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. The initial momentum of the car is $p_c = m_c v_c$, where $m_c$ is the mass of the car and $v_c$ is its initial velocity. The initial momentum of the truck is $p_t = m_t v_t$, where $m_t$ is the mass of the truck and $v_t$ is its initial velocity. The final momentum of the system is $(m_c + m_t) v_f$, where $v_f$ is the final velocity of the system.In the x direction, we have:$$p_{ci} + p_{ti} = (m_c + m_t) v_f \cos 15^\circ$$$$m_c v_c + m_t 0 = (m_c + m_t) 28 \cos 15^\circ$$$$v_c = \frac{(m_c + m_t) 28 \cos 15^\circ}{m_c}$$$$v_c = \frac{28 \cos 15^\circ}{1}$$$$v_c = 26.6 \mathrm{~m/s}$$Therefore, the car was initially traveling at 26.6 m/s.

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