Question

a 4.0-kg object initially moving at 5.0 m/s has a perfectly inelastic collision with a 2.0-kg object initially moving in the opposite direction at 1.0 m/s. what is the absolute value of the change in mechanical energy of the two-object system as a result of this collision?

Explanation:

Step1: Find final velocity (momentum conservation)

Momentum before collision equals momentum after collision. Let $m_1=4.0\ \text{kg}$, $v_1=5.0\ \text{m/s}$, $m_2=2.0\ \text{kg}$, $v_2=-1.0\ \text{m/s}$ (negative for opposite direction), $v_f$ = final velocity.
$$m_1v_1 + m_2v_2 = (m_1+m_2)v_f$$
$$(4.0)(5.0) + (2.0)(-1.0) = (4.0+2.0)v_f$$
$$20 - 2 = 6v_f$$
$$18 = 6v_f \implies v_f = 3.0\ \text{m/s}$$

Step2: Calculate initial kinetic energy

Initial KE is sum of KE of both objects.
$$KE_i = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$
$$KE_i = \frac{1}{2}(4.0)(5.0)^2 + \frac{1}{2}(2.0)(1.0)^2$$
$$KE_i = 0.5(4.0)(25) + 0.5(2.0)(1) = 50 + 1 = 51\ \text{J}$$

Step3: Calculate final kinetic energy

Final KE is KE of combined mass.
$$KE_f = \frac{1}{2}(m_1+m_2)v_f^2$$
$$KE_f = \frac{1}{2}(6.0)(3.0)^2 = 0.5(6.0)(9) = 27\ \text{J}$$

Step4: Find change in mechanical energy

Change in energy is $|KE_f - KE_i|$.
$$|\Delta KE| = |27 - 51| = 24\ \text{J}$$

Answer:

24 J