Question

use the table to answer the question. the table shows the mass and velocity of two objects before and after collision. given the data, what is the kinetic energy of the system before and after in this elastic collision? (1 point) 6 j 3 j -1 j 0 j

Explanation:

Step1: Calculate kinetic energy before collision

The formula for kinetic energy is \(K = \frac{1}{2}mv^{2}\).
For object 1 before collision: \(K_{1i}=\frac{1}{2}m_{1}v_{1i}^{2}=\frac{1}{2}(1.0\space kg)(2.0\space m/s)^{2}= 2\space J\)
For object 2 before collision: \(K_{2i}=\frac{1}{2}m_{2}v_{2i}^{2}=\frac{1}{2}(2.0\space kg)(- 1\space m/s)^{2}=1\space J\)
Total kinetic energy before collision \(K_{i}=K_{1i}+K_{2i}\)
\(K_{i}=2 + 1=3\space J\)

Step2: Calculate kinetic energy after collision

For object 1 after collision: \(K_{1f}=\frac{1}{2}m_{1}v_{1f}^{2}=\frac{1}{2}(1.0\space kg)(-2.0\space m/s)^{2}=2\space J\)
For object 2 after collision: \(K_{2f}=\frac{1}{2}m_{2}v_{2f}^{2}=\frac{1}{2}(2.0\space kg)(1\space m/s)^{2}=1\space J\)
Total kinetic energy after collision \(K_{f}=K_{1f}+K_{2f}\)
\(K_{f}=2 + 1=3\space J\)

Answer:

3 J