Question

select the correct answer.
object a has of mass 7.20 kilograms, and object b has a mass of 5.75 kilograms. the two objects move along a straight line toward each other with velocities +2.00 meters/second and -1.30 meters/second respectively. what is the total kinetic energy of the objects after the collision, if the collision is perfectly elastic?
a. 19.3 joules
b. 21.9 joules
c. 38.5 joules
d. 43.0 joules
e. 50.8 joules

Explanation:

Step1: Recall Kinetic Energy Formula

The kinetic energy of an object is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. For a perfectly elastic collision, total kinetic energy is conserved, so we can calculate the total kinetic energy before the collision (which equals the total after).

Step2: Calculate KE for Object A

For Object A: \( m_A = 7.20 \, \text{kg} \), \( v_A = 2.00 \, \text{m/s} \)
\( KE_A = \frac{1}{2}m_Av_A^2 = \frac{1}{2}(7.20)(2.00)^2 \)
\( KE_A = 0.5 \times 7.20 \times 4.00 = 14.4 \, \text{J} \)

Step3: Calculate KE for Object B

For Object B: \( m_B = 5.75 \, \text{kg} \), \( v_B = -1.30 \, \text{m/s} \) (speed is \( |v_B| = 1.30 \, \text{m/s} \))
\( KE_B = \frac{1}{2}m_Bv_B^2 = \frac{1}{2}(5.75)(1.30)^2 \)
\( KE_B = 0.5 \times 5.75 \times 1.69 \approx 0.5 \times 5.75 \times 1.69 \approx 4.90625 \, \text{J} \) Wait, no—wait, \( 1.30^2 = 1.69 \), \( 5.75 \times 1.69 = 9.7175 \), then \( 0.5 \times 9.7175 = 4.85875 \)? Wait, no, I made a mistake. Wait, no: \( v_B \) magnitude is 1.30, so \( KE_B = 0.5 5.75 (1.30)^2 \). Let's recalculate: \( 1.30^2 = 1.69 \), \( 5.75 * 1.69 = 5.75*1.7 - 5.75*0.01 = 9.775 - 0.0575 = 9.7175 \), then \( 0.5 * 9.7175 = 4.85875 \)? Wait, that can't be right. Wait, no, maybe I messed up the velocity sign? No, kinetic energy depends on speed (magnitude of velocity). Wait, but let's check again. Wait, maybe I miscalculated Object A: \( 0.5 * 7.2 * 4 = 14.4 \), correct. Object B: \( 0.5 5.75 (1.3)^2 \). \( 1.3^2 = 1.69 \), \( 5.75 1.69 = 5.75(1 + 0.6 + 0.09) = 5.75 + 3.45 + 0.5175 = 9.7175 \), then \( 0.5 * 9.7175 = 4.85875 \). Then total KE before collision is \( 14.4 + 4.85875 = 19.25875 \approx 19.3 \, \text{J} \). Since KE is conserved in elastic collision, total KE after is also ~19.3 J.

Answer:

A. 19.3 joules