Respuesta :
Answer: a) v2f = 0.83 m/s b) The distance will be 0.3738 m.
Explanation: a) When two bodies collide with each other with definite velocity, the system is a closed system, which means, there are no other forces acting on it, and the linear momentum is constant. So, for a system of two bodies:
Qi = Qf
m1v1i + m2v2i = m1v1f + m2v2f
2.3 + 4.(-2) = 2.(-3.66) + 4v2f
v2f = [tex]\frac{2.66}{2}[/tex]
v2f = 0.83
The final velocity of mass m2 is v2f = 0.83 m/s.
b) Since m1 slides with coefficient of friction, there is a force of friction acting on it:
[tex]F_{f}[/tex] = μ·[tex]F_{N}[/tex], where [tex]F_{N}[/tex] is a normal force acting on m1.
As there is no up or down movement, [tex]F_{N}[/tex] = [tex]F_{g}[/tex] = m.g = 2*9.8
[tex]F_{f}[/tex] = 0.6*2*9.8
Only the force of friction is acting on m1, so:
[tex]F_{f}[/tex] = m.a
0.6*2*9.8 = 2*a
a = 5.88 m/s²
Using [tex](v_{f}) ^{2} = (v_{i}) ^{2} + 2.a.d[/tex]
[tex](-3.66)^{2} = 3^{2} + 2.5.88.d[/tex]
d = [tex]\frac{4.3956}{11.76}[/tex]
d = 0.3738
The mass m1 traveled 0.3738 m before it came to rest.
