# Hockey puck striking a wall

## Homework Statement

A 259 g hockey puck is sliding on ice with a speed of 5.34 m/s hits a wall at an angle of 34.8° to the wall and bounces back at the same angle with the same speed.

If the wall and the hockey puck are in contact for 11.3 ms, find the magnitude of the impulse on the puck.

## Homework Equations

p=mv
change in p = J
J=F(change in t)

J is impulse force

## The Attempt at a Solution

1st attempt: I tried just multiplying my velocity and mass to get momentum, that failed

2nd attempt: I tried resolving my components
X: (0.259)(5.34)(sin34.8)=0.789 kg m/s
Y: (0.259)(5.34)(cos34.8)=1.14 kg m/s

vf = sqrt((0.789)^2 + (1.14)^2)
= 1.39 m/s

p=ma
= (0.259kg)(1.39m/s)
= 0.360 kg m/s

p=ma = J therefore J=0.360 kg m/s

That FAILED!

I'm really not sure what I'm doing wrong, could someone pls help?

gneill
Mentor
In your second attempt, when you calculated the X and Y components, you found the components of the momentum in the two directions. That's fine. But I don't see what you were doing in calculating vf as the square root of the sum of squares of the momenta. That should just give you back the momentum, not a velocity.

When the puck hits the wall, the velocity and momentum in the y-direction remains unchanged (because the angle of reflection is equal to the angle of incidence, and the overall speed is said to be unchanged). But what happens to the velocity in the x-direction?

Whoops, I have no idea why i did that, vf squared thing. I see that now.
But i assumed that the momentum in the x direction remained constant... should i not have assumed that?

gneill
Mentor
The puck is hitting the wall and changing direction, so one or both of the momenta components is changing.

So, what are the two components of the velocity (or momentum) before and after the collision?

So i can do it either way right?
If it's in terms of momentum, x will be: mv cos34.8 & y will be: mv sin34.8
right?

gneill
Mentor
That'll be the momenta before collision. How about after?

Since the velocity after is the same wouldn't the momentum be the same?

gneill
Mentor
The speed is the same (so stated in the problem). But the velocity has changed. Otherwise, the wall wasn't there and no reflection took place!

Hmm... So does that mean that the components of velocity have changed?

vXf: vcos34.8
vYf: vsin34.8 but I still get the same components?

where v is 5.34 m/s

gneill
Mentor
Suppose you were to throw a ball straight at a wall. It bounces back with the same speed. Is the velocity the same?

No, the velocity is negative... Sry, but i still don't see what you're getting at, sure the velocity would be negative, but when i resolve it... wouldn't it still be the same components?

OH! Wait a minute! Now my angle is going to be 90-34.8 and then
vXf is :5.34sin(90-34.8)
vYf is : 5.34cos(90-34.8)

gneill
Mentor
Yes, the same *magnitude* of components. But the sign of one of them has changed. This is important. Why is it important? Because you're looking to find the impulse, which is the total change in momentum, ∆p, that occurs during the impact of the puck and wall.

Momentum is a vector quantity, just as velocity is, which means that it has magnitude and direction. Momentum has the same direction as velocity. You are going to want to find the difference between the initial (incoming) momentum vector and the final (outgoing) momentum vector.

Now, only one of the velocity (and momentum) components is changing over the collision. So the change in momentum will be confined to that component that is expressing the change. I'm aiming to get you to identify which component that is, and to calculate the change in momentum.

In that case, wouldn't both components of the velocity have changed? So now it would be
vYf: 1.18 m/s
vXf: -5.21 m/s

That's what I've got so far... Erm... would i put it in terms of x and y now?

gneill
Mentor
Attached is a diagram of what is happening. Note that the Y-component isn't changed, and the angle to use is the 34.8° that was given -- it was stated that the angle was with respect to the wall, not the normal to the wall.

#### Attachments

• Puck Wall.jpg
2.8 KB · Views: 378
after calculating the momentum in both x and y directions just think this:
in which direction is the momentum changing- then after simple maths you shall get your answer

these calculations were very wrong conceptually:
[vf = sqrt((0.789)^2 + (1.14)^2)
= 1.39 m/s

p=ma
= (0.259kg)(1.39m/s)
= 0.360 kg m/s]

firstly sqrt((0.789)^2 + (1.14)^2) will not give you velocity. these would give you the net momentum . then you multiplied this with mass which is of no use.

Okay, I've got up to there... Now I have my diff velocities for x which is changing from 3.05 m/s to -3.05 m/s after collision. Do i just take the difference in momenta between the 2 components... or do i have to resolve again?

Oh wow... I can't believe it took me so long. Thanks for being so patient guys! :)

gneill
Mentor
No need to resolve now, since these two velocities are both lie along the same line parallel to the x-axis; they're x-components.

Just form ∆p from m∆v.