A car heading north suddenly collides at an intersection with a truck of the same mass heading east.If they lock together and travel at $28 \mathrm {~m} / \mathrm { s }$ at 15° north of east just after the collision,how fast was the car initially traveling?
A)14 m/s
B)7 m/s
C)28 m/s
D)11 m/s

Question

Quizplus · Accepted Answer

The momentum of the system is conserved in both the x and y directions. In the x direction, the initial momentum is zero, and the final momentum is the mass of the two vehicles times their velocity in the x direction. In the y direction, the initial momentum is the mass of the car times its velocity in the y direction, and the final momentum is the mass of the two vehicles times their velocity in the y direction. Setting these two equations equal to each other and solving for the velocity of the car, we get:$$v_c = \frac{m_c v_{ce}}{m_c + m_t}$$where $m_c$ is the mass of the car, $v_{ce}$ is the velocity of the car in the x direction, $m_t$ is the mass of the truck, and $v_t$ is the velocity of the truck in the y direction.We are given that $m_c = m_t$, $v_{ce} = 28 \mathrm{~m/s}$, and $v_t = 15^\circ$. Substituting these values into the equation above, we get:$$v_c = \frac{(28 \mathrm{~m/s})\cos(15^\circ)}{2} = 14 \mathrm{~m/s}$$Therefore, the car was initially traveling at 14 m/s.

A Car Heading North Suddenly Collides at an Intersection with a Truck