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26 January, 23:14

A 0.5 kg ball is dropped vertically onto a floor, hitting with a speed of 20 m/s. It rebounds with a speed of 15 m/s immediately after the collision. (a) What is the impulse in N-s acting on the ball during the contact

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  1. 27 January, 00:03


    Impulse is defined as change I'm momentum of a body. It is expressed as;

    According to Newton's second law;

    F = ma

    F = m (v-u) / t

    Ft = m (v-u) = Impulse

    Impulse = Ft

    Impulse = m (v-u)

    F is the force acting on the ball

    t is the time taken

    m is the mass of the body

    v is the final velocity

    u is the initial velocity

    To get impulse, we need to get time t first using the equation of motion v = u+gt

    20 = 15 + (9.8) t

    20-15 = 9.8t

    5 = 9.8t

    t = 5/9.8

    t = 0.51secs

    Since Impulse = Ft

    F = mg = 0.5 (9.8)

    F = 4.9N

    Impulse = 4.9*0.51

    Impulse acting on the ball during contact = 2.499Ns
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