/*! 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 33 A \(2.0\) -kg block of wood rest... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2.0\) -kg block of wood rests on a tabletop. A \(7.0\) -g bullet is shot straight up through a hole in the table beneath the block. The bullet lodges in the block, and the block flies \(25 \mathrm{~cm}\) above the tabletop. How fast was the bullet going initially?

Short Answer

Expert verified
The initial speed of the bullet was approximately 636 m/s.

Step by step solution

01

Define the Problem

The problem involves a collision between the bullet and the block of wood. We need to find the initial speed of the bullet before it lodges into the block and causes it to move upwards. The bullet and block system goes up to a height of 25 cm after collision.
02

Apply Conservation of Momentum

Use the conservation of linear momentum for the collision. Before the collision, only the bullet has momentum. After the collision, the bullet and the block move together. Given that the mass of the bullet (m1) is 0.007 kg and the mass of the block (m2) is 2.0 kg, the conservation of momentum can be written as: \[ m_1 u + m_2 imes 0 = (m_1 + m_2) v \] where \(u\) is the initial velocity of the bullet and \(v\) is the velocity of the bullet-block system just after the collision.
03

Use Energy Conservation for the Upward Motion

After the collision, the bullet-block system moves upwards and comes to a stop at its highest point. We use conservation of energy to relate the velocity after the collision to the height achieved.The initial kinetic energy is converted into gravitational potential energy at the highest point. \[ \frac{1}{2} (m_1 + m_2) v^2 = (m_1 + m_2) g h \]where \(h = 0.25\) m is the height, and \(g = 9.8\) m/s² is the acceleration due to gravity.
04

Solve for the Initial Velocity of the Bullet

First, solve for \(v\) from the energy conservation equation:\[ v^2 = 2gh \]\[ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 0.25} \]\[ v \approx 2.214 m/s \]Next, substitute \(v\) into the momentum conservation equation:\[ 0.007u = (2.007)v \]\[ u = \frac{(2.007) imes v}{0.007} \]\[ u = \frac{2.007 \times 2.214}{0.007} \]\[ u \approx 635.97 m/s \]
05

Conclusion

The initial speed of the bullet that resulted in the block rising to the height of 25 cm is approximately 636 m/s.

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Key Concepts

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

Collision
In physics, a collision is an event where two or more bodies exert forces on each other for a relatively short time. Collisions can be broadly categorized into elastic and inelastic types.

In elastic collisions, both momentum and kinetic energy are conserved. However, in inelastic collisions – like our exercise situation where a bullet hits and becomes lodged in a block of wood – only momentum is conserved.

During a collision, such as the one described in the problem, it's essential to consider the conservation of linear momentum. This means that the total momentum of the system (the bullet and the block) just before the collision equals the total momentum immediately afterward.

This principle allows us to connect the speeds of the objects before and after their interaction, making it a powerful tool for solving many physics problems involving collisions.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It is given by the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where \( m \) is the mass and \( v \) is the velocity of the object.

In the exercise described, kinetic energy is a crucial factor both before and after the collision. Initially, the bullet has kinetic energy owing to its high velocity, whereas the block is stationary and thus has no kinetic energy.

Post-collision, the combined bullet and block system possesses kinetic energy, which propels it upwards. However, as it rises to the maximum height, the system’s kinetic energy converts entirely to potential energy.

Understanding how kinetic energy functions in this context helps us solve how high the block and bullet system can rise based on their speed after the collision.
Potential Energy
Potential energy refers to the stored energy in an object due to its position relative to some zero position. The gravitational potential energy, specifically for vertical motion, is given by:
  • \( PE = mgh \)
where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.

In this physics problem, after the bullet lodges in the block, the system moves upwards. As it reaches its peak height of 0.25 meters, all of the system’s kinetic energy converts into potential energy.

This conversion is crucial because it allows us to calculate the velocity of the block-bullet system immediately after the collision by using the potential energy at the highest point. Potential energy and kinetic energy interplay provide a pathway to understand variable energies in the system during different phases of the problem.
Physics Problems
Physics problems often require analyzing complex systems to understand how different physical laws apply. These problems can include a variety of elements like forces, momentum, energy, and motion.

