about physics mechanics

2007-03-19 1:42 am
two trolleys A and B are initiallly at rest. The man on Trolley A pulls the other trolley with a horizontal force of 20N for a short time. Total mass of Trolley A is 80 kg while that of trolley B is 120kg.When the two trolleys meet, the speed of trolley A is 1.2 ms-1.At first, the two trolleys have 9 m apart.

Find (a) the speed of the trolley B
(b)the distance travelled by trolley B

回答 (2)

2007-03-19 2:15 am
✔ 最佳答案
(a) By F = ma, a of Trolley A = 0.25ms^-2
By v = u + at, t = 4.8s
By F = ma, a of trolley B = 0.167ms^-2
By v = u + at, v = 0.8ms^-1
so, speed of the trolley B is 0.8ms^-1
(b) By v^2 - u^2 = 2as
s = 1.92m
so, distance travelled by trolley B is 1.92m


2007-03-18 19:09:04 補充:
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2007-03-19 2:49 am
(a)
Friction subjected by A =20N
Friction subjected by B =20N
Let F₁ & F₂ be the net force of A & B respectively,
So,
F₁=ma
20=(80)(a)
a=0.25 ms^(-2)This is acceralation of A

F₂=ma
20=(120)(a)
a=0.167 ms^(-2) (3 sig.fig) This is acceralation of B

by the equation of motion,
v=u+at
since u=0 ms^(-1) ,v=at
they move @ the same time, let the time used to meet each other be t,
put v=1.2 ms^(-1), a=0.25 ms^(-2)
1.2=(0.25)t
t=4.8 s

put a=0.167 ms^(-2), t=4.8 s
v=(0.167)(4.8)
v=0.802 ms^(-1)this is the velocity of B

(b)
since v^2+ u^2=2as
put v=0.802 ms^(-1), a=0.167 ms^(-2)
s=(0.802^2)/(2*0.167)
s=1.93 m this is the distance travelled by B

2007-03-18 18:52:06 補充:
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