A car, traveling at a speed of 10 m/s (22 mph) makes a head-on collision with a

Question

A car, traveling at a speed of 10 m/s (22 mph) makes a head-on collision with a tree. Thanks to the "crumple zone" engineered in modem cars, the car decelerates (from 10 m/s to zero) as the front of the car crumples over an approximate 1-meter distance. What is the change in the momentum of the 75-kg driver? Assuming uniform deceleration, how much time elapses during the deceleration? What is the average force acting on the driver, assuming the driver is wearing his seatbelt, so he also decelerates from 10 m/s to zero over the same 1-meter distance. What would be the average force on the driver if he did not wear his seatbelt, and was stopped in a 0.5-cm distance by the tree itself?

Explanation / Answer

Here .

speed of car , u = 10 m/s

distance ,d = 1 m

a)

change in momentum = mass * change in speed

change in momentum = 75 * (0 - 10)

change in momentum = -750 Kg.m/s

b)

let the time taken is t

deceleration is d

Using third equation of motion

v^2 - u^2 =2 * a * d

- 10^2 = 2 * a * 1

a = -50 m/s^2

Using first equation of motion

v = u +a *t

0 = 10 - 50 * t

t = 0.2 s

the time taken is 0.2 s

c)

Using second law of motion

average force acting on driver = mass * average deceleration

average force acting on driver = 50 * 75

average force acting on driver = 3750 N

the average force acting on driver is 3750 N

d)

for distance , d = 0.5 cm

d = 0.005 m

Using work energy therum

work done by tree force = change in kinetic energy of driver

0.5 * 75 * 10^2 = 0.005 * F

F = 7.5 *10^5 N

the force acting on driver in this case is 7.5 *10^5 N

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