Two vehicles approach a right angle intersection and then suddenly collide.After the collision,they become entangled.If their mass ratios were 1:4 and their respective speeds as they approached the intersection were both 13 m/s,find the magnitude and direction of the final velocity of the wreck.
A)11 m/s at 14° with respect to the original direction of the lighter car
B)16 m/s at 76° with respect to the original direction of the lighter car
C)11 m/s at 76° with respect to the original direction of the lighter car
D)26 m/s at 76° with respect to the original direction of the lighter car
E)16 m/s at 14° with respect to the original direction of the lighter car

Question

Quizplus · Accepted Answer

To solve this problem, we use the principle of conservation of momentum. Since the vehicles become entangled, we treat them as a single system post-collision. The total momentum before the collision equals the total momentum after the collision.Let's denote the mass of the lighter vehicle as $m$ and the heavier vehicle as $4m$, given their mass ratio of 1:4. Both vehicles have a speed of 13 m/s before the collision. The momentum of each vehicle before the collision is the product of its mass and velocity. The total momentum in the system can be broken down into its components along the x and y axes (assuming one vehicle moves along the x-axis and the other along the y-axis).The momentum along the x-axis before the collision is $m \times 13$, and along the y-axis, it's $4m \times 13$.After the collision, the combined mass is $5m$, and let $v$ be the final velocity of the entangled wreck. The total momentum is conserved, so the magnitude of the total momentum before the collision equals the magnitude of the total momentum after the collision.The magnitude of the total momentum before the collision can be found using Pythagoras' theorem, considering the momentum components as perpendicular vectors:$$ \sqrt{(m \times 13)^2 + (4m \times 13)^2} = 13m \sqrt{1^2 + 4^2} = 13m \sqrt{17} $$Since the total momentum is conserved, the magnitude of the momentum after the collision is also $13m \sqrt{17}$. The final velocity $v$ of the entangled wreck is found by dividing the total momentum by the total mass ($5m$):$$ v = \frac{13m \sqrt{17}}{5m} = \frac{13 \sqrt{17}}{5} $$To find the angle, we note that the tangent of the angle ($\theta$) with respect to the original direction of the lighter car (x-axis) is given by the ratio of the y-component to the x-component of the momentum:$$ \tan(\theta) = \frac{4m \times 13}{m \times 13} = 4 $$Thus, $\theta = \arctan(4)$, which is approximately 76°.The magnitude of the final velocity can be calculated as:$$ v = \frac{13 \sqrt{17}}{5} \approx 11 \, \text{m/s} $$Therefore, the correct answer is 11 m/s at 76° with respect to the original direction of the lighter car.

Two Vehicles Approach a Right Angle Intersection and Then Suddenly