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

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach