Question

A 5.18-g bullet moving at \(672 \mathrm{~m} / \mathrm{s}\) strikes a \(715-\mathrm{g}\) wooden block at rest on a frictionless surface. The bullet emerges with its speed reduced to \(428 \mathrm{~m} / \mathrm{s}\). Find the resulting speed of the block.

Short answer

The resulting speed of the block can be found by solving the above equation which is obtained using the conservation of momentum principle.

Step-by-step solution

Understand The Problem

A bullet passes through a wooden block, the bullet slows down as a result. As the bullet and wooden block system is an isolated system, we can apply the law of conservation of momentum.

Initial Momentum Calculation

Firstly, calculate the initial momentum in the system. In this system, before collision, only bullet is moving so the initial momentum will be the momentum of the bullet. In momentum formula, momentum = mass * velocity. For the bullet, mass( \(m_{1}\) )= 5.18 g = 0.00518 kg (Convert grams into kilograms making sure SI units are used) and the velocity ( \(v_{1i}\)) = 672 m/s, so the initial momentum ( \(P_{i}\)) = \(m_{1} * v_{1i} \).

Final Momentum Calculation

The final momentum is the sum of the momentum of the bullet after passing through the block and the momentum of the wooden block. For the bullet its mass ( \(m_{1}\) )= 0.00518 kg and its final velocity ( \(v_{1f}\)) = 428 m/s. Therefore, the final momentum of the bullet equals \(m_{1} * v_{1f}\). The block is now moving at a velocity( \(v_{2f}\) ) which we are asked to find, and has a mass ( \(m_{2}\) ) = 715 g = 0.715 kg. Hence the final momentum of the block equals \(m_{2} * v_{2f}\). So, the final total momentum = momentum of bullet after passing + momentum of block = \(m_{1} * v_{1f} + m_{2} * v_{2f}\)

Solve For Unknown

Now, use conservation of momentum to equate the initial and final momentum and solve for the unknown velocity. Therefore, \(m_{1} * v_{1i} = m_{1} * v_{1f} + m_{2} * v_{2f}\)

Key concepts

Collision

When discussing collisions, it's essential to understand how objects interact with each other during an impact. In physics, a collision occurs when two or more objects hit each other, exchanging energy and momentum. This interaction can significantly change their speed and direction.
In our exercise, a bullet collides with a wooden block. A unique aspect of this collision is that the bullet passes through the block. This kind of interaction is referred to as a partially inelastic collision since not all momentum is retained within a single combined object post-impact.
The bullet loses some speed as it goes through, and the block starts moving, illustrating how both objects' velocities change due to the collision. Collisions follow the fundamentals of physics, where forces between the objects cause these changes. Understanding collisions helps us delve into how kinetic energy and momentum transfer from one body to another.

Momentum Calculation

Momentum is the product of an object's mass and velocity and is a crucial concept in physics. It provides insight into the motion of objects and how they behave in various scenarios.
In our exercise, we calculate momentum to find the motion outcome after the bullet strikes the block. Initially, before the collision, only the bullet possesses momentum. We calculate it using the formula:
\( \text{Initial Momentum (P)}_i = m_{1} \times v_{1i} \)
where \(m_{1}\) is the mass of the bullet and \(v_{1i}\) its initial velocity. Simplifying units is vital here, ensuring mass is in kilograms.
For momentum post-collision, we sum the momentum of both the now slower bullet and the moving block. This calculation allows us to apply the conservation of momentum to find out how fast the block must be moving. The final momentum incorporates both remaining components of motion:

• \( \text{Momentum of bullet after }: m_{1} \times v_{1f} \)
• \( \text{Momentum of block after }: m_{2} \times v_{2f} \)

Calculating momentum is how we make predictions about motion after an impact, helping us understand outcomes based on initial conditions.

Isolated System

An isolated system is a crucial concept in understanding how the conservation of momentum works. In physics, an isolated system is one where no external forces act on the objects within it. This means that the total momentum before and after any collision must be the same.
In our bullet-block exercise, the system (bullet + block) is assumed to be free from external influences like friction or air resistance. The absence of external forces allows us to apply the conservation of momentum principle. Here, this principle states:

• The total momentum before the collision must equal the total momentum after the collision.
• \( m_{1} \times v_{1i} = m_{1} \times v_{1f} + m_{2} \times v_{2f} \)

This equation lets us solve for unknown variables, like the speed of the wooden block post-collision. Understanding isolated systems aids in predicting and calculating results in dynamic physics scenarios. It provides a framework that assumes only internal interactions affect total momentum, crucial for analyzing situations where collision dynamics are at play.