/*! 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 \(2 \mathrm{~kg}\) block is mo... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2 \mathrm{~kg}\) block is moving at a speed of \(10 \mathrm{~m} / \mathrm{s}\) and makes a perfectly elastic collision with a second block of mass \(M\), which is initially at rest. After the collision, the \(2 \mathrm{~kg}\) block bounces straight back at \(3 \mathrm{~m} / \mathrm{s}\). (a) Determine the mass \(M\) of the second block. (b) Determine the speed of the second block after the collision.

Short Answer

Expert verified
The mass of the second block, \(M\), is 4.2 kg. The speed of the second block after the collision is \(5.1 \text{ m/s}\).

Step by step solution

01

Analyze Conservation of Momentum

In a perfectly elastic collision, the total momentum before the collision is equal to the total momentum after the collision. The initial momentum of the system is the momentum of the moving block, since the second block is initially at rest. The formula for momentum is given by the equation \( p = mv \). Initially, we have:\[p_{ ext{initial}} = (2 \, ext{kg})(10 \, ext{m/s}) = 20 \, ext{kg m/s}.\]
02

Set Up the Equation for Conservation of Momentum

After the collision, the first block (2 kg) has reversed direction with speed 3 m/s, and the second block has some final speed \( v_2 \). Using conservation of momentum:\[20 \, ext{kg m/s} = (2 \, ext{kg})(-3 \, ext{m/s}) + (M \, ext{kg})(v_2) \]Simplifying gives:\[20 = -6 + Mv_2\]
03

Analyze Conservation of Kinetic Energy

In a perfectly elastic collision, kinetic energy is also conserved. Therefore, the total kinetic energy before the collision equals the total kinetic energy after the collision. The initial kinetic energy is:\[KE_{ ext{initial}} = \frac{1}{2} (2 \, ext{kg})(10 \, ext{m/s})^2 = 100 \, ext{J}.\]
04

Set Up the Equation for Conservation of Kinetic Energy

After the collision, the kinetic energies are as follows:For the 2 kg block: \[KE_{1, ext{final}} = \frac{1}{2}(2 \, ext{kg})(3 \, ext{m/s})^2 = 9 \, ext{J}.\]For the second block:\[KE_{2, ext{final}} = \frac{1}{2} M v_2^2.\]Using conservation of kinetic energy:\[100 = 9 + \frac{1}{2} M v_2^2\]
05

Solve Simultaneous Equations for M and v2

We have two equations:1. \(20 = -6 + Mv_2\) 2. \(91 = \frac{1}{2} M v_2^2\) From equation 1:\[M v_2 = 26\]Substitute \(M v_2 = 26\) into equation 2:\[91 = \frac{1}{2}(26v_2/ M) \cdot v_2\]Rearrange to find \(v_2\):\[v_2^2 = 7 \, ext{m/s}\]Find \(v_2\):\[v_2 = 5.1 \, ext{m/s}\]Substitute back to find \(M\):\[91 = \frac{1}{2} M (5.1)^2\]\[M=4.2\, ext{kg}\]
06

Verify Solution

Check both the momentum and energy conservation to ensure calculations are correct. Substitute values \(M = 4.2 \, ext{kg}\) and \(v_2 = 5.1 \, ext{m/s}\) back into the original equations to confirm both momentum and kinetic energy equations are satisfied.

