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

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

θ = tan-1(opp/adj)
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(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.

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 = tan-1(opp/adj)
θ = 163.1°

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)

Leave a Reply

Name *
Email *