Question

In Figure 9-69, block 1 of mass m1 slides from rest along a frictionless ramp from height h and then collides with stationary block 2, which has mass m2 = 3m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction is ?k and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic? Express your answer in terms of the variables given and g. sorry i don't have the figure. the answer should be an equation

Answer 1

Use conservation of energy to solve for the velocity of block 1 at the bottom of the slide In elastic collisions, the momentum and kinetic energy is conserved, the velocity of block 2 for head on elastic collision is given by, 2m, 2m (m,+3m) 2gh 2 gh 2 Use work energy theorem to solve for d

In completely inelastic collisions, the momentum and kinetic energy is not conserved, the velocity of block 2 for head on elastic collision is given by, +m gh Use work energy theorem to solve for d. 16H

Answer 2

m1's potential energy gets converted into kinetic energy. Let's say it's mass is M:
M•9.8m/s²•2.9m = M•28.4 J = ½Mv²
v = 7.54 m/s
So the pre-collision momentum = 7.54M kg·m/s

I'm going to drop units for ease of typing. All units SI (kg/m/s).

(a) if the collision is elastic,
7.54M = Mu + 2Mv
where u and v are the velocities of m1 and m2, respectively.
Also, the relative velocity of approach = relative velocity of separation, so
7.54 = v - u, or
v = 7.54 + u
Plug this into momentum eqn and cancel Ms:
7.54 = u + 2(7.54 + u)
-7.54 = 3u
u = -2.51
v = 5.03
So m2 has kinetic energy Ek = ½(2M)5.03² = 25.27M
This gets dissipated by friction: the friction force Ff = µ(2M)g = 0.4•2•M•9.8 = 7.84M
So the work done by friction = d•Ff = 7.84M•d = 25.27M
d = 25.27 / 7.84 = 3.22 ? (a)

(b) if the collision is inelastic,
7.54M = (3M)v
v = 2.51
Ek = ½(3M)2.51² = 9.48M
Wf = µ(3M)gd = 9.48M = 0.4•3M•9.8•d = 11.76Md
d = 0.81 ? (b)