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Chapter 8: Problem 72

A small wooden block with mass \(0.800 \mathrm{~kg}\) is suspended from the lower end of a light cord that is \(1.60 \mathrm{~m}\) long. The block is initially at rest. A bullet with mass \(12.0 \mathrm{~g}\) is fired at the block with a horizontal velocity \(c_{0}\) - The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of \(0.800 \mathrm{~m}\), the tension in the cord is \(4.80 \mathrm{~N}\). What was the initial speed \(t_{0}\) of the bullct?

Short Answer

Expert verified
The initial speed of bullet is \(270 \mathrm{~m/s}\).

Step by step solution

01

Finding the total mass

Find the total mass \(m_{total}\) of the block and the bullet system after the collision. Convert the bullet's mass from grams to kilograms and add it to the block's mass: \(m_{total} = 0.800 \mathrm{~kg} + 12.0 \mathrm{~g} = 0.812 \mathrm{~kg}\).
02

Calculating the speed of the pendulum after collision

Find the speed \(v_{after}\) of the combined block and bullet system after the collision using the principle of conservation of energy, where the potential energy at the highest point \(m_{total} \cdot g \cdot h\) equals the kinetic energy of the pendulum at the bottom, \(\frac{1}{2} \cdot m_{total} \cdot v_{after}^{2}\). Solving for \(v_{after}\) gives, \( v_{after} = \sqrt{2 \cdot g \cdot h} = \sqrt{2 \cdot 9.8 \mathrm{~m/s^2} \cdot 0.800 \mathrm{m}} = 4.0 \mathrm{~m/s}\).
03

Applying momentum conservation

Apply the principle of conservation of momentum. Initial momentum before the bullet strikes the block is the product of the bullet鈥檚 mass and initial speed \(m_{bullet} \cdot v_0\). After the collision, the total momentum is \(m_{total} \cdot v_{after}\). Thus, \(m_{bullet} \cdot v_0 = m_{total} \cdot v_{after}\). Solving for \(v_0\) gives, \(v_0 = \frac{m_{total} \cdot v_{after}}{m_{bullet}} = \frac{0.812 \mathrm{~kg} \cdot 4.0 \mathrm{~m/s}}{12.0 \mathrm{~g}} = 270 \mathrm{~m/s}\).

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Key Concepts

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

Kinetic and Potential Energy
Understanding the interplay between kinetic and potential energy is essential to solving problems like the one with the wooden block and bullet. Kinetic energy is the energy an object possesses due to its motion. It can be calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \(m\) is the mass of the object and \(v\) is its velocity.

Potential energy, on the other hand, is stored energy based on an object's position or configuration. Gravitational potential energy, represented as \( PE = mgh \), depends on the mass of the object (\(m\)), the acceleration due to gravity (\(g\)), and the height (\(h\)) above a reference point.

In the provided exercise, the bullet鈥檚 kinetic energy just before collision transforms into both kinetic energy of the combined system (bullet plus wooden block) and potential energy as the block-bullet system rises to its highest point post-collision.
Collisions in Physics
Collisions in physics embody the principle of momentum conservation. The total momentum of a closed system remains constant before and after a collision.

In an inelastic collision, such as the one described between the bullet and the wooden block, the objects stick together after impact. The initial momentum, coming exclusively from the bullet, is transferred to the combined mass system of both the block and the bullet. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, which can be mathematically expressed as \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_{after} \), where subscript 1 refers to the bullet and subscript 2 to the block.

Calculating the speed of the bullet involves using this principle. The initial speed \(v_0\) of the bullet is found by equating the momentum of the bullet before the collision to the momentum of the combined block and bullet system afterward.
Pendulum Motion
Pendulum motion plays a crucial role in this problem, as the block swings up to a certain height after the collision. A pendulum's motion is governed by gravitational forces, which convert potential energy back and forth into kinetic energy. When the block with the bullet embedded reaches its highest point, its kinetic energy is temporarily zero and all the energy in the system is potential.

The highest point reached by the pendulum corresponds to the maximum gravitational potential energy, which can be equated to the kinetic energy of the system just after the collision. This allows us to calculate the velocity of the block-bullet system post-collision, and ultimately reveal the initial velocity of the bullet by applying momentum conservation principles.

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Most popular questions from this chapter

In a shipping company distribution center, an open cart of } mass \(50.0 \mathrm{~kg}\) is rolling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) (lig. \(\mathrm{P} 8.87\) ). Ignore friction between the cart and the floor. \(A 15.0 \mathrm{~kg}\) package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of \(4.00 \mathrm{~m}\) above the bottom of the cart, what are (a) the speed of the packagc just bcfore it lands in the cart and (b) the final speed of the cart?

A block with mass \(0.500 \mathrm{~kg}\) sits at rest on a light but not long vertical spring that has spring constant \(80.0 \mathrm{~N} / \mathrm{m}\) and one end on the floor. (a) How much clastic potential cnergy is stored in the spring when the block is sitting at rest on it? (b) A second identical block is dropped onto the first from a height of \(4.00 \mathrm{~m}\) above the first block and sticks to it. What is the maximum elastic potential energy stored in the spring during the motion of the blocks after the collision? (c) What is the maximum distance the first block moves down after the second block has landed on it?

Starting at \(t=0\) a net external force \(F(t)\) in the \(+x\) -direction is applied to an object that has mass \(3.00 \mathrm{~kg}\) and is initially at rest. The force is zero at \(t=0\) and increascs lincarly to \(5.00 \mathrm{~N}\) at \(t=4.00 \mathrm{~s}\) The force then decreases lincarly until it becomes zero at \(t=10.0 \mathrm{~s}\) (a) Draw a graph of \(F\) versus \(t\) from \(t=0\) to \(t=10.0 \mathrm{~s}\). (b) What is the spced of the object at \(t=10.0 \mathrm{~s} ?\)

A rifle bullet with mass \(8.00 \mathrm{~g}\) strikes and embeds itself in a block with mass \(0.992 \mathrm{~kg}\) that rests on a frictionless, horizontal surface and is attached to an ideal spring (Fig. \(\mathrm{P} 8.79\) ). The impact compresses the spring \(15.0 \mathrm{~cm} .\) Calibration of the spring shows that a force of 0.750 \(\mathrm{N}\) is required to compress the spring \(0.250 \mathrm{~cm}\). (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullct?

Animal Propulsion. Squids and octopuses propel themsclves by expelling water, They do this by kecping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A \(6.50 \mathrm{~kg}\) squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has \(1.75 \mathrm{~kg}\) of watcr in its cavity, at what speed must it expel this water suddenly to achieve a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) to cscape the predator? Ignore any drag cffects of the surrounding water. (b) How much kinetic energy does the squid create by this mancuver?
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