Question

This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. At an intersection car B was traveling south and car A was traveling 30° north of east when they slammed into each other. Upon investigation, it was found that after the crash, the two cars got stuck and skidded off at an angle of 10° north of east. Each driver claimed that he was going at the speed limit of 50 km/h and that he tried to slow down, but couldn’t avoid the crash because the other driver was going a lot faster. The masses of cars A and B were 1800 kg and 1500 kg, respectively. Determine (a) which car was going faster, (b) the speed of the faster of the two cars if the slower car was traveling at the speed limit.

Answer 1

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 $\vec v$ is given by

$\vec p=m\vec v$

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.

$m_a\vec v_a+m_b\vec v_b=\vec v_a'm_a+\vec v_b'm_b$

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

$m_a\vec v_a+m_b\vec v_b=\vec v (m_a+m_b)$

(a)

We have called $\vec v$ to the final common velocity. The car B was traveling south which means its rectangular components of the speed are

$\vec v_b=0\hat i-v_b\hat j$

The car A was traveling 30° north of east. Its components are

$\vec v_a=v_acos30^o\hat i+v_asin30^o v_a\hat j$

After the collision, both cars travel at a velocity

$\vec v=vcos10^o\hat i+vsin10^o\hat j$

Let's replace all the velocities into the above formula

$m_a(v_acos30^o\hat i+v_asin30^o v_a\hat j)+m_b(0\hat i-v_b\hat j)= (vcos10^o\hat i+vsin10^o\hat j)(m_a+m_b)$

Equating the x-components:

$m_av_acos30^o=(m_a+m_b)vcos10^o$

Solving for va

$\displaystyle v_a=\frac{(m_a+m_b)vcos10^o}{m_acos30^o}=2.085v$

Equating the y-components:

$m_av_asin30^o-m_bv_b=(m_a+m_b)vsin10^o$

Solving for vb

$\displaystyle v_b=\frac{m_av_asin30^o-(m_a+m_b)vsin10^o}{m_b}=0.869v$

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

$\displaystyle \frac{v_a}{v_b}=\frac{2.085v}{0.869v}=2.4$

Or equivalently

$v_a=2.4v_b=2.4\cdot 50=120\ Km/h$