A 0.050 kg marble strikes a 5.0 kg wooden block and embeds itself in the block. The block (with the marble embedded) flies off its stand at 9.0m/s What was the original velocity of the marble? a 909m/s b 720m/s d 1004m/s

This problem involves the principle of conservation of momentum. The total momentum of a closed system remains constant if no external forces are acting on it. In this case, we are looking at the event of a marble striking and embedding itself in a wooden block, which is an inelastic collision. Before the collision, the wooden block is at rest, so its initial momentum is zero. The marble is moving, so its initial momentum is the product of its mass and its velocity. After the collision, the marble and the block move together with a common velocity.Let's denote the original velocity of the marble as \(v\). The mass of the marble is \(0.050\) kg, and the mass of the block is \(5.0\) kg. The final velocity of the block with the marble embedded is \(9.0\) m/s.Using the conservation of momentum, we can set up the following equation:Initial momentum of the marble + Initial momentum of the block = Final momentum of the combined systemSince the block is initially at rest, its momentum is zero. Therefore, the equation simplifies to:\((0.050 \, \text{kg}) \cdot v = (0.050 \, \text{kg} + 5.0 \, \text{kg}) \cdot 9.0 \, \text{m/s}\)Now we solve for \(v\):\(v = \frac{(0.050 + 5.0) \, \text{kg} \cdot 9.0 \, \text{m/s}}{0.050 \, \text{kg}}\)\(v = \frac{5.050 \, \text{kg} \cdot 9.0 \, \text{m/s}}{0.050 \, \text{kg}}\)\(v = \frac{45.45 \, \text{kg} \cdot \text{m/s}}{0.050 \, \text{kg}}\)\(v = 909 \, \text{m/s}\)