A 900-kg car traveling $30.0 ^ { \circ }$ south of east at 12.0 m/s suddenly collides with a 750-kg car traveling north at 17.0 m/s. The cars stick together after colliding. What is the speed of the wreckage just after the collision?

Question

A 900-kg car traveling $30.0 ^ { \circ }$ south of east at  12.0 m/s suddenly collides with a 750-kg car traveling north at  17.0 m/s. The cars stick together after colliding. What is the speed of the wreckage just after the collision?

Quizplus · Accepted Answer

To find the speed of the wreckage just after the collision, we use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision. First, we break down the velocities of the two cars into their x (east-west) and y (north-south) components.For the 900-kg car:- The x-component of velocity is $12.0 \cos(30.0^\circ) = 10.39 \, \text{m/s}$.- The y-component of velocity is $12.0 \sin(30.0^\circ) = 6.0 \, \text{m/s}$ south, which we'll take as negative.For the 750-kg car traveling north at 17.0 m/s, its velocity can be considered entirely in the y-direction (positive direction for north).The total momentum in the x-direction before the collision is:$p_{x, \text{total}} = (900 \, \text{kg} \cdot 10.39 \, \text{m/s}) + (750 \, \text{kg} \cdot 0) = 9341 \, \text{kg} \cdot \text{m/s}$.The total momentum in the y-direction before the collision is:$p_{y, \text{total}} = (900 \, \text{kg} \cdot -6.0 \, \text{m/s}) + (750 \, \text{kg} \cdot 17.0 \, \text{m/s}) = -5400 + 12750 = 7350 \, \text{kg} \cdot \text{m/s}$.After the collision, the cars stick together, having a combined mass of $900 \, \text{kg} + 750 \, \text{kg} = 1650 \, \text{kg}$.The velocity components of the wreckage can be found by dividing the total momentum in each direction by the total mass:- $v_{x, \text{wreckage}} = \frac{9341}{1650} = 5.66 \, \text{m/s}$- $v_{y, \text{wreckage}} = \frac{7350}{1650} = 4.45 \, \text{m/s}$The magnitude of the velocity of the wreckage is found using the Pythagorean theorem:$v_{\text{wreckage}} = \sqrt{(5.66)^2 + (4.45)^2} = \sqrt{32.04 + 19.80} = \sqrt{51.84} = 7.20 \, \text{m/s}$, which rounds to 7.21 m/s.

A 900-Kg Car Traveling 30.0∘30.0 ^ { \circ }30.0∘ South of East at 12

A 900-Kg Car Traveling $30.0 ^ { \circ }$ South of East at 12