Question

The city police are in pursuit of Robin Banks after his recent holdup at the savings and loan. The high-speed police chase ends at an intersection as a 2080-kg Ford Explorer (driven by Robin) traveling north at 22.6 m/s collides with a 18400-kg garbage truck moving east at 10.4 m/s. The Explorer and the garbage truck entangle together in the middle of the intersection and move as a single object. Determine the post-collision speed and direction of the two entangled vehicles.

Answer 1

Answer:

$v=9.6215m/s\\\theta=13.8^o$

Explanation:

This problem is an example of a perfectly inelastic collision. After the collision, the two cars merge into a single object. So in order to find the velocity of that object, we need to find the velocity of the center of mass of the given system of 2 cars.

Mass of Ford = $m_F$ = 2080 kg

Mass of truck = $m_T$ = 18400 kg

Velocity of Ford = $v_F$ = 22.6 m/s north

Velocity of truck = $v_T$ = 10.4 m/s east

Let us break up the velocity vector into 2 components, one along north ($v_N$) and one along east ($v_E$).

Therefore,

$v_N=\frac{m_Fv_F_{north}+m_Tv_T_{north}}{m_F+m_T} =\frac{(2080\times22.6)+(18400\times0)}{2080+18400} =2.2953m/s$

(speed of truck along the north is zero)

Similarly,

$v_E=\frac{m_Fv_F_{east}+m_Tv_T_{east}}{m_F+m_T} =\frac{(2080\times0)+(18400\times10.4)}{2080+18400} =9.3437m/s$

(speed of Ford along the east is zero)    

Hence, the resultant speed of the entangled cars is given by,

$v=\sqrt{v_N^2+v_E^2} =\sqrt{2.2953^2+9.3437^2}m/s=9.6215 m/s$

and the direction is given by (please refer to the figure attached),

$\theta=tan^{-1}(\frac{v_N}{v_E} )=tan^{-1}(\frac{2.2953}{9.3437} )=13.8^o$