Question

A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together and move 1.93 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

Answer 1

Answer:

The value  is $\mu = 0.72$

Explanation:

From the question we are told that

   The mass of the first car is  $m_1 = 1870\ kg$

    the initial  speed of the car  is  $u = 13.5 \ m/s$

    The  mass of the second car is  $m_2 = 2970\ kg$

    The distance move by both cars is  s =  1.93  m

Generally from the law of momentum conservation

    $m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f$

Here $u_2 = 0$ because the second car is at rest

and  $v_f$ is the final  velocity of the the two car

So

    $1870* 13.5+ 0= ( 1870 + 2970 ) * v_f$      

=> $v_f = 5.22\ m/s$

Generally from kinematic equation

    $v_f^2 = u_2^2 + 2as$

here a is the deceleration

So

    $5.22^2 = 0 + 2 *a * 1.93$

=> $a = 7.06 \ m/s^2$

Generally the frictional  force is equal to the force propelling the car , this can be mathematically represented as

   $F_f = F$

Here  F is mathematically represented as

$F = (m_1 + m_2) * a$

$F = (1870 + 2970) * 7.06$    

$F =34170.4 \ N$  

and

$F_f = \mu * (m_1 + m_2 ) * g$

$F_f = 47432 * \mu$

So

$47432 * \mu = 34170.4$

=> $47432 * \mu = 34170.4$

=> $\mu = 0.72$