Question

During a study of accident survivability a crash-test dummy is placed in a car. The vehicle is brought to a speed of 30.0 ??ℎ and then collided with a stationary barrier. During the first part of the study the upper torso of the dummy collides with the steering wheel and rebounds from the wheel at the same speed in1⁄100 ?ℎ of a second. During the second part of the study, in an identical car, an airbag deploys and brings the dummy to a rest against the steering wheel in 1⁄25 ?ℎ of a second. Find the force, in Newtons, acting on the dummy in each parts of the study. Assume the mass of the upper torso is 50.0 ??.

Answer 1

Here we will use the concept of Force as measure of rate of change of momentum. F = dp/dt

Now since it is given that the car is moving with a speed of 30.0 mph,

v = 30.0 mph = 30 x 1609.34/60 x 60 m/s = 13.4 m/s

Mass of dummy = 50.0 kg

So the momentum of the dummy before hitting PWi = mv = 50.0 kgx 13.4 m/s = 670.0 Ns

Assume the initial direction of motion to be positive, it is given in question that after hitting the dummy rebounds with same velocity thus the momentum

PWf = 50.0 kg x (-13.4 m/s) = -670 Ns

P = PWf - PWi = -1340 Ns

Now, the average force can be given by

F =

For first part,

= 1/100 s

F1 = -1340 Ns / 1/100 s = -1.34 x 105N

For second part,

= 1/25 s

F2 = -1340 Ns/ 1/25 s = -3.35 x 104 N

Here the negative forces implies that the forces are applied in the opposite direction of initial motion.