Question

A block of wood of mass \(0.95 \mathrm{kg}\) is initially at rest. A bullet of mass \(0.050 \mathrm{kg}\) traveling at \(100.0 \mathrm{m} / \mathrm{s}\) strikes the block and becomes embedded in it. With what speed do the block of wood and the bullet move just after the collision?

Short answer

Answer: The final speed of the block of wood and bullet just after the collision is 5.0 m/s.

Step-by-step solution

Define initial and final momenta

Before the collision, the block of wood is at rest and the bullet is moving at a given speed. The initial momentum of the system is given by the momentum of the bullet since the wood block's momentum is zero (rest). After the collision, the block of wood and the bullet are moving together at an unknown speed, which we must find. Initial momentum: \(momentum_{initial} = momentum_{bullet}\) Final momentum: \(momentum_{final} = momentum_{wood + bullet}\)

Apply conservation of linear momentum

According to the conservation of linear momentum, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. Therefore, we can write the equation: \(momentum_{initial} = momentum_{final}\)

Calculate initial momentum

First, we calculate the initial momentum of the bullet: \(momentum_{bullet} = mass_{bullet} \times velocity_{bullet}\) \(momentum_{bullet} = (0.050 \mathrm{kg}) \times (100.0 \mathrm{m/s})\) \(momentum_{bullet} = 5.0 \mathrm{kg \cdot m/s}\)

Calculate final momentum

Now we write the expression for the final momentum of the system, which is equal to the total mass (mass of block plus mass of bullet) multiplied by their unknown final velocity (v): \(momentum_{final} = (mass_{wood} + mass_{bullet}) \times v\)

Set up an equation and solve for final velocity

Now we can set up an equation for the conservation of linear momentum and solve for the final velocity: \(momentum_{initial} = momentum_{final}\) \(mass_{bullet} \times velocity_{bullet} = (mass_{wood} + mass_{bullet}) \times v\) Plugging in the values for masses and velocity: \(5.0 \mathrm{kg \cdot m/s} = (0.95 \mathrm{kg} + 0.050 \mathrm{kg}) \times v\) \(v = \frac{5.0 \mathrm{kg \cdot m/s}}{1.0\mathrm{kg}}\) \(v = 5.0 \mathrm{m/s}\)

State final answer

Therefore, the final speed of the block of wood and bullet together just after the collision is 5.0 m/s.