2004年physics mc

For number 5, D is the correct answer. Remember for Newton Second Law, we only consider the force exerted ON the system. Let's first consider m1 + m2 is a SYSTEM (let's call it system a), the only horizontal force applied to system a is F, which are equal in both figures. Thus, by using F=(m1+m2) * a, we conclude a is the same in both figures. Next, consider m1 is a SYSTEM (let's call it system b), in figure 1. T1 is the only horizontal force in system b. Therefore, T1 / m1 is the acceleration. Then, consider m2 is a SYSTEM (let's call it system c) in figure 2. T2 is the only horizontal force in system c. Therefore, T2 / m2 is the acceleration. Since both figures have the same acceleration, Therefore, (2) T1/ m1 = T2 / m2 is the correct answer.

In system b, T1 = m1* a. In system c, T2 = m2*a. Since a is the same but m1 is larger than m2, therefore, T1 > T2. Thus, (1) T1 > T2 is the correct answer.

Next, consider m2 as a system ( let's call it system d) in figure 1. F and T1 are the horizontal forces applied to this system. Thus, F- T1 = m2*a. (let's call it equation1)
Then, consider m1 as a system (let's call it system e) in figure 2. F and T2 are the horizontal forces applied to this system. Thus F - T2 = m1*a. (let's call it equation2).

By adding equation 1 and 2, we get:
F-T1 + F- T2 = m1*a + m2*a
thus, 2F - T1- T2 = a* (m1 + m2)
thus, 2F- a*(m1+m2) = T1 + T2. (let's call it equation 3)
Since a*(m1+m2) = F, the left hand side of equation 3 becomes F. The right hand side is T1 + T2. Thus, F= T1 + T2. Thus, (3) F=T1+ T2 is also correct.

For number 6, the answer is A. Remember Newton's third Law, the reaction force is always equal to the applied force. P1, the pressure applied from Y to X is the weight of Y. Therefore, the reaction force, P2, which X applied to Y is equal to Y. Therefore, P1=P2. Hence, their ratio is 1:1

2007-04-24 22:14:57 補充：
Sorry, i mis-read the question. P2 is the pressure exerted by the two bricks to the surface, not from brick x to y that i thought before. let's say K be the weight of y. And A be the area y contacts x. By the formula: pressure = force / area. p1 = K / A.

2007-04-24 22:15:56 補充：
The weight of the two bricks is 3K ( I think you understand this part). And the only area that the bricks contact the surface is the bottom of x, which is 2A ( I think you understand this part too).

2007-04-24 22:16:22 補充：
So the pressure exerted by the bricks to the surface is 3K/2A ( the value of p2). Hence the ratio of p1: p2 = (K/A) * (2A/3K) = 2:3.

2007-04-24 22:22:42 補充：
I was frustrated because I typed a long answer before the third answer appears. But it doesn't allow me to submit the answer because of too many words.

2007-04-24 22:23:16 補充：
So I have to edit the content and submit them seperately. Now it turns out that I seem like copy the answer from another member. But anyway, hope we helped you.

5.D
Consider a system of blocks c onsisting of m1 and m2 in figure(a).
By Fnet=ma, F=(m1+m2)a
a=F/(m1+m2)
Since the external force to both systems are equal, the acceleration is of the same magnitude.Consider a force diagram of block m1 in figure(a), by Fnet=ma

T1=m1a
Consider a force diagram of block m2 in figure(b),by Fnet=ma,
T2=m2a
(1)correct.T1=m1a>m2a=T2 (since m1>m2)
Hence,T1>T2
(2)correct.T1/m1=a=T1/m2
(3)correct.T1+T2=m1a+m2a=(m1+m2)a=(m1+m2)a=F

2007-04-24 17:46:03 補充：
6.DSince the width and height of Y are equal to the corresponding measurement of X but its length is that half of X, the volume of Y is that half of X. Since they are made of the same material, the mass (and hence weight) is half that of X.

2007-04-24 17:46:18 補充：
Let the weight of Y be FY( and hence the weight of X is 2FY). Moreover , AX=2AY.P1=FY/AYP2=(FX+FY)/AX=(2FY+FY)/2AY =3FY/2AYHence, P1/P2=(FY/AY)/(3FY/2AY)=2/3

2007-04-25 14:43:10 補充：
Sorry, I type wrong, it should be X and Y instead of X and Y. SInce I type wrong at first, I continued to type wrong afterwards.