Question

Two cars A and B slide on an icy road as they attempt to stop at a traffic light. The mass of A is 1100 kg and the mass of B is 1400 kg. The coefficient of kinetic friction between the locked wheels of both cars and the road is 0.130 car A succeeds in coming to rest at the light, but car B cannot stop and collides rear ends of car A. After the collision car A comes to rest 8.20 m ahead of the impactpoint and B 6.10 m ahead (see figure). Both drivers had their brakes locked throughout the incident. (a) Form the distance each car moved after the collision, find the speed of each car immediately after impact. (b) Use conservation of momentum to find the speed at which car B struck car A. On what grounds can the use of momentum conservation be criticized here ?

Answer 1

For an object with initial speed = v

Acceleration = a

Distance = x

t = v/a ________(1)

Average speed = v/2

$x=\frac{1}{2}a{t}^2$

$=\frac{1}{2}\times a\times {\left(\frac{v}{a}\right)}^2$

$2x=\frac{v^2}{a}$

$v=\sqrt{2}ax$

$=\sqrt{2}\left({\mu }_{k}g\right)x$

$=\sqrt{2}{\mu }_{k}gx$

To obtain the speed of car A after collision, substitute 0.13 for ${\mu }_k$ , $9.81m{s}^{-2}$ for of and $8.20m$ for x in the equations.

$v=\sqrt{2}{\mu }_{k}gx$

$=4.57m{s}^{-1}$

To obtain the speed for can B after collision, ${\mu }_k=0.13$, $9.81m{s}^{-1}$, 6.10 m

$v=\sqrt{2}{\mu }_{k}gx$

= 3.94 m

$W_B{v}_0=m_A{V}_A+m_B{V}_B$

$v_0=\frac{m_A{V}_A+m_B{V}_B}{M_B}$

$v_0=\frac{1100\times 4.57+1400\times 3.44}{1400}$

$=7.53m{s}^{-1}$