A 1200-kg pick-up truck traveling south at 15 m/s suddenly collides with a 750-kg car that is traveling east. The two vehicles stick together and slide along the road after colliding. A highway
Patrol officer investigating the accident determines that the final position of the wreckage after the
Collision is 25 m, at an angle o $50 ^ { \circ }$ ° south of east, from the point of impact. She also determines that
The coefficient of kinetic friction between the tires and the road at that location was 0.40. What was
The speed of the car just before the collision?
A) 20 m/s
B) 4.8 m/s
C) 23 m/s
D) 14 m/s
E) 17 m/s

Question

Quizplus · Accepted Answer

To solve this problem, we use the principles of conservation of momentum and the work-energy theorem. The total momentum before the collision must equal the total momentum after the collision since momentum is conserved in collisions. Let's denote the speed of the car before the collision as $v$.Before the collision, the momentum of the truck is $1200 \, \text{kg} \times 15 \, \text{m/s} = 18000 \, \text{kg} \cdot \text{m/s}$ south, and the momentum of the car is $750 \, \text{kg} \times v$ east.After the collision, the combined mass is $1200 \, \text{kg} + 750 \, \text{kg} = 1950 \, \text{kg}$, and they move at an angle of $50^\circ$ south of east. We can use the angle to decompose the final velocity into south and east components, but we first need to find the magnitude of this velocity.The work done by friction is equal to the change in kinetic energy. The force of friction is $f_k = \mu_k \times m \times g$, where $\mu_k = 0.40$ and $g = 9.8 \, \text{m/s}^2$. The distance slid is $25 \, \text{m}$, so the work done by friction is $W = f_k \times d = \mu_k \times m \times g \times d$.This work equals the negative change in kinetic energy: $-W = \frac{1}{2} m v_i^2 - \frac{1}{2} m v_f^2$, where $v_i$ is the initial speed right after the collision and $v_f = 0$ since the vehicles come to a stop. Solving for $v_i$ gives us the speed right after the collision.To find the speed of the car before the collision, we use the conservation of momentum. The eastward momentum before the collision (from the car) must equal the eastward component of the combined momentum after the collision. Similarly, the southward momentum before the collision (from the truck) equals the southward component of the combined momentum after the collision.Solving these equations involves a bit of algebra and trigonometry, using the given angle to relate the components of the final velocity to the initial velocities of the truck and car. The exact calculations are omitted for brevity, but when you work through them, you find that the speed of the car that satisfies these conditions is approximately $23 \, \text{m/s}$, making option C the correct answer.

A 1200-Kg Pick-Up Truck Traveling South at 15 M/s Suddenly 50∘50 ^ { \circ }50∘

A 1200-Kg Pick-Up Truck Traveling South at 15 M/s Suddenly $50 ^ { \circ }$