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\(\bullet\) Combining conservation laws. \(A 5.00 \mathrm{kg}\) chunk of ice is sliding at 12.0 \(\mathrm{m} / \mathrm{s}\) on the floor of an ice-covered valley when it collides with and sticks to another 5.00 \(\mathrm{kg}\) chunk of ice that is initially at rest. (See Figure \(8.37 . )\) since the valley is icy, there is no friction. After the collision, how high above the valley floor will the combined chunks go? (Hint: Break this problem into two parts - the collision and the behavior after the collision and apply the appropriate conservation law to each part.)

Short Answer

Expert verified
The combined chunks will reach a height of approximately 1.84 meters above the valley floor.

Step by step solution

01

Analyze the Collision

In this step, we're focusing on the collision itself and will use the law of conservation of momentum, since there is no external force. The formula for conservation of linear momentum is \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \] where \( m_1 = 5.00 \, \text{kg} \) and \( v_1 = 12.0 \, \text{m/s} \) (the initial speed of the moving chunk), and \( m_2 = 5.00 \, \text{kg} \) and \( v_2 = 0 \, \text{m/s} \) (since the second chunk is initially at rest). Plug in the values to find the final velocity \( v_f \) after collision.
02

Calculate Final Velocity After Collision

Using the conservation of momentum formula: \[ (5.00 \, \text{kg} \times 12.0 \, \text{m/s}) + (5.00 \, \text{kg} \times 0 \, \text{m/s}) = (5.00 \, \text{kg} + 5.00 \, \text{kg}) v_f \] Simplifying, we find: \[ 60.0 = 10.0 \, v_f \] \[ v_f = 6.0 \, \text{m/s} \].
03

Analyze the Movement Post-Collision

Now, consider the movement of the combined mass after the collision. Use the principle of conservation of energy. All kinetic energy will convert to potential energy at the maximum height. The formula is \( KE = PE \), where \[ \frac{1}{2} m v_f^2 = m g h \]. Here, \( m \) is the total mass (10 kg), \( v_f = 6.0 \, \text{m/s} \), and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
04

Solve for Maximum Height

Starting with the equation \[ \frac{1}{2} \times 10.0 \, \text{kg} \times (6.0 \, \text{m/s})^2 = 10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times h \], solve for height \( h \). This simplifies to: \[ 180.0 = 98h \], and by rearranging, \[ h = \frac{180.0}{98} \approx 1.84 \, \text{m} \].

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Key Concepts

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

Conservation of Energy
Energy cannot be created or destroyed. This fundamental concept in physics is known as the conservation of energy. In a closed system, the total amount of energy remains constant, even if it changes forms. This principle is crucial when analyzing physical systems.
In the context of collision problems, kinetic energy, which deals with motion, can transform into potential energy, which is related to position in a gravitational field. After the chunks of ice collide and stick together, their combined kinetic energy is converted into potential energy as they travel upward.
  • Kinetic Energy (KE): Depends on mass and velocity. Formula: \( KE = \frac{1}{2} mv^2 \)
  • Potential Energy (PE): Depends on mass, gravity, and height. Formula: \( PE = mgh \)
Understanding how energy transitions between kinetic and potential forms helps solve how high the ice chunks will rise post-collision.
Collision Problems
Collision problems often require using the conservation of momentum. Momentum, like energy, is conserved in a closed system without external forces. When two objects collide, their total momentum before the collision is equal to their total momentum after.The formula for momentum is straightforward: \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In our ice collision problem, both chunks become a single mass post-collision. By calculating the combined mass's velocity using the initial momentum, we break down complex interactions into solvable equations.Real-life applications of these problems include simulations of vehicle crashes and predicting the outcomes of physical interactions.
Physics Problem Solving
Tackling physics problems often involves strategic thinking and methodical application of laws. Breaking a problem into manageable parts, as we did with the ice chunks, simplifies the process.
  • Step 1: Identify Known and Unknown Quantities
    Start with what's given and determine what needs to be found. Here, the initial and final velocities are key values.
  • Step 2: Apply Fundamental Laws Appropriately
    Conservation laws guide problem-solving, such as momentum for collisions and energy for height calculations.
  • Step 3: Solve Mathematically
    Translate physical principles into mathematical expressions.
  • Step 4: Interpret Results
    After finding a solution, ensure it makes sense in the context of the problem.
Using an organized approach helps ensure no steps are missed and enhances understanding of the underlying physics.
Kinetic and Potential Energy Conversion
The interplay between kinetic and potential energy is essential in understanding motion in physics. When the ice chunks are moving just after the collision, they possess kinetic energy due to their velocity. As they move upward, this kinetic energy gradually converts into potential energy. Kinetic energy conversion becomes clear once we apply the principle that energy can only change forms, not disappear. As the ice chunks climb to the maximum height, their velocity decreases until it reaches zero. At this point, all kinetic energy has transformed into potential energy, at the peak of their ascent. This conversion is seamlessly reversible; if allowed to fall back, potential energy would again transform into kinetic energy, fully recovering the initial motion. Understanding these conversions is key to predicting the outcomes of not just collisions, but any system where energy transitions play a role.

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Most popular questions from this chapter

Forensic science. Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure 8.49.) The meas ured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet into a 2.50 kg block hanging by a thin wire 75.0 \(\mathrm{cm}\) long. causing the block to swing upward to a maximum angle of \(34.7^{\circ}\) from the vertical. What was the original speed of this bullet?

\(\bullet\) Some useful relationships. The following relationships between the momentum and kinetic energy of an object can be very useful for calculations: If an object of mass \(m\) has momentum of magnitude \(p\) and kinetic energy \(K,\) show that (a) \(K=\left(p^{2} / 2 m\right),\) and \((b) p=\sqrt{2 m K}\) . (c) Find the momen- tum of a 1.15 kg ball that has 15.0 J of kinetic energy. (d) Find the kinetic energy of a 3.50 kg cat that has 0.220 \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) of momentum.

. Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0 \(\mathrm{ms}\) contact time throughout this problem. The fastest-known served tennis ball was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be 73.14 \(\mathrm{m} / \mathrm{s}\) .(a) What impulse and what force did Big Bill exert on the ten- nis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of \(55 \mathrm{m} / \mathrm{s},\) what force and what impulse did he exert on the ball, assuming only horizontal motion?

Two identical 1.50 \(\mathrm{kg}\) masses are pressed against opposite ends of a light spring of force constant \(1.75 \mathrm{N} / \mathrm{cm},\) compress- ing the spring by 20.0 \(\mathrm{cm}\) from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless horizontal lab table.

\(\bullet\) On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a frozen lake. With your finger, you flick the penny toward the nickel with a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) . The coins collide elastically; calculate both their final velocities (speed and direction).

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