The bullet's speed before striking the block is 11.9 m/s.

You can use the law of conservation of momentum to solve this problem. According to this law, the total momentum of an isolated system, if not acted upon by external forces, remains constant regardless of the actions of the objects within the system. Here the system is made up of the bullet and the block. Before the bullet hits the block, the block is stationary, so its momentum is zero. The bullet has some initial momentum which is equal to its mass times its velocity, which we have to find. After the bullet hits the block, they move together with a certain velocity. The combined momentum is equal to the total mass (mass of bullet + mass of block) times the combined velocity. By the law of conservation of momentum, these two momenta should be equal. So, you can set up the equation as follows: (mass of bullet) * (velocity of bullet) = (mass of bullet + mass of block) * (combined velocity) Then plug in the given numbers: (11.9 g) * (velocity of bullet) = (11.9 g + 5319 g) * (1.89 m/s) Don't forget that we must convert grams (g) to kilograms (kg) to match with the standard SI unit for momentum (kg*m/s). So divide the gram values by 1000: (0.0119 kg) * (velocity of bullet) = (0.0119 kg + 5.319 kg) * (1.89 m/s) Now solve for (velocity of bullet): (0.0119 kg) * (velocity of bullet) = 6.3308 kg*m/s Therefore, velocity of bullet = 6.3308 kg*m/s / 0.0119 kg = 531.92 m/s The speed of the bullet before striking the block is approximately 531.92 m/s.