Collision Question?

The morning after a winter ice storm in Dallas, a 1400 kg automobile going west on Chestnut Street at 35.0 km/h collides with a 2800 kg truck going south across Chestnut Street at 50.0 km/h.



Part A: If they become coupled on collision, what is the magnitude of their velocity after colliding? You can ignore forces between the vehicles and the icy road.



Part B: What is the direction of their velocity after colliding?



NOTE: I got 3.86 for part A but it doesnt seem right, can someone check my awnser. Maybe I am doing it wrong. Someone explain how you do both part a and b.

Collision Question?
let x = West

y = South

Vx = 1,400*35/4,200 = 11.667 kph

Vy = 50*2,800/4,200 = 33.333 kph

θ = atan(Vx/Vy) = 19.2900°W of S

V = 11.667/sin(19.2900°) = 35.3160 kph
Reply:Using the law of conservation of momentum, momentum before collision= momentum after collision.

momentum before collision= m1v1+ m2v2

= (1400*35) + (2800*50)

= 189,000kgkm/hr

momentume after collision. in this case the bodies stack together, this is inelastic collision and hence they move as one mass and with the same velocity say V. Total mass = 1400+2800= 4200kgs.

therefore 189,000kgkm/hr= 4200Vkgkm/hr

V= 45km/hr Southwest.
Reply:using the conservation of momentum. Momentum before equals momentum after.



let m1 be the mass of the automobile

let m2 bw the mass of the truck



So, let's look at the initial (before the crash) and final (after the crash) momentum of this problem.



Intially, the automobile goes to the West (or x direction). There is nothing else moving the the x direction except the automobile. We have

Pi(x) = m1*v (automobile)



Intially, the truck is going south. There is nothing else going in the y - direction, except the truck. We have

Pi(y) = m2*v (truck)



Now lets take a look at the final momentum. Let V be the speed of the cars together after the collision. Since both cars moves at an angle after the collsion, there are two components of the velocity of the cars. Vx = cos(θ) and Vy = sin(θ)V.



Since these components come from the velocity of the wreck after the collision, they can be used in finding the final momentum of the wreck.



Final Momentum, x-axis



Pf(x) = (m1 + m2) Vx

Pf(y) (m1 + m2) Vy



we already found Vx and Vy. Plug them in equations above.

Pf(x) = ( m1 + m2) cos(θ)V

Pf(y) (m1 + m2) sin(θ)V



Again, conservation of momentum

Pf = Pi



or:



in x-direction

m1 * v = (m1 + m2)cos(θ)V



in y-direction

m2 * v = (m1 + m2) sin(θ)V







Now plug the numbers in. Remember, the SI unit is in m/s



35 km/hr = 9.722 m/s

50 km/hr = 13.889 m/s



(1400) (-9.722) = (1400 + 2800) cos(θ)V

(2800) (-13.889) = (1400 + 2800) sin(θ)V



-13610.8 = 4200 cos(θ)V

-38889.2 = 4200 sin(θ) V



now divide the second equation by the first equation, you'll get

tan(θ) = -3889.2 / -13610.8



take an tan^-1

θ = 70.71 degrees S of W



use the degrees to find the speed of both cars after the collsion.





-38889.2 = 4200 sin(θ) V

-38889.2 = 4200 sin(70.71)V

V = -38889.2 / (4200 sin(70.71) )

V = -9.81 m/s ( or just 9.81m/s)



convert to km/hr



9.81 m/s = 35.316 km/hr



final answer: 35.316 km/hr at 70.71 degrees S of W



Note: 70.71 degrees S of W is the same thing as 19.29 degrees W of S.
Reply:Right !!

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