Approaching a physics problem typically involves:
  • Identifying the knowns and unknowns
  • Choosing the right principles and equations, such as conservation of energy or momentum.
  • Applying these principles correctly to solve for the unknowns
This exercise demonstrates how breaking down a problem through steps involving conservation principles makes tackling seemingly complex scenarios manageable.

Understanding fundamental concepts, like those related to collisions, kinetic energy, and potential energy, becomes crucial. Consistently applying these principles allows for solutions to a wide range of physics problems involving such interactions and energy transformations.

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Most popular questions from this chapter

A 16 -g mass is moving in the \(+x\) -direction at \(30 \mathrm{~cm} / \mathrm{s}\), while a 4.0-g mass is moving in the \(-x\) -direction at \(50 \mathrm{~cm} / \mathrm{s}\). They collide head on and stick together. Find their velocity after the collision. Assume negligible friction. This is a completely inelastic collision for which \(\mathrm{KE}\) is not conserved, although momentum is. Let the \(16-\mathrm{g}\) mass be \(m_{1}\) and the \(4.0\) -g mass be \(m_{2}\). Take the \(+x\) -direction to be positive. That means that the velocity of the 4.0-g mass has a scalar value of \(v_{2 x}=-50 \mathrm{~cm} / \mathrm{s}\). We apply the law of conservation of momentum to the system consisting of the two masses: Momentum before impact = Momentum after impact $$ \begin{aligned} m_{1} v_{1 x}+m_{2} v_{2 x} &=\left(m_{1}+m_{2}\right) v_{x} \\ (0.016 \mathrm{~kg})(0.30 \mathrm{~m} / \mathrm{s})+(0.0040 \mathrm{~kg})(-0.50 \mathrm{~m} / \mathrm{s}) &=(0.020 \mathrm{~kg}) v_{x} \\ v_{x} &=+0.14 \mathrm{~m} / \mathrm{s} \end{aligned} $$ (Notice that the 4.0-g mass has negative momentum.) Hence, \(\overrightarrow{\mathbf{v}}=0.14 \mathrm{~m} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION

A 6000 -kg truck traveling north at \(5.0 \mathrm{~m} / \mathrm{s}\) collides with a \(4000-\mathrm{kg}\) truck moving west at \(15 \mathrm{~m} / \mathrm{s}\). If the two trucks remain locked together after impact, with what speed and in what direction do they move immediately after the collision?

Two identical railroad cars sit on a horizontal track, with a distance \(D\) between their two centers of mass. By means of a cable between them, a winch on one is used to pull the two together. (a) Describe their relative motion. ( \(b\) ) Repeat the analysis if the mass of one car is now three times that of the other. Keep in mind that the velocity of the center of mass of a system can only be changed by an external force. Here the forces due to the cable acting on the two cars are internal to the two-car system. The net external force on the system is zero, and so its center of mass does not move, even though each car travels toward the other. Taking the origin of coordinates at the center of mass, $$ x_{\mathrm{cm}}=0=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} $$ where \(x_{1}\) and \(x_{2}\) are the positions of the centers of mass of the two cars. (a) If \(m_{1}=m_{2}\), this equation becomes $$ 0=\frac{x_{1}+x_{2}}{2} \quad \text { or } \quad x_{1}=-x_{2} $$ The two cars approach the center of mass, which is originally midway between the two cars (that is, \(D / 2\) from each), in such a way that their centers of mass are always equidistant from it. (b) If \(m_{1}=3 m_{2}\), then we have $$ 0=\frac{3 m_{2} x_{1}+m_{2} x_{2}}{3 m_{2}+m_{2}}=\frac{3 x_{1}+x_{2}}{4} $$ from which \(x_{1}=-x_{2} / 3\). Since \(m_{1}>m_{2}\), it must be that \(x_{1}

(a) What minimum thrust must the engines of a \(2.0 \times 10^{5} \mathrm{~kg}\) rocket have if the rocket is to be able to slowly rise from the Earth when aimed straight upward? \((b)\) If the engines eject gas at the rate of \(20 \mathrm{~kg} / \mathrm{s}\), how fast must the gaseous exhaust be moving as it leaves the engines? Neglect the small change in the mass of the rocket due to the ejected fuel. [Hint: Study Problem 8.20.]

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10 -g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, (c) the coefficient of restitution is \(0.90\).

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