Question

In hockey activities, a warm hockey puck and a frozen hockey puck has a different coefficient of restitution: 0.5 for a warm hockey puck, and 0.35 for a frozen one. NHL requires the frozen pucks to be used in games. To make sure the puck can be used in the game, the referee drops the puck on its side from a height of 2.5 m. How high should the puck bounce if it is a frozen puck

Answer 1

If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

Given the data in the question;

Since the hockey puck was initially in the referee's hands

• Initial velocity; $u = 0m/s$

• Distance or height from which it was dropped;  $h = 2.5m$

• Acceleration due to gravity; $g = 9.8 m/s^2$

• Coefficient of restitution a frozen puck; $0.35$

First we will find the velocity of the Puck when it hits the ground

From the Third Equation of Motion:

$v^2 = u^2 + 2as$

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

$v^2 = u^2 + 2gh$

We substitute in our value and find "v"

$v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s$

Now, Velocity of the hock puck after it hits the ground and bounce back;

We know that; Coefficient of restitution $= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}$

Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision

we substitute in our values;

Relative Velocity after collision $= 0.35 \ *\ 6.85857m/s$

Relative Velocity after collision $= 2.4 m/s$

Now, to determine how high should the puck bounced back

We use the Third Equation of Motion:

$v^2 = u^2 + 2as$

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

$v^2 = u^2 + 2gh$

Now, since the hockey puck bounces back, it is experiencing a negative acceleration

Hence, the equation becomes

$v^2 = u^2 - 2gh$

We substitute our values into the equation and find "h"

$(0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m$

Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

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