Vikas TU
Last Activity: 4 Years ago
Dear student
We can use cross check method.
Let m1 = 1 kg
m2 = 2 kg and
m3 = 3 kg
Suppose the block m1 moves upward with acceleration a1 and the blocks m2 and m3 have relative acceleration a2 due to the difference of weight between them.
Actual acceleration of the blocks m1, m2 and m3 will be
a1, (a1 − a2) and (a1 + a2)
From figure 2, T − 1g − 1a1 = 0 …(i)
From figure 3,
T/2-2g-2(a1-a2)=0 …ii
From figure 4,
T/2-3g-3(a1+a2)=0 …iii
From equations (i) and (ii), eliminating T, we get:
1g + 1a2 = 4g + 4 (a1 + a2)
5a2 − 4a1 = 3g …(iv)
From equations (ii) and (iii), we get:
2g + 2(a1 − a2) = 3g − 3 (a1 − a2)
5a1 + a2 = g …(v)
Solving equations (iv) and (v), we get:
a1=2g/29
a2=g-5a1
a2=g-10g/29
=19g/29
Then a1-a2=2g/29-19g/29
= -17g/ 29
and a1+a2=2g/29+19g/29
=21g/29
So, accelerations of m1, m2 and m3 are
19g/29up, 17g/29 down and 21g/29down, respectively.
Now, u = 0, s = 20 cm = 0.2 m
a2=19g/29
∴ s=ut+1/2at
⇒0.2=1/2×19/29gt²
⇒t=0.25 s