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A 10.0 -g bullet is fired into a stationary block of wood \((m=\) \(5.00 \mathrm{kg}) .\) The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is \(0.600 \mathrm{m} / \mathrm{s} .\) What was the original speed of the bullet?

Short Answer

Expert verified
The initial speed of the bullet is 300.6 m/s.

Step by step solution

01

Formula for the conservation of momentum

Write down the conservation of momentum formula. In this case, it's: \( m_{bullet} \cdot v_{bullet_{initial}} + m_{wood}\cdot v_{wood_{initial}} = (m_{bullet} + m_{wood}) \cdot v_{final}\) Where: \(m_{bullet}\) is the mass of the bullet, \(v_{bullet_{initial}}\) is the initial speed of the bullet, \(m_{wood}\) is the mass of the wood, \(v_{wood_{initial}}\) is the initial speed of the wood (which is 0 because the block was stationary), \(v_{final}\) is the final speed of the bullet and block of wood.
02

Plug given numbers into formula

Now we can input our given values into the formula from step 1. This gives us: \( (10.0 g \cdot v_{bullet_{initial}} + 5.00 kg \cdot 0) = (10.0 g + 5.00 kg) \cdot 0.600 m/s \). Note, it's essential to have all these values in the same units, such as kg and m/s in this case, so convert the mass of the bullet to kg: \( (0.01 kg \cdot v_{bullet_{initial}} + 5.00 kg \cdot 0) = (0.01 kg + 5.00 kg) \cdot 0.600 m/s \). It can be simplified to: \(0.01 kg \cdot v_{bullet_{initial}} = 5.01 kg \cdot 0.600 m/s \).
03

Calculate the initial velocity of the bullet

Now we can solve the simplified equation from step 2, and get the initial speed of the bullet. Divide both sides by \(0.01 kg\), to give: \(v_{bullet_{initial}} = \frac{5.01 kg \cdot 0.600 m/s}{0.01 kg}\) which gives the initial velocity of the bullet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together after impact. In these collisions, the total kinetic energy is not conserved, but the total momentum of the system is. This is a crucial difference from elastic collisions, where both energy and momentum are conserved. In our example, a bullet embeds itself into a block of wood, and together they move as one unit after the collision.

Understanding inelastic collisions involves recognizing that while energy might be lost to factors like sound or heat, momentum remains constant. Momentum conservation enables us to calculate various pre-collision parameters when post-collision metrics are known.
  • The bullet and wood move together after the collision, sharing the final velocity.
  • Kinetic energy transforms or dissipates but does not influence the momentum conservation equations.
Momentum
Momentum is a property of a moving object and is calculated as the product of its mass and velocity. It is a vector quantity, which means it has both magnitude and direction. In a closed system (no external forces acting), momentum is always conserved.

The conservation of momentum can describe many physical events, like the collision described in our example. Here, because the wood block was stationary prior to the collision, its initial momentum was zero. This simplifies our calculations because the only initial momentum comes from the bullet.
  • Calculate momentum by multiplying mass and velocity: \( p = m \times v \).
  • In this example, use the conservation law: the momentum before the collision equals the momentum after.
  • The momentum prior to collision equals the combined momentum of bullet and wood post-collision.
Velocity Calculation
Velocity calculation following a collision involves several steps. To find the initial velocity of the bullet in our exercise, we must rearrange and solve the conservation of momentum formula.

We start by setting up the formula: \[ m_{bullet} \cdot v_{bullet_{initial}} = (m_{bullet} + m_{wood}) \cdot v_{final} \] This equation reflects the fact that all initial momentum is carried by the bullet, and all system momentum afterward is shared by the bullet and wood block together. To solve for the initial bullet velocity, rearrange the equation by dividing both sides by the bullet's mass:\[ v_{bullet_{initial}} = \frac{(m_{bullet} + m_{wood}) \cdot v_{final}}{m_{bullet}} \] Plug the known values into the formula to compute the bullet's initial velocity, keeping units consistent for accuracy.
  • Ensure all units are the same before calculating (e.g., convert grams to kilograms).
  • Perform arithmetic carefully to isolate and solve for the unknown variable.
  • The solution gives insight into how fast the bullet moved initially, before impacting the wood. This is direct application of momentum conservation.

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Most popular questions from this chapter

A ball of mass \(0.200 \mathrm{kg}\) has a velocity of \(150 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} ;\) a ball of mass \(0.300 \mathrm{kg}\) has a velocity of \(-0.400 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} .\) They meet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.

Two blocks of masses \(M\) and \(3 M\) are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig. P9.4). A cord initially holding the blocks together is burned; after this, the block of mass \(3 M\) moves to the right with a speed of \(2.00 \mathrm{m} / \mathrm{s}\) (a) What is the speed of the block of mass \(M ?\) (b) Find the original elastic potential energy in the spring if \(M=0.350 \mathrm{kg}\)

Consider as a system the Sun with the Earth in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system over a period of 6 months. Neglect the influence of other celestial objects. You may obtain the necessary astronomical data from the end papers of the book.

An archer shoots an arrow toward a target that is sliding toward her with a speed of \(2.50 \mathrm{m} / \mathrm{s}\) on a smooth, slippery surface. The \(22.5-\mathrm{g}\) arrow is shot with a speed of \(35.0 \mathrm{m} / \mathrm{s}\) and passes through the 300 -g target, which is stopped by the impact. What is the speed of the arrow after passing through the target?

The first stage of a Saturn \(V\) space vehicle consumed fuel and oxidizer at the rate of \(1.50 \times 10^{4} \mathrm{kg} / \mathrm{s}\), with an exhaust speed of \(2.60 \times 10^{3} \mathrm{m} / \mathrm{s} .\) (a) Calculate the thrust produced by these engines. (b) Find the acceleration of the vehicle just as it lifted off the launch pad on the Earth if the vehicle's initial mass was \(3.00 \times 10^{6} \mathrm{kg} .\) Note: You must include the gravitational force to solve part (b).

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