Question

A 850 kg car initially going 18.0 m/s only in the x-direction runs into a stationary 1400 kg truck. After the collision the car is going 11.0 m/s at an angle of 38 degrees above the x-axis. What is the magnitude and direction of the velocity of the truck right after the collision (give speed and angle)?

Answer 1

According to the conservation of momentum

Mass of the car, m₁ = 850 kg

Mass of the truck m₂ = 1400 kg

The velocity of the car before collision, u1 = 18 i m/s m/s

The velocity of the truck before the collision, u2 = 0

The velocity of the car after the collision, v1 = 11*(Cos38 i + Sin 38j) m/s

From Conservation of momentum equation

850 * 18 i + 1400 * 0 = 850 * 11*(Cos38 i + Sin 38j) m/s + 1400 * v2

15300 i = 7367.8 i + 5755.86 j + 1400 * v2

1400 * v2 = 7932.2 i - 5755.86 j

v2 = (5.66 i - 4.11 j ) m/s

The magnitude of v2 is

v2 = sqrt(5.66^2 + 4.11^2)

v2 = 7 m/s --> Answer

Angle, θ = tan^-1 (-4.11 / 5.66)

= 35.98 deg