A car of mass 1500 kg travels due East with a constant speed of 25.0 m/s. Eventually it turns right, and travels due South with a constant speed of 15 m/s. By making the turn, and reducing its speed, the car's linear momentum changed. What was the direction of the car's change in linear momentum? Give a numerical answer that is an angle, with a qualifier such as North-of-East, or West-of-North, etc

The direction of the car's change in linear momentum is 149.04° West of North

Explanation:

Momentum is defined as the product of mass of a body and its velocity

Momentum = mass * velocity

Change in Momentum = mass * change in velocity

∆P = m∆v

∆P = m (v-u)

Given m = 1500kg

v = 25m/s

u = 15m/s

∆P = 1500 (25-15)

∆P = 1500*10

∆P = 15,000kgm/s

Since the car first travels due East i. e + x direction

x = 25m/s

Travelling due south is negative y direction

y = - 15m/s

Direction of the car change

θ = tan^-1 (y/x)

θ = tan^-1 (-15/25)

θ = tan^-1 (-0.6)

θ = - 30.96°

Since tan is negative in the second quadrant

θ = 180-30.96

θ = 149.04°

The direction of the car's change in linear momentum is 149.04° West of North