Question

A 1210 kg car is driving NE

(at 45.0°) at 15.2 m/s when it is

struck by a moving 1540 kg car.

Afterward, they stick together and

move directly east (at 0°) at 23.3

m/s. What was the x-component

of the second car's initial velocity?

Please help

Answer 1

`V_2_X=33.16m/s`

Step-by-step explanation:

From the question we are told that

Mass of car
`M_1=1210kg \\Angle1=\theta _1 45\textdegree NE`

Velocity of car
`v_1= 15m/s`

Mass of Truck
`M_2= 1540kg \\Angle 2=\theta_2 0\textdegree E`

Final velocity
`v_2= 23.3m/s`

Generally the the equation of the law of conservation of momentum is mathematically given by

Given the x direction

`m_1v_1cos\theta+m_2v_2=(m_1+m_2)v`

`1210*15.2cos45+1540*V_2=(1210+1540)*23.3`

`V_2=((1210+1540)*23.3)/(210*15.2cos45+1540)`

`V_2=33.16m/s`

The x-component of the second car's initial velocity is

`V_2_X=33.16m/s`