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

a) A hockey puck with mass 2*m* 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 4*m* 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?

**2. Relevant equations**

p_{a} + p_{b} = p_{final}

m_{a}v_{a} + m_{b}vb = m_{f}v_{f}

a^{2} + b^{2} = c^{2}

θ = tan^{-1}(opp/adj)

v_{cm,x} = (m_{a}v_{a,x} + m_{b}v_{b,x}) / (m_{a} + m_{b})

v_{cm,y} = (m_{a}v_{a,y} + m_{b}v_{b,y}) / (m_{a} + m_{b})

∑F = Δp/Δt

Δp = J (impulse)

**3. The attempt at a solution**

a) m_{a} = 2*m*

v_{a} = 0.5 m/s (due north)

m_{b} = 4*m*

v_{b} = 1.1 m/s (at 20° west of north)

m_{a}v_{a} + m_{b}vb = m_{f}v_{f}

(2*m*)(0.5) + (4*m*)(1.1) = (2*m*+ 4*m*)v_{final}

5.4*m*/6*m* = 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_{a}v_{a,x} + m_{b}v_{b,x}) / (m_{a} + m_{b})

v_{cm,x} = ((2*m*)(0) + (4*m*)(.38)) / 6*m*

v_{cm,x} = .253 m/s

v_{cm,y} = (m_{a}v_{a,y} + m_{b}v_{b,y}) / (m_{a} + m_{b})

v_{cm,y} = ((2*m*)(0.5) + (4*m*)(1.0)) / 6*m*

v_{cm,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 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)

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