Line c is at rest . line a is going in a positive direction . line b is going in a negative direction . line d is negative too
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

Substituting,


Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,

Where here
is the velocity after the collision.



Answer:
Coefficient of friction = 0.836
Explanation:
If v be the speed after one quarter of the circular path
v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8
tangential acceleration = 5.8 m/s²
radial acceleration = v² /r = 5.8
total acceleration = √2 x 5.8
m x√2 x 5.8 = m x g xμ
μ = √2 x 5.8 / 9.8 = 0.836