Question
A car with a mass of 1250 kg at 2.24 m/s and bumps into a car With a mass of 1300 kg. After the collision the two cars stick together and move forward. How fast will they both move forward
Answers
Answered by
GPT 3.5
To solve this problem, we can use the law of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.
Let's consider the velocity of the first car before the collision as v₁ (2.24 m/s) and the velocity of the second car before the collision as v₂ (unknown). After the collision, the two cars stick together and move forward with a final velocity v.
According to the law of conservation of momentum:
(m₁ * v₁) + (m₂ * v₂) = (m₁ + m₂) * v
Where:
m₁ = mass of the first car (1250 kg)
v₁ = velocity of the first car before the collision (2.24 m/s)
m₂ = mass of the second car (1300 kg)
v₂ = velocity of the second car before the collision (unknown)
m₁ + m₂ = total mass after the collision (m₁ + m₂ = 1250 kg + 1300 kg = 2550 kg)
v = final velocity of both cars after the collision (unknown)
Substituting the given values into the equation:
(1250 kg * 2.24 m/s) + (1300 kg * v₂) = (2550 kg) * v
2800 kg·m/s + (1300 kg * v₂) = 2550 kg * v
Next, we need to use the principle of conservation of kinetic energy to determine the velocity v₂ of the second car before the collision. Kinetic energy is also conserved in an elastic collision.
According to the conservation of kinetic energy,
(1/2 * m₁ * v₁²) + (1/2 * m₂ * v₂²) = (1/2 * (m₁ + m₂) * v²)
Substituting the given values into the equation:
(1/2 * 1250 kg * (2.24 m/s)²) + (1/2 * 1300 kg * v₂²) = (1/2 * 2550 kg * v²)
(1/2 * 1250 kg * 5.0176 m²/s²) + (1/2 * 1300 kg * v₂²) = (1/2 * 2550 kg * v²)
3134.5 kg·m²/s² + (1/2 * 1300 kg * v₂²) = 1275 kg·v²
Now we can solve these two equations simultaneously to find the values of v and v₂.
2800 kg·m/s + (1300 kg * v₂) = 2550 kg * v -----(1)
3134.5 kg·m²/s² + (1/2 * 1300 kg * v₂²) = 1275 kg·v² -----(2)
With some algebraic manipulation, we can rearrange equation (1) to isolate v:
v = (2800 kg·m/s + (1300 kg * v₂)) / 2550 kg
Now substitute this value of v into equation (2) to solve for v₂:
3134.5 kg·m²/s² + (1/2 * 1300 kg * v₂²) = 1275 kg·(2800 kg·m/s + (1300 kg * v₂))^2 / 2550 kg²
Simplifying this equation and solving for v₂ will give us the velocity of the second car before the collision. Once we have v₂, we can calculate the final velocity, v, by substituting the values of v₁, v₂, m₁, and m₂ into equation (1):
v = (2800 kg·m/s + (1300 kg * v₂)) / 2550 kg
Let's consider the velocity of the first car before the collision as v₁ (2.24 m/s) and the velocity of the second car before the collision as v₂ (unknown). After the collision, the two cars stick together and move forward with a final velocity v.
According to the law of conservation of momentum:
(m₁ * v₁) + (m₂ * v₂) = (m₁ + m₂) * v
Where:
m₁ = mass of the first car (1250 kg)
v₁ = velocity of the first car before the collision (2.24 m/s)
m₂ = mass of the second car (1300 kg)
v₂ = velocity of the second car before the collision (unknown)
m₁ + m₂ = total mass after the collision (m₁ + m₂ = 1250 kg + 1300 kg = 2550 kg)
v = final velocity of both cars after the collision (unknown)
Substituting the given values into the equation:
(1250 kg * 2.24 m/s) + (1300 kg * v₂) = (2550 kg) * v
2800 kg·m/s + (1300 kg * v₂) = 2550 kg * v
Next, we need to use the principle of conservation of kinetic energy to determine the velocity v₂ of the second car before the collision. Kinetic energy is also conserved in an elastic collision.
According to the conservation of kinetic energy,
(1/2 * m₁ * v₁²) + (1/2 * m₂ * v₂²) = (1/2 * (m₁ + m₂) * v²)
Substituting the given values into the equation:
(1/2 * 1250 kg * (2.24 m/s)²) + (1/2 * 1300 kg * v₂²) = (1/2 * 2550 kg * v²)
(1/2 * 1250 kg * 5.0176 m²/s²) + (1/2 * 1300 kg * v₂²) = (1/2 * 2550 kg * v²)
3134.5 kg·m²/s² + (1/2 * 1300 kg * v₂²) = 1275 kg·v²
Now we can solve these two equations simultaneously to find the values of v and v₂.
2800 kg·m/s + (1300 kg * v₂) = 2550 kg * v -----(1)
3134.5 kg·m²/s² + (1/2 * 1300 kg * v₂²) = 1275 kg·v² -----(2)
With some algebraic manipulation, we can rearrange equation (1) to isolate v:
v = (2800 kg·m/s + (1300 kg * v₂)) / 2550 kg
Now substitute this value of v into equation (2) to solve for v₂:
3134.5 kg·m²/s² + (1/2 * 1300 kg * v₂²) = 1275 kg·(2800 kg·m/s + (1300 kg * v₂))^2 / 2550 kg²
Simplifying this equation and solving for v₂ will give us the velocity of the second car before the collision. Once we have v₂, we can calculate the final velocity, v, by substituting the values of v₁, v₂, m₁, and m₂ into equation (1):
v = (2800 kg·m/s + (1300 kg * v₂)) / 2550 kg