關於力學的簡單物理問題

Tieria

2009-12-29 6:09 am

1. An object, moving up a smooth inclined plane making an angle θ with horizontal decreases its speed from x to y ms^-1. What is the distance, in metres, traveled in this period?
(Take g = acceleration due to gravity in ms^-2)

A. (x^2 – y^2) / 2g
B. (x^2 – y^2) / 2gcosθ
C. (x^2 – y^2) / 2gsinθ
D. [2(x – y)] / gsinθ

※ 請列明計算步驟※

2. An object passing a mark with a velocity y ms^-1 eastward is subjected to a constant acceleration of x ms^-2 in a westward direction. How long, in seconds, will it take to return to the same mark?

A. y^2 / x
B. y^2 / 2x
C. y / x
D. 2y / x

※請列明計算步驟※

3. Two particles P and Q are allowed to fall freely from the same point, one of which is released a short time before the other. Neglecting air resistance, which of the following statement is / are correct? While falling in air before hitting the ground,

(1) the two particles undergo the same acceleration
(2) their velocities always differ by the same amount
(3) their distance of separation is always the same

A. (1) only
B. (1) and (2) only
C. (1) and (3) only
D. (2) and (3) only
※ 為何statement(3)錯？※

回答 (1)

天同

2009-12-29 6:36 am

✔ 最佳答案

1. decceleration = g.sin(theta)
use equation of motion: v^2 = u^2 + 2.a.s
with u = x v = y a = -g.sin(theta), s =?
y^2 = x^2 + 2(-g.sin(theta)).s
s = (x^2-y^2)/[2g.sin(theta)]
2. Use equation of motion: s = ut + (1/2).at^2
with u = y, a = -x, s = 0 m
0 = yt + (1/2)(-x)t^2
t = 2y/x

3. Assume P is released T seconds before Q
The distance covered by P, s(P), after time t is:
s(P) = (1/2)gt^2

The distance covered by Q t seconds after P is relased
s(Q) = (1/2).g(t-T)^2
Distance between P and Q
= s(P) - s(Q) = (1/2)g[t^2-(t-T)^2]
= (g/2).[2tT - T^2]
= [gT].t - gT^2/2
Since T is a constant, the separation increases as t increases.

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