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1 year ago

How much momentum, in the x-direction, was transferred to the more massive cart, in kilogram meters per second

1 answer:

6 0

The momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.

<h3>Momentum transfered to the more massive cart</h3>

The momentum transfered to the more massive cart is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

m₁ is the mass of the smaller cart
u₁ is the initial velocity of the samller cart
m₂ is the mass of the bigger cart = 3m₁
u₂ is the initial velocity of the bigger cart
v₁ is the final velocity of the smaller cart
v₂ is the final veocity of the bigger cart

⁻ΔP₁ = ΔP₂

ΔP₂ = m₂v₂ - m₂u₂

ΔP₂ = m₂(v₂ - u₂)

ΔP₂ = 3m₁(v₂ - u₂)

ΔP₂ = 3 x 3.8 x (1.7 - 0)

ΔP₂ = 19.38 kgm/s

Thus, the momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.

The complete question is beblow

A cart of mass 3.8 kg is traveling to the right (which we will take to be the positive x-direction for this problem) at a speed of 6.9 m/s. It collides with a stationary cart that is three times as massive. After the collision, the more massive cart is moving at a speed of 1.7 m/s, to the right.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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