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)16 m/s at 14° with respect to the original direction of the lighter car

Question

Quizplus · Accepted Answer

The conservation of momentum principle is used here. 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 masses of the vehicles as $m$ and $4m$, and their speeds as $13 \, \text{m/s}$ each. The momentum of each vehicle before the collision is $m \times 13$ and $4m \times 13$, respectively. The direction of the momentum vectors forms a right angle, so we can use Pythagoras to find the magnitude of the total momentum:$$\sqrt{(m \times 13)^2 + (4m \times 13)^2} = 13m \sqrt{1^2 + 4^2} = 13m \sqrt{17}$$The mass of the combined wreck is $m + 4m = 5m$. The magnitude of the final velocity $v_f$ can be found by dividing the total momentum by the total mass:$$v_f = \frac{13m \sqrt{17}}{5m} = \frac{13\sqrt{17}}{5}$$Calculating the exact value gives $v_f \approx 11 \, \text{m/s}$.For the direction, we use the tangent of the angle formed by the momentum vectors:$$\tan(\theta) = \frac{4m \times 13}{m \times 13} = 4$$Thus, $\theta = \arctan(4)$, which is approximately $76^\circ$ with respect to the original direction of the lighter car. 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