In the figure below, block 1 of mass m1 slides from rest along a frictionless ra

Question

In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.20 m and then collides with stationary block 2, which has mass m2 = 2.00 m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction µk is 0.600 and comes to a stop in distance d within that region. What is the value of distance d if the collision is elastic? m What is the value of distance d if the collision is completely inelastic? m read the eBook Section 9-10 Elastic Collisions in One Dimension

Explanation / Answer

v1 = sqrt(2*9.8*2.2) = 6.56 m/s m*v1 = m*va + 2m*v2 v1 = va + 2*v2 for elastic collision v1 = v2 - va v2 = 4.37 m/sec frictional force = 0.6*9.8*2m = 11.76m N 11.76*d = 0.5*2m*4.37^2 d = 1.624 m {a} for completely inelastic m*v1 = 3m*V2 v2 = 2.187 m/s frictional force = 0.6*9.8*3m 0.6*9.8*3m*d = 0.5*3m*2.187^2 d = 0.4067 m

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---