Inelastic Collision in Two Dimensions

  • Thread starter jaankney
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Homework Statement


A 1300-kg car collides with a 15,000-kg truck at an intersection and they couple together and move off as one leaving a skid mark 5m long that makes an angle of 30.0° with the original direction of the car. If μk = 0.700, find the initial velocities of the car and truck before the collision.
Note: The car is travelling east and the truck is travelling north.


Homework Equations


FIMPACT=FFRICTION
F = μ * m * g
F = m * a
v2=v02 + 2 * a * x
x component of momentum: (m1 + m2) * Vf * cos θ = m1 * v1i
y component of momentum: (m1 + m2) * Vf * cos θ = m2 * v2i


The Attempt at a Solution


The force at impact equals the force of friction: F = (0.700)(16300 kg)(9.8 m/s^2) = 111818 N
Find deceleration of the system: a = 111818 N / 16300 kg = 6.86 m/s^2
Use kinematic equation to find v0: 0 = v02 + (2)(6.86 m/s^2)(5 m); v0 = 8.28 m/s
x component: (16300 kg)(8.28 m/s^2)(cos 30°) = 1300 * v1i; v1i = 89.7 m/s
y component: (16300 kg)(8.28 m/s^2)(sin 30°) = 15000 * v2i; v2i = 4.49 m/s

Obviously the initial velocity for the car is outlandish at 89.7 m/s. I've tried a couple of other equations and I keep getting this same answer. I think the problem is that I am assuming that Vf = v0. When I use tan θ = (m2 * v2i) / (m1 * v1i) I get v2i / v1i = 0.05. If the speed of the car gets much lower the truck will hardly be moving so I am baffled.

Should I try calculating Vf by figuring out the kinetic energy instead?
 

Answers and Replies

  • #2
collinsmark
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Hello jaankney,

Welcome to Physics Forums!

The Attempt at a Solution


The force at impact equals the force of friction: F = (0.700)(16300 kg)(9.8 m/s^2) = 111818 N
Find deceleration of the system: a = 111818 N / 16300 kg = 6.86 m/s^2
Use kinematic equation to find v0: 0 = v02 + (2)(6.86 m/s^2)(5 m); v0 = 8.28 m/s
There's a slightly easier way to get the magnitude of the velocity after the collision by using the work-energy theorem, and [itex] W = \vec F \cdot \vec s [/itex] (for a constant force [itex] \vec F [/itex]). But you get the same answer either way. The way you did it is fine too. :approve: I just though thought I'd point it out for future problems.
x component: (16300 kg)(8.28 m/s^2)(cos 30°) = 1300 * v1i; v1i = 89.7 m/s
y component: (16300 kg)(8.28 m/s^2)(sin 30°) = 15000 * v2i; v2i = 4.49 m/s
'Looks about right to me. :approve: There's some very minor rounding differences compared to the answer that I got, but the differences are minor.
 
  • #3
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Thanks for the response and the suggestion for future problems! I just assumed I did the problem wrong because the velocity was so high. I find these forums really helpful for finding good ways of thinking about problems but this was obviously my first post. Thanks again!
 

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