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aleksklad [387]

4 years ago

7

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

1 answer:

e-lub [12.9K]4 years ago

5 0

Answer:

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

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