Two carts are initially moving to the right on a low-friction track, with cart 1 behind cart 2. Cart 1 has a speed twice that of cart 2 and so moves up and rear-ends cart 2, which has twice the inertia of cart 1. Part A Suppose that the the initial speed of cart 2 is v. What is the speed of cart 1 right after the collision if the collision is elastic?

The Speed of Cart 1 right after the collision if the collision is elastic is;

v₁ = ¹/₃u₁ or expressed as one-third of its' initial speed before the collision

Let the speeds be as follows;

Initial speed of Cart 1 = u₁

final speed of Cart 1 = v₁

Initial speed of Cart 2 = u₂

Final speed of Cart 2 = v₂

Since the collision is elastic, then we can say that;

u₁ + v₁ = u₂ + v₂  ---(eq 1)

Also, we can say that;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂  ---(eq 2)

Let us make v₂ the subject in eq 1 and put in eq 2;

v₂ = u₁ - u₂ + v₁   (eq 3)

Thus;

m₁u₁ + m₂u₂ = m₁v₁ + m₂(u₁ - u₂ + v₁)

making v₁ the subject gives;

v₁ = [(m₁ - m₂)/(m₁ + m₂)]u₁ + [2m₂/(m₁ + m₂)]u₂

We are told that Cart 1 has a speed twice that of cart 2. Thus;

u₂ = ¹/₂v₁ =  ¹/₂u₁

Also, we see that cart 2 has twice the inertia of cart 1.

Thus, m₂ = 2m₁

Plugging in the values gives;

v₁ =  [(m₁ - 2m₁)/(m₁ + 2m₁)]u₁ + [2(2m₁)/(m₁ + 2m₁)]( ¹/₂u₁)

Simplifying this gives;

v₁ = -¹/₃u₁ + ²/₃u₁

v₁ = ¹/₃u₁

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Explanation:

Given

cart 1 has twice the speed of cart 2

cart 2 has twice the mass of cart 1

Suppose m is the mass of cart 1 so mass of cart 2 is 2 m

suppose u is the initial velocity of car 2 so initial velocity of car 1 is u

For Elastic collision velocity after collision is given by

velocity of cart 2 is

velocity of cart 1 is

here

substituting values we get

answer: what are you talking about?

explanation:

15.3 cos 65 degrees trig