/*! 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 50 A \(5.20 \mathrm{~g}\) bullet mo... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 50

A \(5.20 \mathrm{~g}\) bullet moving at \(700 \mathrm{~m} / \mathrm{s}\) strikes a \(700 \mathrm{~g}\) wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to \(450 \mathrm{~m} / \mathrm{s}\). (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Short Answer

Expert verified
(a) The speed of the block is 1.86 m/s. (b) The center of mass speed is 5.16 m/s.

Step by step solution

01

Identify Given Information

We are given the mass of the bullet as \( m_1 = 5.20 \text{ g} = 0.0052 \text{ kg} \), the initial speed of the bullet as \( v_{1i} = 700 \text{ m/s} \), and its final speed after collision as \( v_{1f} = 450 \text{ m/s} \). The mass of the wooden block is \( m_2 = 700 \text{ g} = 0.7 \text{ kg} \), and its initial speed is \( v_{2i} = 0 \text{ m/s} \). We need to find the final speed of the block \( v_{2f} \) and the speed of the bullet-block center of mass.
02

Apply Conservation of Momentum to Find Block's Speed

According to the conservation of momentum, the total momentum before the collision equals the total momentum after the collision. The equation is \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \). Substituting the known values, we get:\[ 0.0052 \times 700 + 0.7 \times 0 = 0.0052 \times 450 + 0.7 \times v_{2f} \]Solving for \( v_{2f} \),\[ 3.64 = 2.34 + 0.7v_{2f} \]\[ 1.30 = 0.7v_{2f} \]\[ v_{2f} = \frac{1.30}{0.7} = 1.857 \text{ m/s} \]The resulting speed of the block is \( 1.86 \text{ m/s} \) when approximated to two decimal places.
03

Calculate Speed of Center of Mass

The speed of the center of mass is calculated using the formula:\[ v_{cm} = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2} \]Inserting the given values:\[ v_{cm} = \frac{0.0052 \times 700 + 0.7 \times 0}{0.0052 + 0.7} = \frac{3.64}{0.7052} \approx 5.16 \text{ m/s} \]Thus, the speed of the bullet-block center of mass is approximately \( 5.16 \text{ m/s} \).

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Key Concepts

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

Bullet and Block Collision
When a bullet hits a stationary block, interesting physics happens. The bullet's motion gives energy to the block, making the block move. In this scenario, the bullet had an initial speed of 700 m/s and, after the collision, it slowed down to 450 m/s. This change in speed is the bullet transferring some of its momentum to the block. The block, which was still at the start, absorbs this momentum and begins to move.

This kind of interaction exemplifies an inelastic collision. An inelastic collision is when two objects collide and do not return to their original shape and size. In most cases, they may even share their motion energy. Note that conservation of momentum still applies dearly, which will be discussed soon. It is crucial to understand that this elegant dance between motion and rest makes it possible to calculate resulting speeds after such collisions.
Center of Mass Velocity
Center of mass is a key concept when studying collision problems. It refers to the point where you can think all mass of a system is concentrated. For the bullet and block, regardless of surrounding activities, their center of mass behaves in predictable ways. Calculating its velocity helps us understand how the two objects move as a single unit.
The velocity of the center of mass is determined from combining the momentum contributions of both objects and dividing by their total mass. In our example, it is calculated using the formula:\[ v_{cm} = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2} \]The value 5.16 m/s was determined. This means if you had a single point representing both the bullet and the block's mass, it would travel at this speed.

The center of mass offers insight into the system's behavior, showing how the collective mass responds to forces like impacts from collisions.
Momentum Calculations
Momentum is a measure of motion of an object, calculated as the product of its mass and velocity. When analyzing collisions, it is vital to understand how momentum before and after the event compares.
First, consider the initial momentum of our bullet and block system: the block initially rests, so its momentum is zero. The only motion contribution comes from the bullet, traveling at a fast speed of 700 m/s. As the collision happens, momentum must remain constant, a principle known as conservation of momentum. This principle is represented by:\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]In short, the sum of the momentum before collision equals the sum of momentum after. After inserting known values and solving, the block's speed was identified as 1.86 m/s.

Momentum conservation holds true in absence of external forces like friction. This helps physicists predict and comprehend how objects behave in closed interactions like our bullet and block collision.

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Most popular questions from this chapter

A \(3.00 \mathrm{~kg}\) block slides on a frictionless horizontal surface, first moving to the left at \(50.0 \mathrm{~m} / \mathrm{s}\). It collides with a spring whose other end is fixed to a wall, compresses the spring, and is brought to rest momentarily. Then it continues to be accelerated toward the right by the force of the compressed spring. The block acquires a final speed of \(40.0 \mathrm{~m} / \mathrm{s}\). It is in contact with the spring for \(0.020 \mathrm{~s}\). Find (a) the magnitude and (b) the direction of the impulse of the spring force on the block. (c) What is the magnitude of the spring's average force on the block?

Block 1 of mass \(3.0 \mathrm{~kg}\) is sliding across a floor with speed \(v_{1}=2.0 \mathrm{~m} / \mathrm{s}\) when it makes a head-on, one-dimensional, elastic collision with initially stationary block 2 of mass \(2.0 \mathrm{~kg}\). The coefficient of kinetic friction between the blocks and the floor is \(\mu_{k}=\) \(0.30\). Find the speeds of (a) block 1 and (b) block 2 just after the collision. Also find (c) their final separation after friction has stopped them and \((\mathrm{d})\) the energy lost to thermal energy because of the friction.

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

Figure 9-28 shows a cubical box that has been constructed from uniform metal plate of negligible thickness. The box is open at the top and has edge length \(L=50 \mathrm{~cm}\). Find (a) the \(x\) coordinate, (b) the \(y\) coordinate, and (c) the \(z\) coordinate of the center of mass of the box.

A \(6090 \mathrm{~kg}\) space probe moving nose-first toward Jupiter at \(120 \mathrm{~m} / \mathrm{s}\) relative to the Sun fires its rocket engine, ejecting \(70.0 \mathrm{~kg}\) of exhaust at a speed of \(253 \mathrm{~m} / \mathrm{s}\) relative to the space probe. What is the final velocity of the probe?
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