[4th repost] What distance will the puck move?

Question

The puck moving with velocity v = 10 m/s hits the floor
at angle 45 deg and continues to move sliding on the
floor.

http://alexandersemenov.tripod.com/puck/...

Coefficient of friction μ = 0.3; g = 9.8 m/s².

Oops, sorry.

http://alexandersemenov.tripod.com/puck/index.htm

"Can't you just find the component of velocity"

My teacher told me that in order to solve
this problem correctly I must find the
horizontal velocity immediately after the impact. When I submitted the solution
v^2 = 2aL, and failed to explain what is v, she treatened to transfer me
to school for kids with special needs.

Just as it happend yesterday with the damned coin.

People, the puck cannot just 'lose' vertical componet of its velocity, some force must act on the puck in order to change its momentum. The real question is what is the magnitude and direction of that force.

It is physics, for heaven sake.

If you do not have a puck, take a book,
and throw it gently on the top of your desk. See what happens.

"This is not a physic's problem but an engineering problem"

The problem states in no uncertain terms that the puck continues to slide along the plane. No bouncing, absolutely inelastic collision, period.

Does it mean that Vh is conserved during the impact or not, is open question.

"many people agreeing on the correct answer"

Many people used to agree that Earth was flat. The Earth did not care.

"I couldn't find a way to make the book "stick" and slide along the floor nicely"

The reporter returned from the wedding and told the editor: "Nothing to write about. The ceremony was cancelled because the groom did not appear".

Answer 1

Edit: EUREKA I've got it. Actually, I'll give Cat credit for figuring out the concept in his PPS.

The naive expression for the horizontal velocity after impact is (10 m/s) cos (45).

But that neglects the fact that the normal force exerted an additional impulse on the puck to kill the downward momentum.
delta p = (masspuck)(10 m/s) sin(45)
So there was an extra normal force of
(delta p)/t for some time t.
So there was an extra friction force of
mu(delta p)/t for some time t.
That friction force exerted an impulse in the minus x direction of
mu(delta p).

So the momentum in the x direction is the naive momentum minus that extra friction impulse
(masspuck)(10 m/s)(cos 45)
- (masspuck)(10 m/s)(sin45)(0.3)

Cancel the mass to get the initial velocity.

Then we can use d = v^2 / 2g(mu) just like everyone knew we should.

:) Fun problem. And everyone who said disparaging things about this problem "not physics, engineering, not enough info, everybody agrees on the answer" should apologize, because the correct answer was there all along staring us in the face. No advanced concepts--just friction and conservation of momentum. We all deserve to get sent to the "special" school for missing it.

Answer 2

I'm watching this question with great interest, because I am wondering if there's going to be an explanation at the end of this. The straightfoward (and probably wrong) treatment of this is to assume that only the horizontal component of the velocity v "survives" the impact, in which case the answer is:

L = (10/√2)² / 2(0.3)(9.8) = 8.5034 meters

At the other extreme, since this is a hard puck landing on a presumably hard floor, all energy being conserved, the entire velocity v survives the impact, in which case the answer is exactly twice that of above, or L = 17.0068 meters.

But real life is something of a mess. If I dropped a puck straight down even on hard ice, it does not skitter straightaway in some direction with the same speed at which it hit the ice. In fact, it will bounce, and it's this bouncing that dissipates the kinetic energy. This problem poses a situation where there IS no bounce at all, somehow the puck has magically stuck to the floor upon impact and continued on some ideal sliding motion, with no explanations given as how this can even occur. In real life, L would be between the two answers given, 8.5 < L < 17 meters, because SOME of the vertical component of the velocity would in fact be converted into horizontal sliding motion. Exactly how much depends critically on the material properties of both the puck and the ground, which was left out.

That's why I'm waiting with waxing curiosity as to what the "right" answer is supposed to be. I can't wait to learn something from this.

Addendum: Some force was responsible for changing the momentum vector of the puck from a 45° to horizontal? But therein lies the rub. More than one kind of "force vector" can result in a horizontal motion. For example, if the force is at 90° to the surface of the floor, then as everybody has been saying, the final horizontal distance travelled would be 8.5 meters. However, if the puck was like a stream of water hitting a bend, so that there's a resultant force at 67.5°, then the distance travelled would be 17 meters. And other force angles are possible, depending on MATERIAL properties of the puck and ground! Will the real force vector please stand up?

