Physics- mechanics (momentum) question--10分

A ball of mass 60 g is released from rest at t = 0 and then rebounds from the ground. The acceleration is given: from 0-0.6 secs, the acceleration is 15 ms-2. From 0.6-0.65 secs, the acceleration is -140 ms-2. From 0.65-1.0 secs, the acceleraion is 15 ms-2.

(a) What is the speed of the ball just before hitting the ground?

(b) What is the speed of the ball when it just leaves the ground?

(c) Find the force and impulse acting on the ball during the rebound.

1 個解答

評分
  • taolo
    Lv 5
    1 十年前
    最愛解答

    i think your question is not reasonable because the speed of the ball when it just leaves the ground is still downward. Guttvennnnnnn's answer is also wrong because it wrongly assumed to be v=0 after 1 secs. I just think the possible question should be change to be = or <10ms-2

    2007-01-28 23:56:58 補充:

    a)v=u at u=0(release from rest), a=15, t=0.6v=0 15(0.6)=9ms-1b) v= u at u=9, a=-140, t=0.05(0.65-0.6)v=9 (-140)*(0.05)=9 (-7)=-2

    2007-01-28 23:58:16 補充:

    a)v=u at u=0(release from rest), a=15, t=0.6v=0 15(0.6)=9ms-1b) v= u at u=9, a=-140, t=0.05(0.65-0.6)v=9 (-140)*(0.05)=9 (-7)=-2

    2007-01-28 23:59:16 補充:

    a)v=u at u=0(release from rest), a=15, t=0.6v=0 15(0.6)=9ms-1b) v= u at u=9, a=-140, t=0.05(0.65-0.6)v=9 (-140)*(0.05)=9 (-7)=-2

    2007-01-28 23:59:59 補充:

    my calculation is that.a)v=u plus at u=0(release from rest), a=15, t=0.6v=0 plus 15(0.6)=9ms-1b) v= u plus at u=9, a=-140, t=0.05(0.65-0.6)v=9 plus (-140)*(0.05)=9 plus (-7)=-2

    2007-01-29 00:02:17 補充:

    c) F=ma a = 140(upward), m=0.06kgF= 0.06*140=8.4N(upward)

    2007-01-29 00:08:49 補充:

    now, i would like to discribe why Guttvennnnnnn's assuming is wrong.I think Guttvennnnnnn's should think that the acceleration is no longer be 15ms-2 in 1sec. It is somehow reasonable but it doesn't mean that v=0.

    2007-01-29 00:08:59 補充:

    For example, you throw a ball upward. it will stop(v=0) when the ball in the highest point but at that time, a is still equal to 9.81(or 10 for HKCEE).

    2007-01-29 20:14:53 補充:

    sorry, i just found out my typing mistakeb) v= u plus at u=9, a=-140, t=0.05(0.65-0.6)v=9 plus (-140)*(0.05)=9 plus (-7)=2m/s

    2007-01-29 20:15:57 補充:

    It is not reasonable for any level if you know what is REBOUND. We could clearly see that the ball is release so it is downward in 0s-0.6s and it reach the ground so the ground give it a 8.4N upward force in 0.6s-0.65s so it should be upward in 0.65s.

    2007-01-29 20:16:08 補充:

    since it leaf the floor so there will be a downward force in 0.65s-1.0s. However, the answer in b) give that the ball is downward in 0.65s.

    2007-01-29 20:17:26 補充:

    although the question is not reasonable, i think my step is correct instead of Guttvennnnnnn one in b and c part

    2007-01-30 11:24:45 補充:

    anyway, the step is that1. v=u at (u=0, a=15, t=0.6)2. v=u at (u=v at Q1, a=-140, t=0.05)3.F=ma(m=0.06, a=-140)

    2007-01-30 11:25:17 補充:

    anyway, the step is that1. v=u plus at (u=0, a=15, t=0.6)2. v=u plus at (u=v at Q1, a=-140, t=0.05)3.F=ma (m=0.06, a=-140)

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