By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m[tex]U^{2}[/tex]
K.E = 1/2 x 0.17 x [tex]6^{2}[/tex]
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/[tex]s^{2}[/tex]
To calculate the final velocity, let us use third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2as
[tex]V^{2}[/tex] = [tex]6^{2}[/tex] + 2 x 0.3 x 61
[tex]V^{2}[/tex] = 36 + 36
[tex]V^{2}[/tex] = 72
V = [tex]\sqrt{72}[/tex]
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: https://brainly.com/question/402617
