Glider A of mass 2.5 kg moves with speed 1.7 m/s on a horizontal rail without friction. It collides elastically with glider B of identical mass 2.5 kg, which is initially at rest. After the collision, what is the value of the speed of glider A, in m/s?

The speed of glider A after the collision is 0 m/s

Explanation:

Hi there!

The two gliders collide elastically. That means that the kinetic energy and momentum of the system are conserved, i. e., they remain constant before and after the collision.

The momentum is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass.

v = velocity.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass.

v = velocity.

The momentum of the system is the sum of the momentum of each glider.

be:

mA = mass of glider A

mB = mass of glider B

vA = velocity of glider A before the collision.

vB = velocity of glider B before the collision.

vA' = velocity of glider A after the collision.

vB' = velocity of glider B after the collision.

The momentum of the system will be:

mA · vA + mB · vB = mA · vA' + mB · vB'

Replacing with the given dа ta:

2.5 kg · 1.7 m/s + 2.5 kg · 0 m/s = 2.5 kg · vA' + 2.5 kg · vB'

divide both sides of the equation by 2.5 kg:

1.7 m/s = vA' + vB'

1.7 m/s - vA' = vB'

Using the conservation of the kinetic energy of the system we can find vA':

1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · vA'² + 1/2 · mB · (1.7 m/s - vA') ²

Let's replace with the given dа ta:

1/2 · 2.5 kg · (1.7 m/s) ² + 0 = 1/2 · 2.5 kg · vA'² + 1/2 · 2.5 kg · (1.7 m/s - vA') ²

divide both sides of the equation by (1/2 · 2.5 kg):

(1.7 m/s) ² = vA'² + (1.7 m/s - vA') ²

(1.7 m/s) ² = vA'² + (1.7 m/s) ² - 2· 1.7 m/s · vA' + vA'²

0 = 2vA'² - 2· 1.7 m/s · vA'

0 = 2vA' (vA' - 1.7 m/s)

vA' = 0

vA' - 1.7 m/s = 0

vA' = 1.7 m/s

Since the velocity of the glider A after the collision can't be the same as before the collision, the velocity of glider A after the collision is 0 m/s.