A car weighs 1000 kg. It is moving with a uniform velocity of 72 km/h towards a straight road. The driver suddenly presses the brakes. The car stops in 0.2 s. The retarding force applied on the car to stop it is
- A.100 N
- B.1000 N
- C.10 kN
- D.100 kN
Correct answer
Explanation
The correct answer is D, 100 kN. The car loses a momentum of 1000 kg × 20 m/s in 0.2 s, so the force is 20,000 ÷ 0.2 = 100,000 N, that is 100 kN.
First convert the speed: 72 km/h × 5/18 = 20 m/s. By Newton's second law, force equals the rate of change of momentum, F = m(v − u)/t. Here the final speed is zero, so the deceleration is 20 ÷ 0.2 = 100 m/s², and F = 1000 × 100 = 10⁵ N. The very short stopping time is what makes the force so large; the same idea explains why airbags and crumple zones lengthen the time of a crash to cut the force on passengers.
A is wrong because 100 N is the deceleration's value with the mass left out.
B is wrong because 1000 N would stop the car only in 20 s.
C is wrong because 10 kN would need 2 s to stop the car, ten times the time given.
Exam tip: change km/h to m/s by multiplying by 5/18, then use F = mΔv/Δt.
Practice Questions
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