# Inelastic Collision in Two Dimensions

• jaankney
In summary, a 1300-kg car and a 15,000-kg truck collide at an intersection and move off together, leaving a 5m skid mark at a 30 degree angle from the car's original direction. With a coefficient of friction of 0.700, the initial velocities of the car and truck can be calculated to be 89.7 m/s and 4.49 m/s respectively. Using the work-energy theorem or kinematic equations, the magnitude of the velocity after the collision can be found to be 8.28 m/s.
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 μk = 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

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?

Hello jaankney,

Welcome to Physics Forums!

jaankney said:

## 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 * 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. There's some very minor rounding differences compared to the answer that I got, but the differences are minor.

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!

## 1. What is an inelastic collision in two dimensions?

An inelastic collision in two dimensions refers to a type of collision between two objects where there is a loss of kinetic energy. This means that after the collision, the objects will not continue moving in their original directions with the same velocities as before the collision.

## 2. How is the momentum conserved in an inelastic collision in two dimensions?

While the kinetic energy is not conserved in an inelastic collision in two dimensions, the total momentum of the system is still conserved. This means that the total momentum of the two objects before the collision will be equal to the total momentum after the collision.

## 3. What is the formula for calculating the final velocities in an inelastic collision in two dimensions?

The formula for calculating the final velocities in an inelastic collision in two dimensions is:
v1f = ((m1 - m2)v1i + 2m2v2i) / (m1 + m2)
v2f = ((m2 - m1)v2i + 2m1v1i) / (m1 + m2)
Where m1 and m2 are the masses of the two objects, and v1i and v2i are their initial velocities.

## 4. What are some real-life examples of inelastic collisions in two dimensions?

Some real-life examples of inelastic collisions in two dimensions include a car crashing into a wall, a ball hitting the ground and not bouncing back with the same velocity, and two objects colliding and sticking together.

## 5. How does an inelastic collision in two dimensions differ from an elastic collision?

An inelastic collision in two dimensions differs from an elastic collision in that in an elastic collision, there is no loss of kinetic energy and the two objects will bounce off each other, while in an inelastic collision, there is a loss of kinetic energy and the two objects will stick together after the collision.

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