/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 A 0.090-kg bullet is Fired verti... [FREE SOLUTION] | 91Ó°ÊÓ

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A 0.090-kg bullet is Fired vertically at \(200 \mathrm{~m} / \mathrm{s}\) into a \(0.15-\mathrm{kg}\) baseball that is initially at rest. How high does the combined buller and baseball rise after the collision, assuming the bullet embeds itself in the ball?

Short Answer

Expert verified
To find the height at which the bullet and baseball combo stops ascending, we first calculate the velocity right after the collision using the conservation of momentum principle. Then, we substitute this velocity into a kinematic equation along with the acceleration due to gravity to find the time till ascent, and the maximum height reached.

Step by step solution

01

Conservation of Momentum

In the vertical collision, the total initial linear momentum equals to the total final linear momentum according to the conservation of momentum. This gives us: \[\mbox{Initial momentum} = \mbox{Final momentum}\] or \[(mass_bullet * velocity_bullet) + (mass_baseball * velocity_baseball) = (mass_bullet + mass_baseball) * v_f\] where \(v_f\) is the final velocity.
02

Calculate the Final Velocity

Now, plug the given quantities into the equation. That gives us: \[(0.090 kg * 200 m/s) + (0.15 kg * 0) = (0.090 kg + 0.15 kg) * v_f\] Solving this equation gives us the final velocity, \(v_f\).
03

Using the Kinematic Equation

The motion of the bullet-baseball combo can be modeled using the kinematic equation to calculate the maximum height. The kinematic equation is: \[h = vf*t - 0.5* g*t^2\] Here, the time \(t\) when the combination comes to rest is the time it takes for the initial velocity to be reduced to zero due to gravity. Setting \(v_f\) equal to zero we have: \[t = v_i/g\] where \(v_i\) is the initial velocity of the bullet-baseball combo (Which we found in step 2), and \(g\) is acceleration due to gravity.
04

Calculate Time and Substitute Back Into the Kinematic Equation

Substitute \(v_i\) from step 2 into the equation for time. Now, by substituting this \(t\) and \(v_i\) back into the kinematic equation, you can calculate the height \(h\).

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Key Concepts

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

Kinematic Equations
Kinematic equations describe the motion of objects by relating the various components such as velocity, acceleration, time, and displacement. They are very handy for solving problems involving linear motion under constant acceleration, such as in the case of a body moving upwards against gravity after a collision. Knowing how to apply these equations can greatly simplify complex motion problems.

For the problem of the bullet and baseball, after the collision, we used kinematic equations to find out how high the combined object rose. The specific equation applied was:- \[ h = v_f \cdot t - 0.5 \cdot g \cdot t^2 \]- Here, \(h\) is the maximum height, \(v_f\) is the final velocity right after the collision, \(g\) is the acceleration due to gravity (approximately \(9.8 \text{ m/s}^2\)), and \(t\) is the time until the object comes to rest momentarily before falling back down.

These equations help predict how far an object will move over time when subjected to a steady force, like gravity, allowing us to calculate its maximum height after jumping or being propelled upwards. Understanding kinematics is crucial for many physics problems involving projectiles and other moving objects.
Collisions
Collisions occur when two or more objects come into contact with each other, exchanging momentum and potentially altering their velocities. There are different types of collisions—elastic, inelastic, and perfectly inelastic—with the latter type best describing the scenario where a bullet embeds itself in a baseball. In a perfectly inelastic collision like this, the objects stick together post-impact, conserving momentum but not kinetic energy. To solve our problem, we apply the principle of conservation of momentum. The total momentum before the collision must equal the total momentum afterward:- \[ (mass_{bullet} \cdot velocity_{bullet}) + (mass_{baseball} \cdot velocity_{baseball}) = (mass_{bullet} + mass_{baseball}) \cdot v_f \]- As the baseball was initially at rest, its velocity is zero. We solve this equation to find \(v_f\), the final velocity of the bullet-plus-baseball system.Collisions are a staple of classical mechanics and provide insight into how energy and momentum transfer between objects during impact.
Projectile Motion
Projectile motion involves the motion of an object that is subject to gravity, allowing it to follow a curved trajectory. In the world of physics, understanding projectile motion is vital for analyzing any object launched, thrown, or otherwise propelled into the air. Our combined bullet and baseball, after the collision, behaves like a projectile moving upwards until gravity gradually slows it to a stop right before descending back down. Key features to understand in projectile motion include:
  • The initial velocity: just after the collision, we computed this to be the starting speed of the projectile.
  • The time of flight: this is the duration the projectile stays in the air.
  • The maximum height: this is determined using kinematic equations, describing how high the projectile ascends before arcing back to the ground.
For any projectile, air resistance is often neglected for simplification unless otherwise specified. By solving the motion equations considering these parameters, we get a comprehensive understanding of how high the projectile will rise, as in the baseball colliding with the bullet at a high speed.

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Most popular questions from this chapter

A rifle with a weight of \(30 \mathrm{~N}\) fires a \(5.0\)-g bullet with a speed of \(300 \mathrm{~m} / \mathrm{s}\). (a) Find the recoil speed of the rifle. (b) If a \(700-\mathrm{N}\) man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle.

A \(45.0-\mathrm{kg}\) girl is standing on a \(150-\mathrm{kg}\) plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of \(1.50 \mathrm{~m} / \mathrm{s}\) to the right relative to the plank. (a) What is her velocity relative to the surface of the ice? (b) What is the velocity of the plank relative to the surface of the ice?

A billiard ball mowing at \(5.00 \mathrm{~m} / \mathrm{s}\) strikes a stationary ball of the same mass. After the collision, the first ball moves at \(4.39 \mathrm{~m} / \mathrm{s}\) at an angle \(0 \mathrm{f} 30^{\circ}\) with respect to the original line of motion. (a) Find the velocity (magnitude and direction) of the second ball after collision. (b) Was the collision inelastic or elastic?

A \(6.5 .0-\mathrm{kg}\) person throws a \(0.0450-\mathrm{kg}\) snowball forward with a ground speed of \(30.0 \mathrm{~m} / \mathrm{s} .\) A second person, with a mass of \(60.0 \mathrm{~kg}\), catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of \(2.50 \mathrm{~m} / \mathrm{s}\), and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

BIO A \(75.0-\mathrm{kg}\) ice skater mowing at \(10.0 \mathrm{~m} / \mathrm{s}\) crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at \(5.00 \mathrm{~m} / \mathrm{s}\). Suppose the average force a skater can experience withour breaking a bone is \(4500 \mathrm{~N}\). If the impact time is \(0.100 \mathrm{~s}\), does a bone break?

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