Find the Angle between Due East and the System after the Collision

1. Jun 13, 2014

Art_Vandelay

Hi, everyone! This is my first post on this website. I answered the first part of the problem correctly, but I have gotten Part B wrong many times (after trying different approaches) and haven't attempted Part C yet. I would really appreciate any and all advice and assistance with this problem! Thank you!

1. The problem statement, all variables and given/known data

a) A hockey puck with mass 2m sliding on frictionless ice at an initial speed of va=0.5 m/s due north again collides with an small penguin of mass 4m sliding 20° west of north with a velocity of vb=1.1 m/s. Again, there are no injuries as the penguin hops onto the puck. Use the vector model and the Pythagorean theorem to determine their speed.

b) Find the angle between due east and the direction of the puck-penguin system after the collision. (Be sure to include the units on your answer.)

c) Our hapless 15-kg penguin is still sliding on the ice in the due east direction with a speed of 5.6 m/s. A friendly skater uses a broom to push the penguin gently off the ice. The skater exerts a force of 10 N to the south for 11 s to help the penguin off the ice. What is the direction of the penguin's momentum vector at the end of the push?

2. Relevant equations

pa + pb = pfinal
mava + mbvb = mfvf
a2 + b2 = c2

vcm,x = (mava,x + mbvb,x) / (ma + mb)
vcm,y = (mava,y + mbvb,y) / (ma + mb)

∑F = Δp/Δt
Δp = J (impulse)

3. The attempt at a solution

a) ma = 2m
va = 0.5 m/s (due north)
mb = 4m
vb = 1.1 m/s (at 20° west of north)

mava + mbvb = mfvf
(2m)(0.5) + (4m)(1.1) = (2m+ 4m)vfinal
5.4m/6m = vfinal
vfinal = 0.9 m/s [CORRECT]

b) For this, I used θ + 90° because I assumed that the puck-penguin system would have moved to the west, and if it's measured from the east x-axis, I think it would require an additional 90° added to the answer.

Attempt 1

I broke the velocities into component vectors: vxa = 0 m/s, vya = 0.5 m/s; vxb = 0.38 m/s, vyb = 1.0 m/s

θ = tan-1((1.0 m/s + 0.5 m/s) / (0.38 m/s))
θ = 75.8°
θ + 90° = 165.8° [INCORRECT]

Attempt 2

Next, I tried using the center of mass components.

vcm,x = (mava,x + mbvb,x) / (ma + mb)
vcm,x = ((2m)(0) + (4m)(.38)) / 6m
vcm,x = .253 m/s

vcm,y = (mava,y + mbvb,y) / (ma + mb)
vcm,y = ((2m)(0.5) + (4m)(1.0)) / 6m
vcm,y = .833

θ = 163.1°

θ = tan-1((2*0.38 m/s) / ((2*1.0 m/s) + 0.5 m/s))
θ = 107° (Thank you, Nathanael!)

c) I'm really not sure how to start on this part.
So far, I have v1x = 5.6 m/s, v1y = 0 m/s; v2x = 0 m/s, v2y = 7.3 m/s

The v2y was found by using the following equation: ∑F = Δp/Δt
F = mv / t
((10 N)(11 s)) / 15 kg = v
v = 7.3 m/s (south)

Last edited: Jun 13, 2014
2. Jun 13, 2014

Nathanael

Your first attempt was close, but you forgot to account for the mass. Your answer would be almost correct if they had the same mass.
(It would still be wrong, because you made one more small mistake: the "opposite" would actually be the component in the west direction, and the "adjacent" would be the North direction, this just comes from how you decided to define your angle.)

Do you know how to involve the masses of the objects to determine the direction?

P.S.
Welcome to Physics Forums!

EDIT:
Actually, if you wanted, the west direction could be the adjacent and the north could be the opposite. But if you did it this way, you would have to make the 0.38 negative (and then you'd have to add 180 since the range of arctangent is [-90°, 90°] so your angle is out of range)

Sorry if I'm making this confusing, it's hard to clearly say, but it's just trig

Last edited: Jun 13, 2014
3. Jun 13, 2014

Nathanael

10N*11s would be the impulse right? Which, you said, is equal to Δp.

So if 10N*11s is the change in momentum, when you divide it by 15kg what number do you get?
(I know you get 7.3, but I mean, what does that number mean? What does it represent?)

4. Jun 13, 2014

Art_Vandelay

Thank you! :)

In this case, since mb is twice as massive as ma, could I multiply the vb value by 2?

Or would I need to utilize a momentum equation?

5. Jun 13, 2014

Art_Vandelay

Since N is kg*m/s2, then dividing that by kg would result in the change in velocity (m/s), so the total of velocity components sum to 7.3 m/s, not only the v2y?

6. Jun 13, 2014

Nathanael

Yes, exactly. That would be the change in velocity (7.3 m/s south)

So the final velocity would be 7.3 m/s south and 5.6 m/s east (Because $V_f=V_i+ΔV$)

So what angle would the penguin be moving in?

Last edited: Jun 13, 2014
7. Jun 13, 2014

Nathanael

You could utilize a momentum equation if you want, but the result would just be what you said (multiply vb by 2)

The reason is simply that if it's twice as massive, it's going to contribute twice the effect on the final velocity

(If that's not intuitive to you, then you could use a momentum equation and derive it that way.)

Note that this intuition only applies because we only care about direction. If you wanted to find the speed, you would have to do it the way you did it in part A

Last edited: Jun 13, 2014
8. Jun 13, 2014

Art_Vandelay

That makes so much sense!
So, θ = -tan-1(7.3/5.6)
θ = -52.5°

Thank you so much for all of your help, Nathanael! :)

9. Jun 13, 2014

Nathanael

You are welcome :)