# Thread: hockey puck speed.

1. ## hockey puck speed.

a hockey player gives it an initial speed of 2.4 m/s.
However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice.

As a result, the puck travels only one-half the distance between the players before sliding to a halt.

So, What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?
in m/s

need help setting this up please.

2. Originally Posted by rcmango
a hockey player gives it an initial speed of 2.4 m/s.
However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice.

As a result, the puck travels only one-half the distance between the players before sliding to a halt.

So, What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?
in m/s

need help setting this up please.
$(v_{f})^2=(v_{i})^2+2ad$

$0=2.4^2+2(a)\frac{1}{2}$

$a=-5.76$

$0=(v_{i})^2+2(-5.76)\frac{2}{2}$

$0=(v_{i})^2-11.52$

$11.52=(v_{i})^2$

$\sqrt{11.52}=v_{i}$

$3.39=v_{i}$

3. Okay, I see how we can use a kinematic equation to solve this. I didn't realise we could just plug in half as it was given in the equation.

So i think, what is KEY here is that you add 1/2 to 1/2 (2/2) so that it makes the entire distance with that acceleration right? which gives the minimum velocity?

another thing, when using acceleration for kinematics equations, I know that I needed the acceleration. However, how do i know when i need to find the acceleration, or when to just plug in acceleration due to gravity (9.81) ?

thanks alot for the help, excellent solution.

4. Originally Posted by rcmango
another thing, when using acceleration for kinematics equations, I know that I needed the acceleration. However, how do i know when i need to find the acceleration, or when to just plug in acceleration due to gravity (9.81) ?
Ooooooooooooouch!! (Mommy, my head hurts!)

NEVER assume an unknown acceleration is equal to g. The only time you can take the acceleration to be equal to g is if you have an object in free fall. This is the ONLY time you may assume that.

And, among other reasons, the acceleration of the hockey puck is horizontal, not vertical...

Originally Posted by rcmango
Okay, I see how we can use a kinematic equation to solve this. I didn't realise we could just plug in half as it was given in the equation.

So i think, what is KEY here is that you add 1/2 to 1/2 (2/2) so that it makes the entire distance with that acceleration right? which gives the minimum velocity?
$W_{nc} = \Delta E$

$W_{fric} = \Delta KE$ (I am assuming the ice is horizontal and flat.)

$-f_k s = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2$
Where s is the magnitude of the displacement between the starting and ending points..

$-f_k s = - \frac{1}{2}mv_0^2$ since the hockey puck comes to rest at the end.

Now, in the first part, we are given that the puck only goes half the distance (d) when given an initial speed $v_0$, so
$-f_k \left ( \frac{1}{2} d \right ) = - \frac{1}{2}mv_0^2$

$f_k = \frac{\frac{1}{2}mv_0^2}{\left ( \frac{1}{2} d \right )} = \frac{mv_0^2}{d}$

For the second part, we know the puck goes the whole distance d, but at a new initial speed. (Let's call this one $u_0$ to distinguish it from the other one.)

So
$-f_k d = - \frac{1}{2}mu_0^2$

$u_0 = \sqrt{\frac{f_k d}{\frac{1}{2}m}} = \sqrt{\frac{2 f_k d}{m}}$

and we know $f_k$ from the first part of the problem...
$u_0 = \sqrt{\frac{2 \left ( \frac{mv_0^2}{d} \right ) d }{m}}$

$u_0 = \sqrt{\frac{2 mv_0^2}{m}}$

$u_0 = \sqrt{2 v_0^2}$

Here's a challenge for you: See if you can understand what's going on well enough that you can do this whole problem in something like two lines...

-Dan

5. Originally Posted by rcmango
a hockey player gives it an initial speed of 2.4 m/s.
However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice.

As a result, the puck travels only one-half the distance between the players before sliding to a halt.

So, What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?
in m/s

need help setting this up please.
The friction is independent of speed, so if the puck goes twice as far
the friction does twice as much work, which equals the loss of KE, so
the KE must be doubled, but KE is proportional to the square of speed,
so the initial speed needs to be a factor of sqrt(2) larger for the puck
to just reach the second player.

That is it needs to be hit at 2.4 x sqrt(2) ~= 3.394 m/s.

(If you understand what is going on you don't need that forest of algebra seen above)

RonL