25. a 1383-kg car moving south at 11.2 m/s is struck by a 1732-kg car moving east at 31.3 m/s. after the…

Explanation:

Step1: Set coordinate system

Let east = positive x-direction, south = positive y-direction.

Step2: Calculate initial x-momentum

Momentum: $p_{x,i} = m_2 v_{2x} = 1732\ \text{kg} \times 31.3\ \text{m/s} = 54211.6\ \text{kg·m/s}$

Step3: Calculate initial y-momentum

Momentum: $p_{y,i} = m_1 v_{1y} = 1383\ \text{kg} \times 11.2\ \text{m/s} = 15489.6\ \text{kg·m/s}$

Step4: Find total mass post-collision

Total mass: $m_{total} = 1383\ \text{kg} + 1732\ \text{kg} = 3115\ \text{kg}$

Step5: Find final x-velocity

By conservation of momentum: $v_{x,f} = \frac{p_{x,i}}{m_{total}} = \frac{54211.6}{3115} \approx 17.4\ \text{m/s}$

Step6: Find final y-velocity

By conservation of momentum: $v_{y,f} = \frac{p_{y,i}}{m_{total}} = \frac{15489.6}{3115} \approx 4.97\ \text{m/s}$

Step7: Calculate final speed

Speed: $v_f = \sqrt{v_{x,f}^2 + v_{y,f}^2} = \sqrt{(17.4)^2 + (4.97)^2} \approx \sqrt{302.76 + 24.70} \approx \sqrt{327.46} \approx 18.1\ \text{m/s}$

Step8: Calculate direction angle

Angle from east (toward south): $\theta = \tan^{-1}\left(\frac{v_{y,f}}{v_{x,f}}\right) = \tan^{-1}\left(\frac{4.97}{17.4}\right) \approx \tan^{-1}(0.2856) \approx 15.9^\circ$

Answer:

The two cars move at approximately 18.1 m/s, at an angle of 15.9 degrees south of east immediately after the collision.