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Question / Instructions :

╰┈➤ A 3.0 x 10^2 gram cart moves on an air truck at 1.5 m/s it collides with and sticks to another cart of mass 5.0 x 10^2 grams, which was stationary before collision? What is the velocity of combined cart after collision?

Answer / Step-by-step eplanation :

╰┈➤ To solve this problem, we can apply the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.

The momentum (p) of an object is calculated by multiplying its mass (m) by its velocity (v):

p = m * v

Let's denote the velocity of the combined cart after the collision as V (unknown). We can set up the equation as follows:

(m1 * v1) + (m2 * v2) = (m1 + m2) * V

where:

m1 = mass of the first cart = 3.0 x 10^2 grams = 3.0 x 10^(-1) kg

v1 = velocity of the first cart = 1.5 m/s

m2 = mass of the second cart = 5.0 x 10^2 grams = 5.0 x 10^(-1) kg

v2 = velocity of the second cart (initially stationary) = 0 m/s

Substituting the given values into the equation, we have:

(3.0 x 10^(-1) kg * 1.5 m/s) + (5.0 x 10^(-1) kg * 0 m/s) = (3.0 x 10^(-1) kg + 5.0 x 10^(-1) kg) * V

(4.5 x 10^(-1) kg * m/s) = (8.0 x 10^(-1) kg) * V

Dividing both sides of the equation by (8.0 x 10^(-1) kg), we get:

(4.5 x 10^(-1) kg * m/s) / (8.0 x 10^(-1) kg) = V

V ≈ 0.5625 m/s

Therefore, the velocity of the combined cart after the collision is approximately 0.5625 m/s.