A380-g air track cart traveling at 1.25 m/s collides elastically with a 340-g cart traveling in the opposite direction at 1.33 m/s. what is the speed of the 340-g cart after the collision?

The speed of the 340-g cart after the collision is 1.39 m/s

Explanation:

Hi there!

Since the carts collide elastically, the kinetic energy and momentum of the system remain constant.

The momentum of the system is calculated as the sum of the momentums of each cart. This momentum has to be the same after the collision.

be:

m1 = mass of the 380-g cart.

v1 = velocity of the 380-g cart.

m2 = mass of the 340-g cart.

v2 = velocity of the 340-g cart.

v1´= velocity of the 380-g cart after the collision.

v2´= velocity of the 340-g cart after the collision.

Then, the momentum of the system will be:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Replacig with the given data:

0.380 kg · 1.25 m/s + 0.340 kg · (-1.33 m/s) = 0.380 kg · v1´ + 0.340 kg · v2´

0.0228 kg · m/s =  0.380 kg · v1´ + 0.340 kg · v2´

Solving for v1´:

(0.0228 kg · m/s -  0.340 kg · v2´)/0.380 kg = v1´

The kinetic energy of the system will be the sum of the kinetic energies of each cart. This kinetic energy will be the same after the collision:

1/2 · m1 · v1² + 1/2 · m2 · v2² = 1/2 · m1 · v1´²+ 1/2 · m2 · v2´²

divide both sides of the equation by 1/2:

m1 · v1² + m2 · v2² = m1 · v1´²+ m2 · v2´²

Replacing with the given data:

(I will omit units in the calculation for clarity)

0.380 · 1.25² + 0.340 · 1.33² = 0.380 · v1²´+ 0.340 · v2´²

replacing v1´ = (0.0228  -  0.340 · v2´)/0.380  

1.195176  = 0.380 · [(0.0228 -  0.340 · v2´)/0.380]² + 0.340 · v2´²

1.195176 =[( 5.1984 × 10⁻⁴ - 0.015504 · v2´ + 0.1156 · v2´² ) / 0.380] + 0.340 · v2´²

1.195176 = 1.368 × 10⁻³ - 0.0408  · v2´ + 0.304210526 · v2´² + 0.340 · v2´²

0 = -1.195176 + 1.368 × 10⁻³ - 0.0408  · v2´ + 0.644210526 · v2´²

0 = -1.193808 - 0.0408  · v2´ + 0.644210526 · v2´²

Solving the quadratic equation:  

v2´ = 1.39 m/s

v2´ = -1.33

Since the speed of the 340-g cart can´t be the same as before the collision, the speed of the 340-g cart after the collision is 1.39 m/s

momentum of soft ball three times less than baskey ball