91Ó°ÊÓ

Identify the given mass and velocities for bodies A and B before and after the collision. \( m_A = 2.0 \, \mathrm{kg} \) \( m_B = 2.0 \, \mathrm{kg} \) \( \vec{v}_{A 1} = (15 \, \mathrm{m/s}) \hat{i} + (30 \, \mathrm{m/s}) \hat{j} \) \( \vec{v}_{B 1} = (-10 \, \mathrm{m/s}) \hat{i} + (5 \, \mathrm{m/s}) \hat{j} \) \( \vec{v}_{A 2} = (-5 \, \mathrm{m/s}) \hat{i} + (20 \, \mathrm{m/s}) \hat{j} \)

According to the law of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision. \( \vec{p}_{\text{before}} = \vec{p}_{\text{after}} \) Calculate the total momentum before the collision: \[ \vec{p}_{\text{total, before}} = m_A \vec{v}_{A 1} + m_B \vec{v}_{B 1} \] \[ \vec{p}_{\text{total, before}} = 2.0 \times ((15 \hat{i} + 30 \hat{j})) + 2.0 \times ((-10 \hat{i} + 5 \hat{j})) \] \[ \vec{p}_{\text{total, before}} = (30 \hat{i} + 60 \hat{j}) + (-20 \hat{i} + 10 \hat{j}) \] \[ \vec{p}_{\text{total, before}} = 10 \hat{i} + 70 \hat{j} \] Calculate the total momentum after the collision: \[ \vec{p}_{\text{total, after}} = m_A \vec{v}_{A 2} + m_B \vec{v}_{B 2} \] Since we know \( \vec{v}_{A 2} \) but need \( \vec{v}_{B 2} \), set up the equation to solve: \[ 10 \hat{i} + 70 \hat{j} = 2.0 \times (-5 \hat{i} + 20 \hat{j}) + 2.0 \vec{v}_{B 2} \] \[ 10 \hat{i} + 70 \hat{j} = (-10 \hat{i} + 40 \hat{j}) + 2.0 \vec{v}_{B 2} \] Isolate \( \vec{v}_{B 2} \): \[ 2.0 \vec{v}_{B 2} = (10 + 10) \hat{i} + (70 - 40) \hat{j} \] \[ 2.0 \vec{v}_{B 2} = 20 \hat{i} + 30 \hat{j} \] \[ \vec{v}_{B 2} = 10 \hat{i} + 15 \hat{j} \] So, the final velocity of B is \( \vec{v}_{B 2} = 10 \hat{i} + 15 \hat{j} \).

Kinetic energy is given by \( KE = \frac{1}{2} m v^2 \). First, calculate the kinetic energy before the collision: \[ KE_{\text{before}} = \frac{1}{2} m_A \| \vec{v}_{A 1} \|^2 + \frac{1}{2} m_B \| \vec{v}_{B 1} \|^2 \] \[ KE_{\text{before}} = \frac{1}{2} 2.0 ((15)^2 + (30)^2) + \frac{1}{2} 2.0 ((-10)^2 + (5)^2) \] \[ KE_{\text{before}} = (1.0) (225 + 900) + (1.0) (100 + 25) \] \[ KE_{\text{before}} = 1125 + 125 \] \[ KE_{\text{before}} = 1250 \, \mathrm{J} \] Next, calculate the kinetic energy after the collision: \[ KE_{\text{after}} = \frac{1}{2} m_A \| \vec{v}_{A 2} \|^2 + \frac{1}{2} m_B \| \vec{v}_{B 2} \|^2 \] \[ KE_{\text{after}} = \frac{1}{2} 2.0 ((-5)^2 + (20)^2) + \frac{1}{2} 2.0 ((10)^2 + (15)^2) \] \[ KE_{\text{after}} = (1.0) (25 + 400) + (1.0) (100 + 225) \] \[ KE_{\text{after}} = 425 + 325 \] \[ KE_{\text{after}} = 750 \, \mathrm{J} \]

The change in kinetic energy is given by the difference between the initial and final kinetic energies. \[ \Delta KE = KE_{\text{after}} - KE_{\text{before}} \] \[ \Delta KE = 750 \, \mathrm{J} - 1250 \, \mathrm{J} \] \[ \Delta KE = -500 \, \mathrm{J} \] The kinetic energy lost in the collision is 500 J.