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Question

1. Mass A of 10 kg and mass B of 15 kg areconnected by a light inelastic string passing
over a smooth pulley . Intially mass A is on the ground while mass B is held 10 m above the ground .Mass B is then released.

a)What is the acceleration of mass B?
ans 2 ms-1.

b) What is the greatest height reached by mass A?

A.10m
B.11m
C.12m
D.13m

Ans C I only 唔明呢題點計~ Plz 列steps ~ thz

Answer 1

Consider mass A,
u = 0 m/s, a = 2 m/s, s = 10 m, v=?
using v^2 = u^2 + 2.a.s
v^2 = 2.(2)(10) (m/s)^2
thus v = 6.325 m/s

this is the velocity attained when A reaches a height of 10 m above ground (i.e. when mass B reaches ground).

Using conservation of mechanical energy to find the height reached further by A ,
(1/2)(10)(6.325)^2 = (10)(10)(h)
thus h = 2 m

The total height reached by mass A = (10+2) m = 12m

Answer 2

(a)
F = ma
15x10-10x10 = (10+15)a
50 = 25a
a = 2 m s -2

(b)
10 m (A)

2007-02-25 06:09:21 補充：
(b)v2 = u2 + 2asv2 = 0 + 2x2x10v2 = 401/2 mv2 = mgh1/2 v2 = ghh = 1/2 v2 /gh = 1/2 x 40 / 10h = 2When B stops, A will reach its max final velocity v, this KE changes to PE with height achieved of 2 m,total height = 10 + 2 = 12 m (answer C)