find the time taken when A collides with B.
11Let Force x be applied on A and 2x
y + 3 x = 15
Acceleration of the Lower block = 15 - 3x/10
Acceleration of the upper block = Acceleration of the Lower block
B:
2x/2 = 15 - 3x /10
10x = 15 - 3x
13x = 15
x = 15/13
This is not the possible max = 1m/s
Therefore Force of friction on B is 2N
Now
y + x + 2 = 15
y + x = 13
y = 13-x
Acceleration = 13-x/10
x= 13 - x/10
11x = 13
x = 13/11 N on A due to friction
Acceleration of A is 1.18m/s2
Acceleration of B is 1m/s2
Sa=1.18 * t2/2
Sb=1* t2/2
Sa - Sb = 3
0.18*t2/2 = 3
t2 = 6/0.18
t2 = 100/3
t = 10/√3 seconds
Am i rite?
