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 μk 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.
A.) h/4μk
B.) h/16μk
The first thing you have to do is compute the velocity of block 2 in each of the two cases: elastic and completely inelastic collision with block 1.
In the elastic case, both momentum and kinetic energy are conserved.
Vcm = Vo/4 is the center of mass velocity. This leads to
V2(final) = Vo/2
V1(final) = -Vo/2
In the inelastic case,
M1 Vo = (M1 + V2) Vfinal
= 4M1*Vfinal
Vfinal = Vo/4 for both blocks
The distance d of the slide is given by
(1/2) M Vfinal^2 = M g ìk X
Use that to solve for X. M cancels out
X = Vfinal^2/(2*g*ìk)
It slides four times farther in the elastic case, because Vfinal is twice as large.
To find the value of distance d, we need to analyze the two cases separately: elastic collision and completely inelastic collision. Let's start with the elastic collision.
(a) Elastic Collision:
In an elastic collision, both the momentum and kinetic energy are conserved. We can use these principles to solve for the final velocities of the two blocks after the collision.
Let v1f be the final velocity of block 1 and v2f be the final velocity of block 2 after the collision.
Momentum Conservation:
Before the collision, only block 1 is moving, so its momentum is given by p1i = m1 * 0 = 0.
After the collision, the total momentum is conserved, so we have p1f + p2f = 0.
p1f + p2f = m1 * v1f + m2 * v2f ...(1)
Kinetic Energy Conservation:
Before the collision, block 1 has potential energy due to its height h, which will be converted into kinetic energy after sliding down the ramp. Block 2 is stationary, so it has no initial kinetic energy. The total kinetic energy is conserved during the collision, so we have (1/2) * m1 * 0^2 + (1/2) * m2 * 0^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2.
0 + 0 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2 ...(2)
Now, let's solve equations (1) and (2) simultaneously to find the final velocities of the blocks.
From equation (1):
m1 * v1f + m2 * v2f = 0
v2f = -(m1 / m2) * v1f
Substituting this into equation (2):
0 = (1/2) * m1 * v1f^2 + (1/2) * m2 * [-(m1 / m2) * v1f]^2
0 = (1/2) * m1 * v1f^2 + (1/2) * m2 * (m1^2 / m2^2) * v1f^2
0 = v1f^2 * [(1/2) * m1 + (1/2) * m1^3 / m2]
0 = v1f^2 * (m1 / 2 + m1^3 / (2 * m2^2))
Note that m2 = 3m1, so the equation simplifies to:
0 = v1f^2 * (m1 / 2 + m1^3 / (2 * (3m1)^2))
0 = v1f^2 * (m1 / 2 + m1^3 / (18m1^2))
0 = v1f^2 * (m1 / 2 + 1 / 18)
Since we are looking for the value of distance d, which is related to the stopping distance of block 2, let's express it in terms of the final velocity v2f.
Using the equation v = u^2 / (2a) (where v is the final velocity, u is the initial velocity, and a is the deceleration), we can write:
v2f^2 = 0^2 - 2 * μk * g * d
d = -v2f^2 / (2 * μk * g)
Therefore, we need to find the final velocity v2f when the collision is elastic, and then substitute it into this equation to solve for d.
Now, let's move on to the completely inelastic collision.
(b) Completely Inelastic Collision:
In a completely inelastic collision, the two blocks stick together after the collision and move as a single unit with a common final velocity.
To calculate the final velocity of the combined blocks, we can use the principle of momentum conservation.
Momentum Conservation:
Before the collision, the momentum of block 1 is m1 * 0 = 0, and the momentum of block 2 is m2 * 0 = 0.
After the collision, the total momentum is conserved, so we have p1f + p2f = (m1 + m2) * vf.
Since the blocks stick together, the final velocity of the combined blocks is the same for both blocks. Let's denote it as vf.
p1f + p2f = (m1 + m2) * vf
m1 * v1f + m2 * v2f = (m1 + m2) * vf ...(3)
Again, let's express the stopping distance d in terms of the final velocity vf.
Using the equation v = u^2 / (2a), we can write:
vf^2 = 0^2 - 2 * μk * g * d
d = -vf^2 / (2 * μk * g)
Therefore, we need to find the final velocity vf when the collision is completely inelastic, and then substitute it into this equation to solve for d.
In summary:
(a) For the elastic collision, solve equations (1) and (2) simultaneously to find the final velocities v1f and v2f. Then substitute v2f into d = -v2f^2 / (2 * μk * g) to get the value of distance d.
(b) For the completely inelastic collision, solve equation (3) to find the final velocity vf. Then substitute vf into d = -vf^2 / (2 * μk * g) to get the value of distance d.