A 1000-kg car approaches an intersection traveling north at 20.0 m/s. A 1200-kg car approaches the same intersection traveling east at 22.0 m/s. The two cars collide at the intersection and lock together. Ignoring any external forces that act on the cars during the collision, what is the velocity of the cars immediately after the collision?

This problem applies momentum balance over the cars collided
m1v1y + m2v2y = (m1 + m2)v'y
0 + (1500 kg)(25.0 m/s) = (2500 kg)v'y
v'y = 15.0 m/s

Find the vector sum of the velocities:

v^2 = v'x^2 + v'y^2 = (8.00 m/s)^2 + (15.0 m/s)^2
v = 17.0 m/s

find the angle: θ = tan^-1(15.0 m/s / 8.00 m/s) = 61.9°

now, the momentum: p = mv = (2500 kg)(17.0 m/s)

p = 4.25 x 10^4 kg x m/s at 61.9° N of E

Answer Link

aristocles aristocles · Answer 2 · 2019-09-10T03:53:23+02:00

Answer:

[tex]\vec v = 12\hat i + 9.09 \hat j m/s[/tex]

Explanation:

During the collision of two cars there is no external force on this system

so the total momentum of two cars will remains conserved

so we will have

[tex]P_1 + P_2 = P_f[/tex]

[tex]P_1 = 1000(20)\hat j[/tex]

[tex]P_2 = 1200(22)\hat i[/tex]

now when two cars collide inelastically then they will move together after collision

so we have

[tex]20000\hat j + 26400\hat i = (1200 + 1000) \vec v[/tex]

[tex]9.09\hat j + 12\hat i = \vec v[/tex]