Find Car B's Initial Velocity: A Physics Puzzle

[cbarker1]

Dear Every one, Two cars approach an extremely icy four-way perpendicular intersection. Car A travels northward at 12 m/s and car B is traveling eastward. They collide and stick together, traveling at 40.1° north of east. What was the initial velocity of car B (in m/s)? (Enter the magnitude. Assume the masses of the cars are equal.)

Work:

Given:
$$u_1=12 m/s$$
$$Car A= <0,1> North $$
$$Car B=<1,0> East $$

$$y: m_a*u_1+m_a*v_1=(m_a+m_b)*v_f*cos (40.1 degrees)$$
$$x:m_b*u=(m_a+m_b)*v_f*sin (40.1 degrees)$$

 

[MarkFL]

What we want to do is apply conservation of momentum ("extremely icy" is our cue to neglect any friction as negligible). Let's orient our coordinate axes such that north points in the positive $y$ direction and east points in the positive $x$ direction.

Momentum of car A:

$$m\cdot\vec{v_A}=m\langle 0,12 \rangle$$

Momentum of car B:

$$m\cdot\vec{v_B}=m\langle v_B,0 \rangle$$

And so by addition, we find the total momentum of the two cars before the impact to be:

$$m\langle v_B,12 \rangle$$

Now, the momentum of the two cars stuck together after the impact is given by:

$$2mv_f\langle \cos\left(40.1^{\circ}\right),\sin\left(40.1^{\circ}\right) \rangle$$

Equate the components of the initial and final momenta, and what system results?

 

[MarkFL]

As a followup, we obtain the two equations (after dividing out $m$):

$$v_B=2v_f\cos\left(40.1^{\circ}\right)$$

$$12=2v_f\sin\left(40.1^{\circ}\right)\implies 2v_f=\frac{12}{\sin\left(40.1^{\circ}\right)}$$

Substituting for $2v_f$ into the first equation, there results:

$$v_B=12\cot\left(40.1^{\circ}\right)\approx14.25$$