Answer:
(a) Car A was going faster
(b) Car A was at 120 Km/h
Explanation:
Linear Momentum
The momentum of an object of mass m traveling at a velocity [tex]\vec v[/tex] is given by
[tex]\vec p=m\vec v[/tex]
Both the moment and the velocity are vectors. If a system of particles A and B collide and no external forces act on them, the total momentum is conserved. i.e.
[tex]m_a\vec v_a+m_b\vec v_b=\vec v_a'm_a+\vec v_b'm_b[/tex]
Where ma, mb are the masses of the particles A and B respectively and va, vb, va' and vb' are their respective velocities before and after the collision.
The question states that after the collision, both cars get stuck which means their final velocity is common to them, and our equation becomes
[tex]m_a\vec v_a+m_b\vec v_b=\vec v (m_a+m_b)[/tex]
(a)
We have called [tex]\vec v[/tex] to the final common velocity. The car B was traveling south which means its rectangular components of the speed are
[tex]\vec v_b=0\hat i-v_b\hat j[/tex]
The car A was traveling 30° north of east. Its components are
[tex]\vec v_a=v_acos30^o\hat i+v_asin30^o v_a\hat j[/tex]
After the collision, both cars travel at a velocity
[tex]\vec v=vcos10^o\hat i+vsin10^o\hat j[/tex]
Let's replace all the velocities into the above formula
[tex]m_a(v_acos30^o\hat i+v_asin30^o v_a\hat j)+m_b(0\hat i-v_b\hat j)= (vcos10^o\hat i+vsin10^o\hat j)(m_a+m_b)[/tex]
Equating the x-components:
[tex]m_av_acos30^o=(m_a+m_b)vcos10^o[/tex]
Solving for va
[tex]\displaystyle v_a=\frac{(m_a+m_b)vcos10^o}{m_acos30^o}=2.085v[/tex]
Equating the y-components:
[tex]m_av_asin30^o-m_bv_b=(m_a+m_b)vsin10^o[/tex]
Solving for vb
[tex]\displaystyle v_b=\frac{m_av_asin30^o-(m_a+m_b)vsin10^o}{m_b}=0.869v[/tex]
We can see car A was going faster than car B
(b)
If the slower car was at speed limit (50 Km/h), we will find the speed of the car A. Dividing va/vb, we get
[tex]\displaystyle \frac{v_a}{v_b}=\frac{2.085v}{0.869v}=2.4[/tex]
Or equivalently
[tex]v_a=2.4v_b=2.4\cdot 50=120\ Km/h[/tex]
