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mr_godi [17]
1 year ago
14

A diagram labeled Before Collision shows two train cars moving toward each other. The car on the left is labeled m Subscript 1 B

aseline = 600 kilograms and v Subscript 1 Baseline 4 meters per second. The car on the right is labeled m Subscript 2 Baseline = 400 kilograms and v Subscript 2 Baseline negative 2 meters per second. A second drawing labeled After Collision shows two train cars moving away from each other.
Consider a system to be two train cars traveling toward each other.


What is the total momentum of the system before the train cars collide?



kg •


What must the total momentum of the system be after the train cars collide?



kg •
Physics
1 answer:
OLEGan [10]1 year ago
3 0

(a) The total momentum of the system before the train cars collide is 1,600 kgm/s.

(b) The total momentum of the system be after the train cars collide is 1,600 kgm/s.

<h3>What is the total momentum of the car system before the collision?</h3>

The total momentum of the car system before the collision is determined by applying the formula for linear momentum.

Pi = m₁u₁ + m₂u₂

where;

  • m₁ is the mass of the car on the right
  • m₂ is the mass of the car on the left
  • u₁ is the initial velocity of the right
  • u₂ is the initial velocity of the car on the left

Let the rightward direction = positive

Let the leftward direction = negative

Pi = (600 kg x 4 m/s)  +  (400 kg) x (-2 m/s)

Pi = 2,400 kgm/s  -  800 kgm/s

Pi = 1,600 kgm/s

Based on the law of conservation of linear momentum, the sum of the initial momentum of an isolated system is <u>equal</u> to the sum of the final momentum of the system

Pf = Pi = 1,600 kgm/s.

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

#SPJ1

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Scenario
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Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

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\boxed {\boxed {\sf v_i= 4 \ m/s}}

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

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