A 0.240 kg ball moves at a speed of 3.0m/s strikes a wall and bounces straight back at 2.40m/s. The collision lasts 0.0150s. How much kinetice energy in joules is lost during the collision?
Wouln't you just subtract the final KE from the initial KE? That is my understanding of conservation of energy.
o yea!
thNKS
Yes, you are correct! The change in kinetic energy during a collision can be calculated by subtracting the final kinetic energy from the initial kinetic energy. In this case, the initial kinetic energy is the kinetic energy before the collision and the final kinetic energy is the kinetic energy after the collision.
To calculate the initial kinetic energy (KEi), we can use the formula:
KEi = 0.5 * mass * velocity^2
Given the mass (m) of the ball is 0.240 kg and the initial velocity (v) is 3.0 m/s, we can substitute these values into the formula:
KEi = 0.5 * 0.240 kg * (3.0 m/s)^2
Solving this equation, we find that the initial kinetic energy is:
KEi = 1.08 J
Now, let's calculate the final kinetic energy (KEf) using the same formula, but with the final velocity (v) of 2.40 m/s:
KEf = 0.5 * 0.240 kg * (2.40 m/s)^2
Solving this equation, we find that the final kinetic energy is:
KEf = 0.6912 J
To find the change in kinetic energy during the collision, we subtract the final kinetic energy from the initial kinetic energy:
Change in KE = KEi - KEf
Substituting the values we calculated:
Change in KE = 1.08 J - 0.6912 J
The change in kinetic energy is:
Change in KE ≈ 0.3888 J
Therefore, approximately 0.3888 Joules of kinetic energy are lost during the collision.