Question

on a frictionless horizontal air table, puck a (with mass 0.250 kg) is moving toward puck b (with mass 0.366 kg), which is initially at rest. after the collision, puck a has velocity 0.121 m/s to the left, and puck b has velocity 0.652 m/s to the right. what was the speed ( v_{a1} ) of puck a before the collision? ( v_{a1} = 0.834 ) m/s part b calculate ( delta k ), the change in the total kinetic energy of the system that occurs during the collision. view available hint(s) ( delta k = 0.007 ) j submit previous answers request answer x incorrect; try again; 4 attempts remaining provide feedback

Explanation:

Step1: Define initial total kinetic energy

Initial kinetic energy is only from puck A (puck B is at rest):
$$K_i = \frac{1}{2}m_A v_{Ai}^2$$
Substitute values: $m_A=0.250\ \text{kg}$, $v_{Ai}=0.834\ \text{m/s}$
$$K_i = \frac{1}{2}(0.250)(0.834)^2 = 0.0871485\ \text{J}$$

Step2: Define final total kinetic energy

Final kinetic energy is sum of puck A and B's kinetic energy:
$$K_f = \frac{1}{2}m_A v_{Af}^2 + \frac{1}{2}m_B v_{Bf}^2$$
Substitute values: $m_A=0.250\ \text{kg}$, $v_{Af}=-0.121\ \text{m/s}$ (left is negative), $m_B=0.366\ \text{kg}$, $v_{Bf}=0.652\ \text{m/s}$
$$K_f = \frac{1}{2}(0.250)(-0.121)^2 + \frac{1}{2}(0.366)(0.652)^2$$
$$K_f = 0.001830125 + 0.078700704 = 0.080530829\ \text{J}$$

Step3: Calculate kinetic energy change

$$\Delta K = K_f - K_i$$
$$\Delta K = 0.080530829 - 0.0871485$$

Answer:

$\Delta K = -0.00662\ \text{J}$