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Chapter 7: Problem 55

To view an interactive solution to a problem that is similar to this one, go to and select Interactive Solution \(\underline{7.55}\) A \(0.015-\mathrm{kg}\) bullet is fired straight up at a falling wooden block that has a mass of \(1.8 \mathrm{~kg}\). The bullet has a speed of \(810 \mathrm{~m} / \mathrm{s}\) when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time \(t\) when the collision with the bullet occurred. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time \(t\).

Short Answer

Expert verified
The block fell for approximately 0.689 seconds before the collision.

Step by step solution

01

Define the Conservation of Momentum

The principle of conservation of momentum applies to the collision. Before the collision, the bullet and block have their respective momenta, and the total momentum in the system must be conserved. Let the velocity of the block before the collision be denoted as \(v_{block}\). The initial momentum is \(m_{bullet} \cdot v_{bullet} + m_{block} \cdot v_{block}\). The final momentum (after collision) is \((m_{bullet} + m_{block}) \cdot v_{final}\). Because the block and bullet come to a stop at the top of the building, \(v_{final} = 0\).
02

Set Up Momentum Equation

Apply the formula for conservation of momentum: \[ (m_{bullet} \cdot v_{bullet} + m_{block} \cdot v_{block}) = (m_{bullet} + m_{block}) \cdot v_{final} \]Since \(v_{final} = 0\), simplify the equation to:\[ m_{bullet} \cdot v_{bullet} + m_{block} \cdot v_{block} = 0 \].This gives \(v_{block} = -\frac{m_{bullet} \cdot v_{bullet}}{m_{block}}\).
03

Calculate the Block's Velocity Before Collision

Substitute the known values into the equation from Step 2: \[ v_{block} = -\frac{0.015 \, \text{kg} \times 810 \, \text{m/s}}{1.8 \, \text{kg}} \]This evaluates to \( v_{block} = -6.75 \, \text{m/s} \). The negative sign indicates the block is moving downward.
04

Use Free Fall to Find Falling Time

The block was falling freely under gravity before the collision. Use the equation for free fall: \[ v = u + g t \] where \( u = 0 \) (initial velocity), \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity), and \( v = v_{block} = -6.75 \, \text{m/s} \). Solve for \( t \):\[ -6.75 = 0 + 9.8 t \]\[ t = \frac{-6.75}{9.8} \approx 0.689 \, \text{s} \].
05

Conclusion

The time \(t\) it took for the block to fall to the point of collision is approximately \(0.689 \, \text{seconds}\).

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Key Concepts

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

Physics Problems
Physics problems often challenge our understanding by involving multiple concepts at once. The given exercise is a perfect example, where the principles of momentum and free fall are applied to solve a collision scenario. When handling such problems:
  • Identify all known variables and the ones you need to find.
  • Understand the physical principles governing the problem, like momentum conservation in collisions here.
  • Break down the problem into smaller, manageable parts. For instance, separately consider the momentum before and after collision.
  • Use mathematical equations to link the concepts together, deriving unknown quantities from known ones.
By systematically approaching each part, seemingly complex physics problems can be simplified and solved.
Collision Mechanics
Collision mechanics plays a vital role in understanding how objects interact. In this exercise, we focus on the conservation of momentum during a collision. When two objects collide, the total momentum before the impact is equal to the total momentum after, assuming no external forces act on the system.
  • Before the collision, calculate the momentum of each object (mass multiplied by velocity).
  • Apply the principle of conservation of momentum to set the total initial momentum equal to the total final momentum.
  • Here, because the block and bullet come to a stop after reaching the building's top, their final velocity is zero.
  • This situation results in a momentum equation:\[ m_{bullet} \cdot v_{bullet} + m_{block} \cdot v_{block} = 0 \]from which we find the block's velocity just before collision.
Understanding these steps helps in analyzing and solving collision-related problems in physics.
Free Fall Motion
Free fall motion describes how objects move under the influence of gravity alone. In this problem, the wooden block experiences free fall before the collision. Knowing the block's velocity (\[v_{block}\]) when hit by the bullet helps determine how long it had been falling.
  • Use the free fall equation:\[ v = u + g t \]where \(u\) is the initial velocity (0 for free fall), \(g\) is the acceleration due to gravity (9.8 m/s虏), and \(v\) is the final velocity.
  • Substituting the given values shows\[ -6.75 = 0 + 9.8 t \]
  • Solving for \(t\) gives the result:\[ t \approx 0.689 \, \text{s} \]
  • This is the time elapsed from when the block started falling until the collision.
Through this, we see how motion equations describe real-world physics phenomena like free-fall.

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Most popular questions from this chapter

Kevin has a mass of \(87 \mathrm{~kg}\) and is skating with in-line skates. He sees his \(22-\mathrm{kg}\) younger brother up ahead standing on the sidewalk, with his back turned. Coming up from behind, he grabs his brother and rolls off at a speed of \(2.4 \mathrm{~m} / \mathrm{s}\). Ignoring friction, find Kevin's speed just before he grabbed his brother.

Multiple-Concept Example 7 outlines the general approach to problems like this one. Concept Simulation 7.2 at provides a view of this elastic collision. Two identical balls are traveling toward each other with velocities of -4.0 and \(+7.0 \mathrm{~m} / \mathrm{s},\) and they experience an elastic head-on collision. Obtain the velocities (magnitude and direction) of each ball after the collision.

ssm A golf ball bounces down a flight of steel stairs, striking several steps on the way down, but never hitting the edge of a step. The ball starts at the top step with a vertical velocity component of zero. If all the collisions with the stairs are elastic, and if the vertical height of the staircase is \(3.00 \mathrm{~m}\), determine the bounce height when the ball reaches the bottom of the stairs. Neglect air resistance.

A \(55-\mathrm{kg}\) swimmer is standing on a stationary \(210-\mathrm{kg}\).oating raft. The swimmer then runs off the raft horizontally with a velocity of \(+4.6 \mathrm{~m} / \mathrm{s}\) relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.

A \(0.500-\mathrm{kg}\) ball is dropped from rest at a point \(1.20 \mathrm{~m}\) above the floor. The ball rebounds straight upward to a height of \(0.700 \mathrm{~m}\). What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?
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