kg bowling ball going 4m/s strikes a 9 kg bowling ball going -6m/s They have ifter the collision the 12 kg ball is going -3m/s What is the speed pf the 9 kg ball -5.5m/s 3.33m/s 5.5m/s

Here's how to solve this collision problem using the principle of conservation of momentum:<br /><br />**Understanding Conservation of Momentum**<br /><br />In a closed system (where no external forces are acting), the total momentum before a collision is equal to the total momentum after the collision. Momentum is calculated as mass times velocity (p = mv).<br /><br />**Setting up the Equation**<br /><br />* **Before collision:**<br /> * Momentum of 12 kg ball: (12 kg) * (4 m/s) = 48 kg m/s<br /> * Momentum of 9 kg ball: (9 kg) * (-6 m/s) = -54 kg m/s<br /> * Total momentum: 48 kg m/s + (-54 kg m/s) = -6 kg m/s<br /><br />* **After collision:**<br /> * Momentum of 12 kg ball: (12 kg) * (-3 m/s) = -36 kg m/s<br /> * Momentum of 9 kg ball: (9 kg) * (v) = 9v kg m/s (where 'v' is the unknown velocity)<br /> * Total momentum: -36 kg m/s + 9v kg m/s<br /><br />**Solving for the Unknown Velocity (v)**<br /><br />Since the total momentum before and after the collision must be equal:<br /><br />-6 kg m/s = -36 kg m/s + 9v kg m/s<br /><br />Now, solve for v:<br /><br />1. Add 36 kg m/s to both sides: 30 kg m/s = 9v kg m/s<br />2. Divide both sides by 9 kg: v = 30/9 m/s<br />3. Simplify: v ≈ 3.33 m/s<br /><br />**Answer:**<br /><br />The speed of the 9 kg ball after the collision is approximately $\boxed{3.33 m/s}$.<br />