1. The problem statement, all variables and given/known data A car with a mass of 1875 kg is travelling along a country road when the driver sees a deer dart out onto the road. The driver slams on the brakes and manages to stop before hitting the deer. The driver of a second car (mass of 2135 kg) is driving too close and does not see the deer. When the driver realizes that the car ahead is stopping, he hits the brakes but is unable to stop. The cars lock together and skid another 4.58 m. Allof the motion is along a straight line. If the coefficient of friction between the dry concrete and rubber tires is 0.750, what was the speed of the second car when it hit the stopped car? 2. Relevant equations Notation: "A" is the stopped car and "B" is the second car. v' is the resulting velocity after the two cars have collided and locked. mA = 1875 kg mB = 2135 kg uF = 0.750 d= 4.58m The velocity of car B before the collision and the final velocity are not given. vB = ? v' = ? So momentum is conserved, and the resulting equation would be: mB * vB = (mA + mB) * v' <--- Eqn 1 Also, the energy is conserved. The kinetic energy of the second car (B) right before the collision should equal the energy of friction (thermal energy?) used to stop the two locked cars (I think this is right...). So: Ek(before) = Ef (after) 0.5 * mB * vB ^2 = uF * Fn * d <--- Eqn 2 3. The attempt at a solution So basically my method was to use Eqn 2 to isolate "vB" , then sub this rearranged equation into Eqn 1: Eqn 2: 0.5 * mB * vB ^2 = uF * Fn * d 0.5 * mB * vB ^2 = uF * (mA + mB) * g * d vB = SQRT ([2* uF * (mA + mB) * g * d] / mB) vB = SQRT ([2* 0.750 * (1875 + 2135) * 9.8 * 4.58] / 2135) vB = 11.25 m/s ---> 40.48 km/h Eqn 1: mB * vB = (mA + mB) * v' v' = (mB * vB)/ (mA + mB) v' = (2135 * 40.48) / (2135 + 1875) v' = 21.5 km/h The answer is 55.5km /h ...can someone check over my method for me please?