Block 1, with mass m1 and speed 3.6 m/s, slides along an x axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass m2 = 0.40m1. The two blocks then slide into a region with a coefficient of kinetic friction of 0.50 where they stop. How far into that region do the two blocks slide? (a) block 1 m (b) block 2 m

a) The block 1 slides 0.24 m into the rough region.

b) The block 2 slides 2.7 m

Explanation:

Hi there!

First, let's find the final velocity of each block. With that velocities, we can calculate the kinetic energy of each block. The kinetic energy of the blocks will be equal to the work done by friction to stop them. From the equation of work, we can calculate the distance traveled by the blocks.

Since the collision is elastic, the momentum and kinetic energy of the system composed of the two blocks is constant.

The momentum of the system is calculated as the sum of the momenta of each block:

m1 · v1 + m2 · v2 = m1 · v1' + m2 · v2'

Where:

m1 and m2 = mass of blocks 1 and 2 respectively.

v1 and v2 = velocity of blocks 1 and 2 respectively.

v1' and v2' = final velocity of blocks 1 and 2 respectively.

Using the data we have, we can solve the eqaution for v1':

m1 · 3.6 m/s + 0.40 m1 · 0 = m1 · v1' + 0.40 m1 · v2'

3.6 m/s · m1 = m1 · v1' + 0.40 m1 · v2'

3.6 m/s = v1' + 0.40 v2'

v1' = 3.6 m/s - 0.40 v2'

The kinetic energy of the system also remains constant:

1/2 m1 · (v1) ² + 1/2 m2 · (v2) ² = 1/2 m1 · (v1') ² + 1/2 m2 · (v2') ²

Multiply by 2 both sides of the equation:

m1 · (v1) ² + m2 · (v2) ² = m1 · (v1') ² + m2 · (v2') ²

Let's replace with the dа ta:

m1 · (3.6 m/s) ² + 0.40 m1 · 0 = m1 · (v1') ² + 0.40 m1 (v2') ²

divide by m1:

(3.6 m/s) ² = (v1') ² + 0.40 (v2') ²

Replace v1' = 3.6 m/s - 0.40 v2'

(3.6 m/s) ² = (3.6 m/s - 0.40 v2') ² + 0.40 (v2') ²

Let's solve for v2':

(3.6 m/s) ² = (3.6 m/s) ² - 2.88 v2' + 0.16 (v2') ² + 0.40 (v2') ²

0 = 0.56 (v2') ² - 2.88 v2'

0 = v2' (0.56 v2' - 2.88) v2' = 0 (the initial velocity)

0 = 0.56 v2' - 2.88

2.88/0.56 = v2'

v2' = 5.1 m/s

Now let's calculate v1':

v1' = 3.6 m/s - 0.40 v2'

v1' = 3.6 m/s - 0.40 (5.1 m/s)

v1' = 1.56 m/s

Now, let's calculate the final kinetic energy (KE) of each block:

a) Block 1:

KE = 1/2 · m1 · (1.56 m/s) ² = m1 · 1.2 m²/s²

The work done by friction is calculated as follows:

W = Fr · s

Where:

Fr = friction force.

s = traveled distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction.

And the normal force is calculated in this case as:

N = m1 · g

Where g is the acceleration due to gravity.

Then, the work done by friction will be:

W = m1 · g · μ · s

The kinetic energy of an object is the negative work that must be done on that object to bring it to stop. Then:

m1 · 1.2 m²/s² = m1 · g · μ · s

Solving for s:

s = m1 · 1.2 m²/s² / m1 · g · μ

s = 1.2 m²/s² / 9.8 m/s² · 0.50

s = 0.24 m

The block 1 slides 0.24 m into the rough region.

b) For block 2 the kinetic energy will be the following:

KE = 1/2 · 0.4 · m1 · (5.1 m/s) ² = m1 · 5.2 m²/s²

The friction force will be:

Fr = 0.4 m1 · g · μ

And the work done will be:

W = 0.4 m1 · g · μ · s

Since W = ΔKE,

Then:

m1 · 5.2 m²/s² = 0.4 m1 · g · μ · s

Solving for s:

5.2 m²/s² / (0.4 · g · μ) = s

s = 5.2 m²/s² / (0.4 · 9.8 m/s² · 0.50)

s = 2.7 m

The block 2 slides 2.7 m