Anonymous
Anonymous asked in Science & MathematicsPhysics · 9 years ago

Could you explain me how to do this, please. I really wanna learn to do it.?

Bowling machine with mass 40.1kg and a cricket ball has mass of 0.16kg.

The gun fires the ball at 150 km/h ? a)Speed of the machine?

b) by the the the balls has reacher the batsman it has slowed to 110km/h. the batsman hits the ball with the 4kg bat and causes it to move 90km/h back to the machine? what is the average force exerted during the collision if the bat and the ball are in contact for 0.0008s?

c) the batsman hits the ball at an angle of 39? no air resistance how far will the ball travel?

d) totalk mechanical energy of the ball at the top of its trajectory?

1 Answer

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  • 9 years ago
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    Momentum = mass * velocity = Impulse = Force * time

    a) The ball and the machine are assumed at rest before the ball is thrown ====> Momentum before = Momentum after ====> The machine recoils ===> Momentum of machine = - Momentum of ball so the sum of the 2 is still 0

    We need consistant units ===> 150000 m/hr / 3600 s/hr = 41.67 m/s

    Mo_ Ball = Mo_machine

    0.16 kg * 41.67 m/s = 40.1 kg * V

    V = 0.17 m/s (Asked for speed or I woukld call it negative

    b) Mass of bat does not matter

    We need the Change in momentum of the ball

    Ball_in + Ball_out = total momentum change

    110,000 m/hr / 3600 s/hr = 30.56 m/s

    90,000 m/s / 3600 s/hr = 25 m/s

    0.16 kg * 30.56 + 0.16 * 25 =

    4.89 + 4,0 = 8.89 kg*m/s

    Im = force * time

    8.89 = F * 0.0008

    F = 11,112.5 N

    c) Projectile motion

    Range = V^2 * sin(2*theta)/g

    R = (30.36)^2 * sin(2 * 39)/9.8

    R = 921.73 * sin(78)/9.8

    R = 921.73 * 0.978/9.8

    R = 921.73 * 0.1092.17 m

    d) For this we need to know how high the ball goes ====> Need vertical component of velocity

    sin 39 = y/30.36

    y = 19.11 m/s

    Now Newton's Equations of motion

    d = 0.5at^2 + Vo*t (Vo = 0 and that term drops out)

    d = 0.5 * 9.8 * t^2 (Now we need t)

    V = at + Vo

    19.11 m/s = 9,8*t

    t = 1.95 s

    d = 0.5 * 9.8 * 1.95^2

    d = 18.63 m up

    Now we need the Kinetic Energy and the Potential Enmergy

    KE = 0.5 * M * Vx (Now we need the horizontal component of ve,ocity

    cos 38 = x/30.36

    Vx = 23.92

    KE = 0.5 * 0.16 kg * 23.92^2 = 45.77 J

    PE = MGH = 0.16 * 9.8 * 18.63 = 29.21 J

    Total energy = KE + PE = 45.77 + 29.21 = 74.98 J

    QED

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