Question

select the correct answer.

objects a and b each have a mass of 25 kilograms. object a has a velocity of 5.98 meters/second. object b is stationary. they undergo a perfectly elastic collision in one dimension. what is the total kinetic energy of the system after the collision?

a. \\(1.2 \times 10^2\\) joules

b. \\(4.5 \times 10^2\\) joules

c. \\(5.0 \times 10^2\\) joules

d. \\(9.5 \times 10^2\\) joules

e. \\(1.1 \times 10^3\\) joules

Explanation:

Step1: Recall Kinetic Energy in Elastic Collision

In a perfectly elastic collision, kinetic energy is conserved. So total KE after collision equals total KE before collision.

Step2: Calculate Initial KE of System

The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \). Object B is stationary (\( v_B = 0 \)), so its KE is 0. For Object A: \( m = 25 \, \text{kg} \), \( v = 5.98 \, \text{m/s} \).

$$ KE_A = \frac{1}{2} \times 25 \times (5.98)^2 $$

Step3: Compute the Value

First, calculate \( (5.98)^2 \approx 35.76 \). Then:

$$ KE_A = 0.5 \times 25 \times 35.76 = 12.5 \times 35.76 \approx 447 \, \text{J} \approx 4.5 \times 10^2 \, \text{J} $$

Since KE is conserved, total KE after collision is also \( \approx 4.5 \times 10^2 \, \text{J} \).

Answer:

B. \( 4.5 \times 10^2 \) joules