35-kg child rides a relatively massless sled down a  ?

 A 35-kg child rides a relatively massless sled down a hill and then coasts along the flat section at the bottom, where a second 35-kg child jumps on the sled as it passes by her. If the speed of the sled is 3.5 m/s before the second child jumps on, what is its speed after she jumps on? 

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  • Anonymous
    3 weeks ago

    1.75 hkhjhkk dgdgf

  • oubaas
    Lv 7
    3 weeks ago

    Momentum is conserved, therefore :

    V = m*Vi / 2m = Vi/2 = 3.5/2 = 1.75 m/sec

  • 3 weeks ago

    momentum is conserved, so

    m1v1 = (m1+m2)v2

    since the mass doubles, the velocity halves

    v2 = 1.75m/s

  • 3 weeks ago

    1.75 m/s  ( ie half of the velocity just before she got on )

    MOMENTUM is conserved 35 * 3.5 = 70 * v -> v = 3.5 * 35/70 = 3.5/2 = 1.75 m/s

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  • Anonymous
    3 weeks ago

    < a relatively massless >

    Something is, or isn't, massless.  There's no such thing as 'relatively massless', just like there's no such thing as being slightly pregnant.

  • 3 weeks ago

    Momentum is always conserved.

    Initial momentum = 35 * 3.5 = 122.5 kg * m/s

    Final momentum = 70 * v

    v = 122.5 ÷ 70 = 17.5 m/s

    The velocity of both cars after the collision is one half of velocity because the mass of both cars is twice the  car of the first car. I hope this is helpful for you!

  • m1 * v1 + m2 * v2 = m3 * v3

    m3 = m1 + m2

    v1 = 3.5

    v2 = 0

    m1 = 35

    m2 = 35

    35 * 3.5 + 35 * 0 = (35 + 35) * v

    35 * 3.5 = 2 * 35 * v

    3.5 = 2 * v

    1.75 = v

    1.75 m/s

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