Inelastic Collision in Two Dimensions

[jaankney]

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 μ_k = 0.700, find the initial velocities of the car and truck before the collision.
Note: The car is traveling east and the truck is traveling north.

Homework Equations

F_IMPACT=F_FRICTION
F = μ * m * g
F = m * a
v²=v₀² + 2 * a * x
x component of momentum: (m₁ + m₂) * V_f * cos θ = m₁ * v_1i
y component of momentum: (m₁ + m₂) * V_f * cos θ = m₂ * v_2i

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 * v_1i; v_1i = 89.7 m/s
y component: (16300 kg)(8.28 m/s^2)(sin 30°) = 15000 * v_2i; v_2i = 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 V_f = v₀. When I use tan θ = (m₂ * v_2i) / (m₁ * v_1i) I get v_2i / v_1i = 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 V_f by figuring out the kinetic energy instead?

 

[collinsmark]

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 $W = \vec F \cdot \vec s$ (for a constant force $\vec F$). But you get the same answer either way. The way you did it is fine too. I just though thought I'd point it out for future problems.

x component: (16300 kg)(8.28 m/s^2)(cos 30°) = 1300 * v_1i; v_1i = 89.7 m/s
y component: (16300 kg)(8.28 m/s^2)(sin 30°) = 15000 * v_2i; v_2i = 4.49 m/s

'Looks about right to me. There's some very minor rounding differences compared to the answer that I got, but the differences are minor.

 

[jaankney]

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!