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  • 1 decade ago
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    1. The answer is B.

    You are correct to illustrate the direction of weight. But the direction of the push is perpendicular to the weight of the bucket, so they are independent of each other. Now, by Newton's 2nd law of motion, F = ma

    Mass of the bucket is a property of matter, that is it will not change over space, no matter it is on Earth or the Moon. So, they have the same value. As the push is the same, the acceleration is the same.

    Hence, they have the same ease to push the bucket because they have the same mass.

    2. The answer is C.

    Let @ be the inclination of the slope, f be the friction on the slope.

    Friction always acts in a direction opposite to the relative motion between the two objects. Let the direction upward the slope be positive

    So, when the block is moving upward, friction acts down the slope.

    By Newton's 2nd law of motion,

    ma = -mgsin@ - f

    a = -gsin@ - f/m, which is a constant

    When the block is moving downward, friction acts up the slope

    again, ma' = -mgsin@ + f

    a' = -gsin@ + f/m

    As the block moves up, then instantaneously stop at the highest point, and move down to the bottom, acceleration of the block is always NEGATIVE.

    Apparently, a < a', so the value of acceleration when it is moving up the slope is MORE negative than when it is moving down the slope.

    3. The answer is A.

    Your equation written in the diagram is valid ONLY when the force acting on the bullet is a constant, but now it is changing.

    So, area under the graph = impulse = -change of momentum of the bullet (because f should be negative, so we have to add a negative to the equation)

    So, (2.0 X 10^-3)(5000) / 2 = -(mv - mu)

    5 = (0.1)(250 - v)

    Speed of bullet after penetration, v = 200 ms^-1

    2009-03-07 17:50:36 補充:

    4. The answer is B.

    The velocity of the vehicle is changing (because the direction is changing, not magnitude, velocity is a vector) throughout the process, so there is a change of momentum during the time interval.

    2009-03-07 17:50:45 補充:

    So (1) is wrong, note that no change in speed does not imply there is no change in velocity.

    After returns to A, its velocity is the same as it was initially. So, the change of momentum is zero at t = 90 s since its velocity is the same. So, (2) is correct.

    2009-03-07 17:50:55 補充:

    For (3), there is always a change in velocity during the process, By Newton's 2nd law of motion, F = ma = m(v - u)/t. As there is a change in velocity, there is a force acting on the vehicle during the process.

    2009-03-07 17:51:05 補充:

    (P.S. In fact there is a force acting on the vehicle pointing towards the centre of the circular path to account for its centripetal force, this is included in AL physics syllabus but not CE... Anyway, you should know the net force acting on the vehicle is not zero.)

    2009-03-07 17:51:18 補充:

    5. The answer is B.

    As the string is inelastic, tension in the string, T = 13 N

    So, net force acting on the block along the plane, F = T - mgsin30* = 13 - (1)(10)sin30* = 8 N

    2009-03-07 17:51:24 補充:

    By Newton's 2nd law of motion,

    F = ma

    8 = 1(a)

    Acceleration, a = 8 ms^-2

    By equation of motion, v^2 = u^2 + 2as

    v^2 = 0 + 2(8)(1)

    Speed gained, v = 4 ms^-1

    2009-03-07 17:52:53 補充:

    Please separate your questions into several posts. I have exceeded the word limit so I cannot explain to you fully. If you can separate them, I can explain more fully.

    Source(s): Physics king
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