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Key Concepts

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

Conservation of Momentum
The conservation of momentum principle states that in a closed system, the total momentum before an event (like a collision) is equal to the total momentum after the event. Momentum itself is a product of mass and velocity, given by the equation \( p = mv \). In the context of our exercise, a moving block collides with another block at rest.To solve such a collision problem, you first calculate the total initial momentum, which here is only due to the moving block: \(p_{\text{initial}} = (2 \, \text{kg})(10 \, \text{m/s}) = 20 \, \text{kg m/s}\). Post-collision, the momentum of the two-block system must still equal this initial value. Balancing the equations helps us determine the unknowns, like the velocity and mass of the second block.
Conservation of Kinetic Energy
In a perfectly elastic collision, not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy is given by the formula \( KE = \frac{1}{2}mv^2 \). Before the collision, only the first block has kinetic energy, calculated as \( KE_{\text{initial}} = \frac{1}{2} (2 \, \text{kg})(10 \, \text{m/s})^2 = 100 \, \text{J}\).After the collision, both blocks have kinetic energy. For the 2 kg block, you recompute its kinetic energy with its new velocity: \( KE_{1,\text{final}} = \frac{1}{2}(2 \, \text{kg})(3 \, \text{m/s})^2 = 9 \, \text{J}\). Conservation equations allow us to find the kinetic energy of the second block, helping us solve for unknown variables like the mass or velocity.
Collision Problems
Collision problems in physics require understanding and applying the laws of conservation of momentum and kinetic energy. In this exercise, a 2 kg block collides elastically with a stationary block, allowing us to use these laws to find unknowns like the mass and velocity of the second block. These types of problems can be tricky, often needing simultaneous equations for an accurate solution. By translating the physical scenario into mathematical equations, even a potentially confusing setup becomes manageable. Remember to always check your solution by putting the calculated values back into the original equations, ensuring both momentum and energy were indeed conserved.
Physics Problem Solving
Approaching physics problems systematically can greatly enhance understanding and accuracy. For collision problems, like the one we are tackling, start by clearly defining what you know and what you need to find.
  • Identify the conservation laws applicable (momentum and kinetic energy for elastic collisions).
  • Write down the initial conditions, such as masses and velocities.
  • Set up equations representing conservation laws.
  • Solve these equations, often simultaneously, for the unknowns.
With practice, this approach becomes a natural part of physics problem solving, allowing you to tackle increasingly complex scenarios efficiently.

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Most popular questions from this chapter

Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the water's surface to knock them into the water, where the fish can eat them. \(\mathrm{A} 65 \mathrm{~g}\) fish at rest just on the water's surface can expel a \(0.30 \mathrm{~g}\) drop of water in a short burst of \(5.0 \mathrm{~ms}\). High-speed measurements show that the water has a speed of \(2.5 \mathrm{~m} / \mathrm{s}\) just after the archerfish expels it. A fish shoots a drop of water at an insect that hovers on the water's surface. Just before colliding with the insect, the drop is still moving at the speed it had when it left the fish's mouth. In the collision, the drop sticks to the insect, and the speed of the insect and water just after the collision is measured to be \(2.0 \mathrm{~m} / \mathrm{s}\). What is the insect's mass? A. \(0.038 \mathrm{~g}\) B. \(0.075 \mathrm{~g}\) C. \(0.24 \mathrm{~g}\) D. \(0.38 \mathrm{~g}\)

A 4.25 g bullet traveling horizontally with a velocity of magnitude \(375 \mathrm{~m} / \mathrm{s}\) is fired into a wooden block with mass \(1.12 \mathrm{~kg},\) initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to \(122 \mathrm{~m} / \mathrm{s} .\) How fast is the block moving just after the bullet emerges from it?

A \(15.0 \mathrm{~kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (See Figure \(8.40 .\) ) Suddenly it is struck by a \(3.00 \mathrm{~kg}\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts the collision and the behavior after the collision -and apply the appropriate conservation law to each part.)

A \(750 \mathrm{~kg}\) car is stalled on an icy road during a snowstorm. A \(1000 \mathrm{~kg}\) car traveling eastbound at \(10 \mathrm{~m} / \mathrm{s}\) collides with the rear of the stalled car. After being hit, the \(750 \mathrm{~kg}\) car slides on the ice at \(4 \mathrm{~m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What are the magnitude and direction of the velocity of the \(1000 \mathrm{~kg}\) car after the collision? (b) Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision. (You may ignore the effects of friction during the collision.)

A \(2 \mathrm{~kg}\) block is moving at \(5 \mathrm{~m} / \mathrm{s}\) along a frictionless table and collides with a second \(2 \mathrm{~kg}\) block that is initially at rest. After the collision, the two blocks stick together and then slide up a \(45^{\circ}\) frictionless inclined plane, as shown in Figure \(8.41 .\) Calculate the maximum distance \(L\) that the two blocks travel up the incline.

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