Question

0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t 0, the puck is moving to the right at 3.00 m/s. (a) Calculate the velocity of the puck ( magnitude and direction) after a force of 25.0 N directed to the right has been applied for 0.050 s. (b) If, instead, a force of 12.0 N directed to the left is applied from 0 tot 0.050s, what is the final velocity of the puck

Answer 1

Answer: $(a)10.812\ m/s\ (b)\ 0.75\ m/s\ \text{left}$

Explanation:

Given

Mass of hockey puck $m=0.160\ kg$

Initial velocity of hockey puck is $u=3\ m/s$

First a horizontal force of $25\ N$ is applied to the right for $0.05\ s$

acceleration associated with it is

$\Rightarrow a=\dfrac{25}{0.160}\\\\\Rightarrow a=156.5\ m/s^2$

Using equation of motion i.e.

$\Rightarrow v=u+at\\\Rightarrow v=3+156.25\times 0.05\\\Rightarrow v=3+7.812\\\Rightarrow v=10.812\ m/s$

(b) When a force of $12\ N$ is applied for 0.05s

Using equation of motion i.e.

$\Rightarrow v=3-\dfrac{12}{0.160}\times 0.05\\\\\Rightarrow v=3-75\times 0.05\\\Rightarrow v=3-3.75\\\Rightarrow v=-0.75\ m/s\\\Rightarrow v=0.75\ \text{towards left}$