Answer:
22.73 m/s or 81.72 kph
Explanation
We can find the combined mass of both cars as
970 kg + 2300 kg = 3270 kg.
Then the normal force of the cars can be calculated as
F(n)= mg
Where g is acceleration due to gravity 9.8m/s^2
3270 kg ×9.8 = 32046 kg*m/s^2.
coefficient of kinetic friction between tires and road to be 0.80 × F(n)
Then the frictional force can be calculated as
= (32046kg*m/s^2 × 0.80 )
= 25636.8 kg*m/s^2
We can now calculate the work done that was used stopping the cars as
Frictional force × distance
(25636.8 kg*m/s^2 ) × 2.9m= 74346.72kg*m^2/s^2
From kinetic energy formula, the combined velocity of the car can be determined
E=0.5 M V²
√(2E/M) = V
√(2*74346.72kg*m^2/s^2 / 3270 kg) = V
V= √ (45.472)
V=6.743293m/s
the momentum of both cars can be determined as
6.743293m/s * 3270 kg
= 22050.57kg*m/s
Now the final momentum of both cars must be equal to the the momentum of
the sports car just prior to the collision. Therefore, the speed of the sports car at impact.
=(22050.57 kg*m/s) / 970 kg = 22.73 m/s
We can convert that to km/h.
22.73 m/s * 3600 s/h / 1000 m/km = 81.72 kph
