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CHW 7: Momentum Problem 8.102 ) 19 of 20 Constants Part A For the Texas Departme

ID: 1884260 • Letter: C

Question

CHW 7: Momentum Problem 8.102 ) 19 of 20 Constants Part A For the Texas Department of Public Safety, you are on a What was the Durangos speed just before the collision? foggy moning in a remote section of the Texas e. A 2012 Prius traveling due north in a highway in with a 2013 Dodge Durango that was traveling due east. After the locked together and skidded across the level ground until it struck a tree. You measure that the tree is 35 ft from the point of impact. The line from the point of impact to the tree is in a direction 39° north of east. From experience, you estimate that the coefficient of kinetic friction between the ground and the wreckage is 0.45. Shortly before the collision, a highway patrolman with a radar gun measured the speed of the Prius to be 50 mph and, according to a witness, the Prius driver made no attempt to slow down. Four people with a total weight of 460 lb were in the Durango. The only person in the Prius was the 150-lb driver. The Durango with its passengers had a weight of 6500 lb, and the Prius with its driver had a weight of 3042 lb. of the two vehicles was mph Submit Part B How fast was the wreckage traveling just before it struck the tree? mph 103 245

Explanation / Answer

Solution,

Initial momentum = 3042 lb *50mph = 1379.828 * 22.352kgm/s = 30842 kgm/s due north

initial momentum of Durango in SI unit = 2948.35 v due east

let the final speed of the system be vf

mass of the system in kg =4328 kg

equating momentum in north direction:

30842 kgm/s = 4328.178 vf sin 39 degree

30842=2724 Vf

vf =11.322 m/s = 25.3 mph..........speed of system after collision

now equating momentum in eastward direction

2948.35 v = 4328.178 vf cos 39 degree

2948.35 v = 4328.178*11.322 *cos 39 degree

v= [4328.178*11.322 *cos 39 degree]/2948.35

v=12.92 m/s = 28.9 mph (Ans)

B)

Weight, f = 0.45(6500 + 3042) = 4294 lb

Acceleration a = 4294/(202.1 + 94.56) = 14.47 ft/s^2

Using kinematic equation

V^2 = u^2 - 2ax

V^2 = (37.15)^2 - (2 x 14.47 x 35)

V= 19.16 ft/s = 13.03 mph (Ans)

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