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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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7 0
3 years ago
A 102 kg football player runs at a speed of 8 m/s to sack the quarterback. What is
leva [86]

The mass of the quarterback is 61.2 kg.

Explanation:

mass of the football player = m1 = 102 kg

mass of the quarterback = m2 = ?

velocity of the football player = v1 = 8 m/s

According to the law of conservation of momentum:

The total momentum of a system before and after the collision remains constant. Assuming the situation as an isolated system which is not affected by any external factors, we have:

m₁v₁ + m₂v₂ = (m₁+m₂)V

Here, we need to find m₂.

We assume that the quarterback is standing still when he is attacked by the football player so v₂ = 0 m/s

After the collision both of them fall to the ground with a velocity of 5 m/s so V = 5 m/s

102(8) + m2(0) = (102 + m2)(5)\\816 + 0 = (102 + m2)(5)\\816/5 = 102 + m2\\163.2 - 102 = m2\\m2 = 61.2 kg

Keywords: momentum, velocity, law of conservation of momentum

Learn more about Law of Conservation of Momentum from brainly.com/question/7538238

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3 0
3 years ago
In the reaction C + O2 → CO2, 18 g of carbon react with oxygen to produce 72 g of carbon dioxide. What mass of oxygen would be n
Mazyrski [523]

Answer:

54 g of oxygen ...

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3 years ago
Question 6 is the answer I need
daser333 [38]

Answer:

B

Explanation:

A bicameral system describes a government that has a two-house legislative system, such as the House of Representatives and the Senate that make up the U.S. Congress.

8 0
3 years ago
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

7 0
3 years ago
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