A body of mass 4.2kg moving with velocity of10m/s due east collides with stationery body of mass 2.8kg.If they stick together after collision and move with velocity (v) due east,calculate kinetic energy before collision and kinetic energy after collision and laws of energy
4.2*10 = (4.2+2.8)v
KE before: 1/2 * 4.2 * 10^2
KE after: 1/2 * (4.2+2.8) * v^2
To calculate the kinetic energy before and after the collision, we can use the formulas:
Kinetic energy (KE) = (1/2) * mass * velocity^2
Before the collision:
Mass of the first body (m1) = 4.2 kg
Velocity of the first body (v1) = 10 m/s
Kinetic energy of the first body before the collision:
KE1 = (1/2) * m1 * v1^2
= (1/2) * 4.2 kg * (10 m/s)^2
= 210 J
Mass of the second body (m2) = 2.8 kg (stationary body)
Velocity of the second body (v2) = 0 m/s
Kinetic energy of the second body before the collision:
KE2 = (1/2) * m2 * v2^2
= (1/2) * 2.8 kg * (0 m/s)^2
= 0 J (since it is stationary)
Total kinetic energy before the collision:
Total KE before = KE1 + KE2
= 210 J + 0 J
= 210 J
After the collision, the bodies stick together and move with a velocity (v).
Total mass after the collision (m_total) = m1 + m2
= 4.2 kg + 2.8 kg
= 7 kg
Kinetic energy after the collision:
KE_after = (1/2) * m_total * v^2
Using the principle of conservation of linear momentum:
m1 * v1 + m2 * v2 = m_total * v
Applying this principle to solve for v:
4.2 kg * 10 m/s + 2.8 kg * 0 m/s = 7 kg * v
42 kg*m/s = 7 kg * v
v = 42 kg*m/s / 7 kg
v = 6 m/s
Now, we can calculate the kinetic energy after the collision:
KE_after = (1/2) * m_total * v^2
= (1/2) * 7 kg * (6 m/s)^2
= 126 J
Laws of energy involved in this collision:
1. Conservation of kinetic energy: The total kinetic energy before the collision (210 J) is equal to the total kinetic energy after the collision (126 J). Some of the initial kinetic energy is converted into other forms of energy, such as heat or sound.
2. Conservation of linear momentum: The total linear momentum before the collision (m1 * v1 + m2 * v2) is equal to the total linear momentum after the collision (m_total * v). Linear momentum is conserved in collisions.
To calculate the kinetic energy before and after the collision, we need to understand the concept of kinetic energy and the law of conservation of energy.
1. Kinetic Energy (KE):
Kinetic energy is the energy possessed by an object due to its motion. It is calculated using the formula: KE = 0.5 * mass * velocity^2.
2. Law of Conservation of Energy:
The law of conservation of energy states that in an isolated system, the total energy remains constant. This means that the total energy before a collision is equal to the total energy after the collision.
Now let's calculate the kinetic energy before and after the collision:
Before Collision:
Mass of the first body (m1) = 4.2 kg
Velocity of the first body (v1) = 10 m/s
Kinetic energy of the first body before collision (KE1) = 0.5 * m1 * v1^2
KE1 = 0.5 * 4.2 kg * (10 m/s)^2
After Collision:
Mass of the combined bodies after collision (m2) = 4.2 kg + 2.8 kg = 7 kg
Velocity of the combined bodies after collision (v2) = v (as given in the question)
Kinetic energy of the combined bodies after collision (KE2) = 0.5 * m2 * v2^2
KE2 = 0.5 * 7 kg * v^2
According to the law of conservation of energy, KE1 = KE2. So,
0.5 * 4.2 kg * (10 m/s)^2 = 0.5 * 7 kg * v^2
This equation can be solved to find the value of v. Once you have the value of v, you can calculate the kinetic energy after the collision (KE2) using the above equation.
The laws of energy that apply in this scenario are:
1. Law of Conservation of Energy: The total energy before the collision (sum of kinetic energies) is equal to the total energy after the collision.
2. Law of Conservation of Linear Momentum: The total linear momentum before the collision is equal to the total linear momentum after the collision.