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On a low-friction track, a 0.66-kg cart initially going at 1.85 m/s to the right collides with a cart of unknown inertia initially going at 2.17 m/s to the left. After the collision, the 0.66-kg cart is going at 1.32 m/s to the left, and the cart of unknown inertia is going at 3.22 m/s to the right. The collision takes 0.010 s.

What is the unknown inertia?
What is the average acceleration of the heavier cart?
What is the average acceleration of the lighter cart?

1 Answer

9 votes

Answer:

(a) the unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart is 317 m/s²

(c) the average acceleration of the lighter cart is 539 m/s²

Step-by-step explanation:

Given;

mass of the first cart, m₁ = 0.66 kg

initial speed of the first cart, u₁ = 1.85 m/s

let the mass of the cart with unknown inertia be m₂

initial velocity of the second cart, u₂ = 2.17 m/s to the left

velocity of the first cart after collision, v₁ = 1.32 m/s to the left

velocity of the second cart after collision, v₂ = 3.22 m/s

time of collision, t = 0.010 s

(a) What is the unknown inertia?

Apply the principle of conservation of linear momentum, to determine the unknown inertia.

let leftward direction be negative direction

let rightward direction be positive direction

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)

1.221 - 2.17m₂ = -0.8712 + 3.22m₂

1.221 + 0.8712 = 3.22m₂ + 2.17m₂

2.0922 = 5.39m₂

m₂ = 2.0922 / 5.39

m₂ = 0.388 kg

The unknown inertia is 0.388 kg

(b) the average acceleration of the heavier cart

the heavier cart has a mass of 0.66 kg


a = (v_1 - u_1)/(t) \\\\a = (-1.32 - 1.85)/(0.01) \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2

(c) the average acceleration of the lighter cart;

the lighter cart has a mass of 0.388 kg


a = (v_2 - u_2)/(t) \\\\a = (3.22 - (-2.17))/(0.01) \\\\a =(3.22 \ +\ 2.17)/(0.01) \\\\a= 539\ m/s^2

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User Rnunes
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