A bullet shot into a block of wood acquires a speed of 1.89 m/s.

The scenario describes a bullet impacting a stationary block of wood, and both then moving together. This is an example of a perfectly inelastic collision. The law of conservation of momentum can be applied here; the total momentum before the collision is equal to the total momentum after the collision. We can denote: mb and vb as the mass and initial velocity of the bullet respectively, mbw and vbw as the total mass and velocity of the bullet+wood block system after the bullet is lodged, The respective momenta would be mb*vb and mbw*vbw. Equating both we get: mb * vb = mbw * vbw The velocity of the bullet before it strikes the block vb can be calculated by manipulating the equation: vb = mbw * vbw / mb Now, just insert the given values into the equation: = (5319 g + 11.9 g) * 1.89 m/s / 11.9 g = 5330.9 g * 1.89 m/s / 11.9 g What we need to remember here is to convert the mass from grams to kilograms (as the standard unit for mass in physics is kilograms): 1 g = 0.001 kg So the equation will be = 5.3309 kg * 1.89 m/s / 0.0119 kg = 892.35 m/s So, the speed of the bullet before striking the block would be approximately 892.35 m/s.