Ever willing to be the empiricist, I got a book, "The Road to Reality" (by Roger Penrose), and tossed it on different kinds of floors, at roughly 45° trajectories, and sure enough, in every case the book bounced. I couldn't find a way to make the book "stick" and slide along the floor nicely.

Answer 3

Okay, here we go.
First, find the x component of the puck velocity. (y component becomes 0 when it hits the floor in this problem)
10m/s @ 45deg
Draw the vector triangle, then use trig to get the x component.
cos(45) = x/10 (remember to use radians, 45 deg = pi/4)
x = 7.07m/s

To find the distance traveled, use the formula:
2aD = vf^2 - vi^2
where:
a = acceleration
D = distance traveled
vf = final velocity = 0 (looking for when it stops)
vi = initial velocity = 7.07 (we just calculated)

Acceleration is found by using newton's law (F=ma) and the formula for frictional force (Ff = uN, where N is the normal force).
N = m*g
Ff = u*m*g
F=m*a = u*m*g (cancel out mass)
a=u*g
for this problem, acceleration is actually deceleration, or negative acceleration, so call a, (-a)
a = -ug

Insert a, vf, and vi back into the equation above to get
2*(-u*g)*D = 0 - 7.07^2
2*(-.3*9.8)*D = -49.98
-5.88*D = -49.98
D = 8.5m

And that is your answer.
The puck slides 8.5m before stopping.

I hope this helped, and was clear.


NOTE:
Some people have said that the puck cannot loose its vertical velocity.
They are wrong, that is exactly what is happening in this situation.
This is what vectors explain. The velocity in the x direction is independent from velocity in the y direction.
The collision in this problem is assumed to be inelastic, where the puck will deform to absorb the impact. No energy is lost, just converted to heat.

The help demonstrate how x and y velocities are independent, consider this example.
A bullet is fired horizontally at 300m/s. After it leaves the barrel, it is pulled down by gravity at a rate of 9.8m/s^2. How does this affect the horizontal velocity?


SECOND NOTE:
Looking through your first post, repost, 3rd repost, and 4th repost, you have many people agreeing on the correct answer. Who is telling you that this answer is wrong?

At the start of any physics problem, you have to state your assumptions. The information given in this problem leads us to the assumption that the puck sticks to the ground (called a purely inelastic collision, COR=0). There is not enough information for us to assume that is bounces. If I were to assume that it bounces, I can assume that it is a perfectly elastic collision, in which case it would bounce forever.
Another assumption that was made (which no one has complained about yet) is neglecting air resistance. In reality a sliding puck will be slowed down by air resistance.

Answer 4

Ok. The puck has vertical and horizontal motion. The vertical motion will be lost somehow (bouncing, or the floor starts moving... whatever).

The horizontal motion is

v = 10m/s* sin(45) or cos(45) = 7.07m/s

The kinetic energy 1/2 mv^2 is dissipated into friction work
Wf = 0.3mg*d

so d = 1/2v^2/(0.3*g) = 8.50m
______

P.S.: This is not a physic's problem but an engineering problem. It requires experimenting and developing an approximate model of the results. In reality the book IS GOING TO BOUNCE, once, maybe twice. While it bounces, it will feel no friction, and thus will travel further than 8.5m. What we need to know, from experiment, is how high the book rebounds, what was the time it spent in contact with the table while bouncing.
_____________________________

P.P.S: If one assumes the book doesn't bounce, then one must consider that for a small duration of time, the book presses on the table with more than just its weight, so the friction force is higher in this initial stage, and the book will travel less than 8.5m.

Answer 5

Kinetic energy=work done to slow the puck down
.5mv^=(u mg) d

d=.5 m v^2 / (u m g )
since v=10m/s cos(45)=7.07m/s
d= (7.07)^2/(2 x .3 x 9.8) = 8.5m (vertical component = 0)

The vertical component had to contribute to the normal force.. we have a bouncing puck or if the energy was absorbed we have these 8.5 m.
Also I wonder if the collision was elastic or inelastic.

So the puck landed on its flat surface This is physics for haven sake! LOL.

I wonder!