/*! 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 127 A block of mass \(0.50 \mathrm{~... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 127

A block of mass \(0.50 \mathrm{~kg}\) is moving with a speed of \(2.00 \mathrm{~ms}^{-1}\) on a smooth surface. It strikes another mass of \(1.00 \mathrm{~kg}\) and then they move together as a single body. The energy loss during the collision is (A) \(0.16 \mathrm{~J}\) (B) \(1.00 \mathrm{~J}\) (C) \(0.67 \mathrm{~J}\) (D) \(0.34 \mathrm{~J}\)

Short Answer

Expert verified
The initial kinetic energy is \(K_{initial}=\frac{1}{2}(0.50)(2.00)^2=1.00 \mathrm{~J}\). Using conservation of momentum, the final velocity is \(v_f=\frac{(0.50)(2.00)}{(0.50 + 1.00)}=0.67 \mathrm{~ms}^{-1}\). The final kinetic energy is \(K_{final}=\frac{1}{2}(1.50)(0.67)^2=0.34 \mathrm{~J}\). The energy loss during the collision is 螖E = K_initial - K_final = 1.00 J - 0.34 J = \(0.66 \mathrm{~J}\) approximately, which is closest to option (C) $0.67 \mathrm{~J}$.

Step by step solution

01

Find the initial kinetic energy

Before the collision occurs, the first block has a mass of 0.50 kg and an initial velocity of 2.00 ms鈦宦�. The second block is initially stationary, so its velocity is 0 ms鈦宦�. We can calculate the kinetic energy of each block using the following formula: \[K=\frac{1}{2}mv^2\] Then, we will add the individual kinetic energies to find the total initial kinetic energy.
02

Apply conservation of momentum

To find the final velocity of the two blocks after the collision, we will use the conservation of linear momentum. For a collision in one dimension, the conservation of momentum states that: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v_f\) where \(m_1\) and \(m_2\) are the masses of the blocks, \(u_1\) and \(u_2\) are their initial velocities (u鈧� being 0, as mentioned), and \(v_f\) is the final velocity of the combined blocks after the collision. Notice that \(u_2\)=0 ms鈦宦�, since the second block is initially stationary. Solve for \(v_f\).
03

Calculate the final kinetic energy

Once we find the final velocity, we can calculate the kinetic energy of the combined blocks after the collision using the mass and final velocity. Since the blocks are now moving together, their combined mass is the sum of their individual masses, and we can find the final kinetic energy as: \[K_f=\frac{1}{2}(m_1 + m_2)v_f^2\]
04

Determine the energy loss

Now that we have both the initial and final kinetic energies, we can find the energy loss during the collision. The energy loss will be the difference between the initial kinetic energy and the final kinetic energy: 螖E = K_initial - K_final Calculate the energy loss and compare it to the given options to find the correct answer.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is an essential concept in physics because it helps us understand how objects move and interact. To calculate the kinetic energy
  • The formula used is: \(K = \frac{1}{2} mv^2\).
  • \(m\) is the mass of the object, and \(v\) is its velocity.
For example, in the exercise, the initial kinetic energy is calculated for a block with mass 0.50 kg moving at a speed of 2.00 m/s. Using the formula, we substitute the values:
  • \(K = \frac{1}{2} \times 0.50 \times (2.00)^2 = 1.00 \text{ J}\).
Understanding the kinetic energy helps in determining how much work is required to bring the object to a stop, or conversely, how much effort is needed to get it moving again.
Collision Mechanics
Collisions are interactions between two or more bodies that results in an exchange of energy and momentum. The mechanics of collisions is all about understanding these exchanges. A fundamental tool in collision mechanics is the conservation of linear momentum.When two objects collide, their combined momentum before the collision equals their combined momentum after the collision. This concept is crucial in solving the exercise. The momentum before the collision is calculated by multiplying the masses of the blocks by their velocities:
  • Initial momentum = \( m_1 u_1 + m_2 u_2 \)
  • For our problem: \(0.50 \times 2.00 + 1.00 \times 0 = 1.00 \text{ kg m/s}\)
After collision, the blocks move as one body, hence:
  • \((m_1 + m_2) v_f = 1.00\)
  • The new speed \(v_f\) can be calculated using this equation.
The principle of momentum conservation allows us to predict the final state of the colliding bodies.
Energy Loss
In real-world collisions, some energy is usually lost due to factors like sound, heat, or deformation. This loss of kinetic energy can be calculated once we know the initial and final kinetic energies.
  • Initial kinetic energy = energy of the moving block before collision.
  • Final kinetic energy = energy of the combined mass moving together.
The energy loss \(\Delta E\) is found by:
  • \(\Delta E = K_{\text{initial}} - K_{\text{final}}\)
It's important because it tells us how much energy is "lost" during the interaction. For instance, in the exercise, the calculated initial and final kinetic energies allow us to find the energy difference, which reveals the energy dissipated during the process. This understanding helps explain why, after a collision, the objects do not continue moving with the same level of energy as before.

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Most popular questions from this chapter

One end of a light spring of spring constant \(k\) is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is \(\frac{1}{2} k x^{2}\). The possible cases are: (A) The spring was initially compressed by a distance \(x\) and was finally in its natural length. (B) It was initially stretched by a distance \(x\) and finally was in its natural length. (C) It was initially in its natural length and finally in a compressed position. (D) It was initially in its natural length and finally in a stretched position.

When a body moves in a circle, the work done by the centripetal force is always \((\mathrm{A})>0\) (B) \(<0\) (C) Zero (D) None of these

A locomotive of mass \(m\) starts moving so that its velocity varies as \(v=\alpha s^{2 / 3}\), where \(\alpha\) is a constant and \(s\) is the distance traversed. The total work done by all the forces acting on the locomotive during the first \(t\) second after the start of motion is (A) \(\frac{1}{8} m \alpha^{4} t^{2}\) (B) \(\frac{m \alpha^{6} t^{4}}{162}\) (C) \(\frac{m \alpha^{6} t^{4}}{81}\) (D) \(\frac{m \alpha^{4} t^{2}}{2}\)

The relationship between force and position is shown in Fig. \(4.19\) (in one- dimensional case). The work done by the force in displacing a body from \(x=1 \mathrm{~cm}\) to \(x=\) \(5 \mathrm{~cm}\) is (A) 20 ergs (B) \(60 \mathrm{ergs}\) (C) 70 ergs (D) 700 ergs

A block of mass \(2 \mathrm{~kg}\) is hanging over a smooth and light pulley through a light string. The other end of string is pulled by a constant force \(F\). The kinetic energy of block increases by \(16 \mathrm{~J}\) in \(2 \mathrm{~s}\), then (A) force \(F\) may be \(24 \mathrm{~N}\). (B) force \(F\) must be \(24 \mathrm{~N}\). (C) potential energy must be increase. (D) potential energy may be increase.
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