Question

A car with a mass of \(1700 \mathrm{kg}\) is traveling directly northeast $\left(45^{\circ} \text { between north and east) at a speed of } 14 \mathrm{m} / \mathrm{s}\right.\( \)(31 \mathrm{mph}),$ and collides with a smaller car with a mass of \(1300 \mathrm{kg}\) that is traveling directly south at a speed of \(18 \mathrm{m} / \mathrm{s}\) \((40 \mathrm{mph}) .\) The two cars stick together during the collision. With what speed and direction does the tangled mess of metal move right after the collision?

Short answer

Solution: 1. Understand the momentum conservation principle: The total momentum of the system is conserved in a collision if there are no external forces acting on the system. 2. Resolve velocities into components: Break down the velocities of both cars into their x and y components. 3. Calculate initial and final momenta: Determine the initial momenta along the x and y directions for each car and conserve them after the collision. 4. Calculate the magnitude and direction of the final velocity: Use the Pythagorean theorem to find the magnitude, and the arctangent function to find the direction. Calculations: 1. For larger car: - \(v_{1x} = 14 \cos(45^{\circ}) = 9.90 \frac{\mathrm{m}}{\mathrm{s}}\) - \(v_{1y} = 14 \sin(45^{\circ}) = 9.90 \frac{\mathrm{m}}{\mathrm{s}}\) 2. For smaller car: - \(v_{2x} = 0 \frac{\mathrm{m}}{\mathrm{s}}\) - \(v_{2y} = -18 \frac{\mathrm{m}}{\mathrm{s}}\) 3. Initial momenta: - \(p_{1ix} = 1800 \times 9.90 = 17820 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{1iy} = 1800 \times 9.90 = 17820 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{2ix} = 0 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{2iy} = 1200 \times (-18) = -21600 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) 4. Final momenta and velocity components: - \(p_{fx} = 17820 + 0 = 17820 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(p_{fy} = 17820 - 21600 = -3780 \frac{\mathrm{kg \cdot m}}{\mathrm{s}}\) - \(v_{fx} = \frac{17820}{3000} = 5.94 \frac{\mathrm{m}}{\mathrm{s}}\) - \(v_{fy} = \frac{-3780}{3000} = -1.26 \frac{\mathrm{m}}{\mathrm{s}}\) 5. Magnitude and direction: - \(v_{f} = \sqrt{5.94^2 + (-1.26)^2} = 6.07 \frac{\mathrm{m}}{\mathrm{s}}\) - \(\theta = \arctan \frac{-1.26}{5.94} = -12.05^{\circ}\) (measured clockwise from east) The final velocity of the combined wreckage after the collision is 6.07 m/s, 12.05° south of east.

Step-by-step solution

Understand the momentum conservation principle

Momentum conservation is a fundamental principle in physics. In a collision, the total momentum of the system before the collision is equal to the total momentum after the collision if there are no external forces acting on the system. Mathematically, this can be expressed as: \(\vec{p}\_{i} = \vec{p}\_{f}\) where \(\vec{p}\_{i}\) and \(\vec{p}\_{f}\) are the initial and final momenta of the colliding objects. In this exercise, we'll focus on the momentum conservation in two dimensions (2D), as the velocities are given in terms of north, east, and south directions.

Resolve velocities into components

We'll break down the velocities of both cars into their components, as this makes it more manageable when working with the conservation of momentum. The velocity components of the larger car moving northeast are: \(v\_{1x} = v\_{1} \cos(45^{\circ}) = 14 \frac{\mathrm{m}}{\mathrm{s}} \cos(45^{\circ})\) \(v\_{1y} = v\_{1} \sin(45^{\circ}) = 14 \frac{\mathrm{m}}{\mathrm{s}} \sin(45^{\circ})\) The velocity components of the smaller car moving south are: \(v\_{2x} = 0 \frac{\mathrm{m}}{\mathrm{s}}\) (as it's moving in the north-south direction only) \(v\_{2y} = -18 \frac{\mathrm{m}}{\mathrm{s}}\) (since it's moving in the negative y-direction)

Calculate initial and final momenta

For each car, we can calculate the initial momenta along the x and y directions using their masses and velocity components: \(p\_{1ix} = m\_{1}v\_{1x}\) \(p\_{1iy} = m\_{1}v\_{1y}\) \(p\_{2ix} = m\_{2}v\_{2x}\) \(p\_{2iy} = m\_{2}v\_{2y}\) According to the conservation of momentum, the total momentum before the collision must equal the total momentum after the collision. Since the cars stick together after the collision, we can say: \(p\_{fx} = p\_{1ix} + p\_{2ix}\) \(p\_{fy} = p\_{1iy} + p\_{2iy}\)

Calculate the magnitude and direction of the final velocity

To find the final velocity components (\(v\_{fx}\) and \(v\_{fy}\)), we divide the final momentum components by the total mass (m1 + m2). \(v\_{fx} = \frac{p\_{fx}}{m\_{1} + m\_{2}}\) \(v\_{fy} = \frac{p\_{fy}}{m\_{1} + m\_{2}}\) Now, we can find the magnitude of the final velocity using the Pythagorean theorem: \(v\_{f} = \sqrt{v\_{fx}^2 + v\_{fy}^2}\) And the direction can be calculated using the arctangent function: \(\theta = \arctan \frac{v\_{fy}}{v\_{fx}}\) Finally, we'll have the magnitude and direction of the final velocity, which are the speed and direction that the tangled mess of metal moves right after the collision.