

User Vikas Gautam
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