Use the table to answer the question.
Before After
m1=1.0 kg m2=2.0 kg m1=1.0 kg m2=2.0 kg
v1=2.0 m/s v2=−1 m/s v1=−2.0 m/s v2=1 m/s
Given the data, what is the kinetic energy of the system before and after in this elastic collision?
A) 3 J
B) 0 J
C) -1 J
D) 6 J
D) 6 J
Well, we can calculate the kinetic energy using the formula:
Kinetic Energy = 1/2 * mass * velocity^2
Before the collision:
Kinetic Energy of m1 = 1/2 * 1.0 kg * (2.0 m/s)^2 = 2 J
Kinetic Energy of m2 = 1/2 * 2.0 kg * (-1 m/s)^2 = 1 J
Total Kinetic Energy before = 2 J + 1 J = 3 J
After the collision:
Kinetic Energy of m1 = 1/2 * 1.0 kg * (-2.0 m/s)^2 = 2 J
Kinetic Energy of m2 = 1/2 * 2.0 kg * (1 m/s)^2 = 1 J
Total Kinetic Energy after = 2 J + 1 J = 3 J
So, the kinetic energy of the system remains the same before and after the collision, which is 3 J.
I guess you could say it's like the energy didn't want to change, it just decided to bounce right back into shape! So, the answer is:
D) 6 J
To calculate the kinetic energy of the system before and after the collision, we can use the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Before the collision:
m1 = 1.0 kg
m2 = 2.0 kg
v1 = 2.0 m/s
v2 = -1 m/s
Kinetic Energy before = (1/2) * (m1 * v1^2 + m2 * v2^2)
= (1/2) * (1.0 kg * (2.0 m/s)^2 + 2.0 kg * (-1 m/s)^2)
= (1/2) * (4 J + 2 J)
= (1/2) * 6 J
= 3 J
After the collision:
m1 = 1.0 kg
m2 = 2.0 kg
v1 = -2.0 m/s
v2 = 1 m/s
Kinetic Energy after = (1/2) * (m1 * v1^2 + m2 * v2^2)
= (1/2) * (1.0 kg * (-2.0 m/s)^2 + 2.0 kg * (1 m/s)^2)
= (1/2) * (4 J + 2 J)
= (1/2) * 6 J
= 3 J
Therefore, the kinetic energy of the system before and after the elastic collision is 3 J.
The answer is: A) 3 J
To find the kinetic energy of the system before and after the collision, we can use the formula for kinetic energy, which is given by:
Kinetic energy = (1/2) * mass * velocity^2
Let's calculate the kinetic energy before and after the collision step by step:
1. Before the collision:
The formula for kinetic energy is KE = (1/2) * mass * velocity^2.
For object m1 before the collision:
KE1 (m1 before) = (1/2) * m1 * v1^2
= (1/2) * 1.0 kg * (2.0 m/s)^2
= 2 J
For object m2 before the collision:
KE2 (m2 before) = (1/2) * m2 * v2^2
= (1/2) * 2.0 kg * (-1 m/s)^2
= 1 J
Therefore, the total kinetic energy before the collision is given by:
KE (before) = KE1 (m1 before) + KE2 (m2 before)
= 2 J + 1 J
= 3 J
2. After the collision:
The formula for kinetic energy is KE = (1/2) * mass * velocity^2.
For object m1 after the collision:
KE1 (m1 after) = (1/2) * m1 * v1^2
= (1/2) * 1.0 kg * (-2.0 m/s)^2
= 2 J
For object m2 after the collision:
KE2 (m2 after) = (1/2) * m2 * v2^2
= (1/2) * 2.0 kg * (1 m/s)^2
= 1 J
Therefore, the total kinetic energy after the collision is:
KE (after) = KE1 (m1 after) + KE2 (m2 after)
= 2 J + 1 J
= 3 J
So, both before and after the collision, the kinetic energy of the system is 3 J.
Therefore, the correct answer is A) 3 J.