Elastic Collision of two blocks

[chiurox]

Homework Statement

Block A of mass, mA = 2.0 kg, is moving on a frictionless surface with a velocity of 5.0 m/s to the right and another block B of mass, mB = 8.0 kg, is moving with a velocity of 3.0 m/s to the left, as shown in the diagram below. The two block eventually collide.

a. If the collision is elastic, what are the final velocities of the two blocks?

The Attempt at a Solution

(2.0)(5.0) + (8.0)(-3.0) = (2.0)vaf + (8.0)vbf
10 – 24 = (2.0)vaf + (8.0)vbf
-14 = (2.0)vaf + (8.0)vbf divide the equation by 2
-7 = vaf + 4.0vbf //equation 1
So, I don't know now how to solve for these two final variables.

 

[Marco_84]

Homework Statement

Block A of mass, mA = 2.0 kg, is moving on a frictionless surface with a velocity of 5.0 m/s to the right and another block B of mass, mB = 8.0 kg, is moving with a velocity of 3.0 m/s to the left, as shown in the diagram below. The two block eventually collide.

a. If the collision is elastic, what are the final velocities of the two blocks?

The Attempt at a Solution

(2.0)(5.0) + (8.0)(-3.0) = (2.0)vaf + (8.0)vbf
10 – 24 = (2.0)vaf + (8.0)vbf
-14 = (2.0)vaf + (8.0)vbf divide the equation by 2
-7 = vaf + 4.0vbf //equation 1
So, I don't know now how to solve for these two final variables.

you need to put also the kinetic energy conservation law.

bye marco

 

[ogb p]

I would approach this problem by first understanding the concept of elastic collisions. In an elastic collision, both kinetic energy and momentum are conserved. This means that the total kinetic energy and total momentum before the collision must be equal to the total kinetic energy and total momentum after the collision.

Using this concept, I would set up the equations for conservation of momentum and conservation of kinetic energy:

Conservation of momentum:
mAvAi + mBvBi = mAvAf + mBvBf

Conservation of kinetic energy:
½mAvAi^2 + ½mBvBi^2 = ½mAvAf^2 + ½mBvBf^2

Where:
mA and mB are the masses of blocks A and B, respectively
vAi and vBi are the initial velocities of blocks A and B, respectively
vAf and vBf are the final velocities of blocks A and B, respectively

Using the given values, we can substitute and solve for the final velocities:

Conservation of momentum:
(2.0 kg)(5.0 m/s) + (8.0 kg)(-3.0 m/s) = (2.0 kg)vAf + (8.0 kg)vBf
10 kg m/s - 24 kg m/s = (2.0 kg)vAf + (8.0 kg)vBf
-14 kg m/s = (2.0 kg)vAf + (8.0 kg)vBf

Conservation of kinetic energy:
½(2.0 kg)(5.0 m/s)^2 + ½(8.0 kg)(-3.0 m/s)^2 = ½(2.0 kg)vAf^2 + ½(8.0 kg)vBf^2
25 J + 36 J = (2.0 kg)vAf^2 + (8.0 kg)vBf^2
61 J = (2.0 kg)vAf^2 + (8.0 kg)vBf^2

Now, we have two equations with two unknowns (vAf and vBf). We can solve this system of equations to find the final velocities:

-14 kg m/s = (2.0 kg)vAf + (8.0 kg)vBf
61 J = (2.0 kg)vAf^2 + (8.0 kg)v