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 7 5-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

a) Change in momentum = Mv2 - mV1 = (75)(10-0) = 750

b) v2= u2+aS

0 = 102+2*a*1

a= 50 m/s2

c) Force = change in momentum/time

time = 10/50 = 1/5 second

Force = 750/(1/5) = 3750 N

D)

if S = 0.005 m v2= u2+aS 0 = 10^2+2*a*0.005 a= 10000 m/s2 and Force = change in momentum/time time = 10/10000 = 10^-3 second Force = 750/(10^-3) = 750000 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.