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 s

Question

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

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Khải Quang 4 years 2021-07-21T16:31:33+00:00 1 Answers 118 views 0

Answers ( )

  1. Xavia
    0
    2021-07-21T16:33:20+00:00
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    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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