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

[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!

Homework Statement

a) A hockey puck with mass 2m sliding on frictionless ice at an initial speed of v_a=0.5 m/s due north again collides with an small penguin of mass 4m sliding 20° west of north with a velocity of v_b=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?

Homework Equations

p_a + p_b = p_final
m_av_a + m_bvb = m_fv_f
a² + b² = c²

θ = tan^-1(opp/adj)
v_cm,x = (m_av_a,x + m_bv_b,x) / (m_a + m_b)
v_cm,y = (m_av_a,y + m_bv_b,y) / (m_a + m_b)

∑F = Δp/Δt
Δp = J (impulse)

The Attempt at a Solution

a) m_a = 2m
v_a = 0.5 m/s (due north)
m_b = 4m
v_b = 1.1 m/s (at 20° west of north)

m_av_a + m_bvb = m_fv_f
(2m)(0.5) + (4m)(1.1) = (2m+ 4m)v_final
5.4m/6m = v_final
v_final = 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: v_xa = 0 m/s, v_ya = 0.5 m/s; v_xb = 0.38 m/s, v_yb = 1.0 m/s

θ = tan^-1(opp/adj)
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.

v_cm,x = (m_av_a,x + m_bv_b,x) / (m_a + m_b)
v_cm,x = ((2m)(0) + (4m)(.38)) / 6m
v_cm,x = .253 m/s

v_cm,y = (m_av_a,y + m_bv_b,y) / (m_a + m_b)
v_cm,y = ((2m)(0.5) + (4m)(1.0)) / 6m
v_cm,y = .833

θ = tan^-1(opp/adj)
θ = 163.1°

Attempt 3: CORRECT ANSWER

θ = tan^-1(opp/adj)
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 v_1x = 5.6 m/s, v_1y = 0 m/s; v_2x = 0 m/s, v_2y = 7.3 m/s

The v_2y 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)

 

[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

 

[Nathanael]

((10 N)(11 s)) / 15 kg = v
v = 7.3 m/s (south)

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?)

 

[Art_Vandelay]

Thank you! :)

In this case, since m_b is twice as massive as m_a, could I multiply the v_b value by 2?

Or would I need to utilize a momentum equation?

 

[Art_Vandelay]

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?)

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 v_2y?

 

[Nathanael]

Since N is kg*m/s², then dividing that by kg would result in the change in velocity

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?

 

[Nathanael]

Thank you! :)

In this case, since m_b is twice as massive as m_a, could I multiply the v_b value by 2?

Or would I need to utilize a momentum equation?

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

 

[Art_Vandelay]

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?

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! :)

 

[Nathanael]

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

You are welcome :)