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Sagot :
The momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.
Momentum transfered to the more massive cart
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: https://brainly.com/question/7538238
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