select the correct answer.\na car with a mass of 1.5×10³ kilograms is traveling west at a velocity of 22…

select the correct answer.\na car with a mass of 1.5×10³ kilograms is traveling west at a velocity of 22 meters/second. it hits a stationary car with a mass of 9.0×10² kilograms. if the collision is inelastic, what is the final direction and approximate velocity of the two cars?\na. 14 meters/second to the west\nb. 14 meters/second to the east\nc. 22 meters/second to the east\nd. 22 meters/second to the west

select the correct answer.\na car with a mass of 1.5×10³ kilograms is traveling west at a velocity of 22 meters/second. it hits a stationary car with a mass of 9.0×10² kilograms. if the collision is inelastic, what is the final direction and approximate velocity of the two cars?\na. 14 meters/second to the west\nb. 14 meters/second to the east\nc. 22 meters/second to the east\nd. 22 meters/second to the west

Answer

Explanation:

Step1: Apply conservation of momentum formula

The formula for conservation of momentum in an in - elastic collision is $m_1v_1=(m_1 + m_2)v_f$, where $m_1$ is the mass of the first car, $v_1$ is the initial velocity of the first car, $m_2$ is the mass of the second car, and $v_f$ is the final velocity of the combined cars. Given $m_1 = 1.5\times10^{3}\text{ kg}$, $v_1=22\text{ m/s}$ (west), and $m_2 = 9.0\times10^{2}\text{ kg}$.

Step2: Calculate the sum of the masses

$m_1 + m_2=1.5\times 10^{3}+9.0\times 10^{2}=1500 + 900=2400\text{ kg}$

Step3: Solve for the final velocity

$v_f=\frac{m_1v_1}{m_1 + m_2}=\frac{1.5\times 10^{3}\times22}{2400}=\frac{1500\times22}{2400}=\frac{33000}{2400}=13.75\approx14\text{ m/s}$ Since the first car was moving west and momentum is conserved in the direction of the initial motion of the moving car, the direction is west.

Answer:

A. 14 meters/second to the west