Question

A 1445-kg car,c, moving east at 21.2m/s, collides with a 2625-kg car,d,moving south at 17.5 m/s, and the cars stick together. In what direction and with what speed do they move after the collision

Answer 1

Answer:

After the collision, they move with a speed of 13.57 m/s at 56.30° south of east

Explanation:

Conservation Of Momentum

The total momentum of a system of particles with masses m1,m2,...mn that interact without the action of external forces, is conserved. It means that, being $\vec P$ and $\vec P'$ the initial  and final momentums respectively:

$\vec P = \vec P'$

Or equivalently, for a two-mass system

$m_1\vec v_1+m_2\vec v_2=m_1\vec v_1'+m_2\vec v_2'$

All the velocities are vectors. Let's express the velocities in magnitude v and direction $\theta$ as $(v,\theta)$. It's rectangular components will be

$v_x=vcos\theta$

$v_y=vsin\theta$

The first car is moving east at 21.2 m/s. Its velocity is

$\vec v_1=(21.2,0^o)$

Recall that East is the zero-degree reference for angles

Expressing in rectangular form:

$\vec v_1==$

The second car is moving south at 17.5 m/s. Its velocity is

$\vec v_2=(17.5,270^o)$

$\vec v_2==$

The total initial momentum is

$\vec P=m_1\vec v_1+m_2\vec v_2$

$\vec P=1445+2625$

$\vec P=\ Kg.m/s$

They collide and stick together in a common mass and velocity \vec v', thus

$\vec P'=(m_1+m_2)\vec v'$

It must be equal to the initial momentum, thus

$(m_1+m_2)\vec v'=\ Kg.m/s$

Solving for $\vec v'$

$\displaystyle \vec v'=\frac{\ Kg.m/s}{m_1+m_2}$

$\displaystyle \vec v'=\frac{\ Kg.m/s}{4070}$

$\displaystyle \vec v'=\ m/s$

The magnitude of \vec v' is

$|\vec v'|=\sqrt{7.53^2+(-11.29)^2}=13.57\ m/s$

The direction angle is

$\displaystyle tan\theta=\frac{-11.29}{7.53}$

$\displaystyle \theta=-56.30^o$

After the collision, they move with a speed of 13.57 m/s at 56.30° south of east