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Chapter 8: Problem 96

A blue puck with mass \(0.0400 \mathrm{kg},\) sliding with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass \(m,\) initially at rest. After the collision, the velocity of the blue puck is 0.050 \(\mathrm{m} / \mathrm{s}\) in the same direction as its initial velocity. Find (a) the velocity (magnitude and direction) of the red puck after the collision and (b) the mass \(m\) of the red puck.

Short Answer

Expert verified
Red puck's velocity: 2.64 m/s; mass: 0.00227 kg.

Step by step solution

01

Understand the Problem

The exercise involves a collision between two pucks on a frictionless surface where momentum and energy are conserved, as it is a perfectly elastic collision. The initial conditions for the blue puck are given, and we need to find the post-collision velocity of the red puck and its mass.
02

Write down the conservation of momentum equation

For the system, use conservation of momentum: \( m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} \). Here, \(m_1 = 0.0400 \text{ kg}, v_{1i} = 0.200 \text{ m/s}\) (initial blue puck velocity), \(v_{2i} = 0\) (red puck initially at rest), \(v_{1f} = 0.050 \text{ m/s}\) (blue puck final velocity). Substitute these into the equation to get \(0.0400 \times 0.200 = 0.0400 \times 0.050 + m \cdot v_{2f}\).
03

Solve for final velocity of red puck

Rearrange the equation from Step 2 to solve for \(v_{2f}\): \(v_{2f} = \frac{0.0400 \times 0.150}{m} = \frac{0.0060}{m}\). This gives the expression for the final velocity of the red puck.
04

Write down the conservation of kinetic energy equation

In an elastic collision, kinetic energy is conserved: \( \frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \). Substitute known values \( \frac{1}{2} \times 0.0400 \times 0.200^2 = \frac{1}{2} \times 0.0400 \times 0.050^2 + \frac{1}{2} \times m \times \left(\frac{0.0060}{m}\right)^2 \). Simplify to \(0.0080 - 0.0001 = \frac{0.0060^2}{2m}\).
05

Solve for mass of the red puck

Rearrange the equation from Step 4 to solve for \(m\): \( 0.0079 = \frac{0.000036}{2m} \), which simplifies to \( m = \frac{0.000036}{2 \times 0.0079} = 0.00227 \text{ kg}\).
06

Substitute mass to find the velocity of red puck

Now that we have the mass \(m = 0.00227\text{ kg}\), substitute it back into the equation for \(v_{2f}\): \(v_{2f} = \frac{0.0060}{0.00227} \approx 2.643\text{ m/s}\). This is the velocity of the red puck post-collision in the same direction as the initial velocity of the blue puck.

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Key Concepts

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

Conservation of Momentum
Momentum is like the fairy godmother of physics, ensuring that all is balanced in motion. In a frictionless and closed system, such as the collision scenario described, the total momentum before the collision equals the total momentum after the collision.

This concept is crucial for solving problems related to elastic collisions. Momentum, defined as the product of mass and velocity, is symbolized by the equation: \( p = mv \). During our puck collision, the principle can be expressed using the formula:
  • Initial Momentum = Final Momentum
  • \( m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} \)
Here, \( m_1 \) and \( m_2 \) represent the masses of the blue and red pucks, respectively. The given velocities represent the initial and final states of the pucks.

By plugging in the known values, you can solve for unknown variables such as the final velocity of the red puck after the collision, ensuring the balance of momentum remains intact.
Conservation of Kinetic Energy
Kinetic energy is the energy of motion, and in elastic collisions like ours, it is carefully preserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The equation for kinetic energy (KE) is expressed as:
  • \( KE = \frac{1}{2} mv^2 \)
For a collision between two objects, the conservation of kinetic energy is described by:
  • \( \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \)
In our exercise, this principle helps us determine unknowns like the mass of the red puck.
After expressing the equivalence and substituting known values, you can solve for missing quantities such as the mass \( m \) of the red puck. This equation accounts for every watt of energy, illustrating how it's conserved in elastic collisions.
Head-On Collision
A head-on collision occurs when two objects move directly towards each other along the same line. In our discussion, imagine two hockey pucks on a perfectly smooth ice rink.

Such a collision involves two objects colliding along a single, straight path. These collisions are fundamental to understanding momentum and energy distribution.
Understanding head-on collisions:
  • The total momentum of the system is conserved.
  • Kinetic energy remains constant through the collision.
  • The direction(s) of the velocities after the collision are determined by initial conditions. In perfectly elastic collisions, objects may bounce back or one might gain speed depending on mass and initial velocity profiles.
For the exercise at hand, the head-on nature simplifies complexity because all movements are linear and direct, making calculations straightforward based on known principles of momentum and energy conservation.

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Most popular questions from this chapter

Just before it is struck by a racket, a tennis ball weighing 0.560 \(\mathrm{N}\) has a velocity of \((20.0 \mathrm{m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{m} / \mathrm{s}) \hat{\boldsymbol{J}}\) . During the 3.00 \(\mathrm{ms}\) that the racket and ball are in contact, the net force on the ball is constant and equal to \(-(380 \mathrm{N}) \hat{\boldsymbol{\imath}}+(110 \mathrm{N}) \hat{\boldsymbol{J}}\) . (a) What are the \(x\) - and \(y\) -components of the impulse of the net force applied to the ball? (b) What are the \(x\) - and \(y\) -components of the final velocity of the ball?

A C6-5 model rocket engine has an impulse of 10.0 \(\mathrm{N} \cdot \mathrm{s}\) while burning 0.0125 \(\mathrm{kg}\) of propellant in 1.70 s. It has a maximum thrust of 13.3 \(\mathrm{N} .\) The initial mass of the engine plus propellant is 0.0258 \(\mathrm{kg} .\) (a) What fraction of the maximum thrust is the average thrust? (b) Calculate the relative speed of the exhaust gases, assuming it is constant. (c) Assuming that the relative speed of the exhaust gases is constant, find the final speed of the engine if it was attached to a very light frame and fired from rest in gravity-free outer space.

Automobile Accident Analysis. You are called as an expert witness to analyze the following auto accident: Car \(B,\) of mass \(1900 \mathrm{kg},\) was stopped at a red light when it was hit from behind by car \(A,\) of mass 1500 \(\mathrm{kg}\) . The cars locked bumpers during the collision and slid to a stop with brakes locked on all wheels. Measurements of the skid marks left by the tires showed them to be 7.15 \(\mathrm{m}\) long. The coefficient of kinetic friction between the tires and the road was 0.65 . (a) What was the speed of car A just before the collision? (b) If the speed limit was 35 \(\mathrm{mph}\) , was car \(A\) speeding, and if so, by how many miles per hour was it exceeding the speed limit?

Energy Sharing. An object with mass \(m,\) initially at rest, explodes into two fragments, one with mass \(m_{A}\) and the other with mass \(m_{B},\) where \(m_{A}+m_{B}=m .\) (a) If energy \(Q\) is released in the explosion, how much kinetic energy does each fragment have immediately after the explosion? (b) What percentage of the total energy released does each fragment get when one fragment has four times the mass of the other?

At a classic auto show, a \(840-\mathrm{kg} 1955\) Nash Metropolitan motors by at 9.0 \(\mathrm{m} / \mathrm{s}\) , followed by a \(1620-\mathrm{kg} 1957\) Packard Clipper purring past at 5.0 \(\mathrm{m} / \mathrm{s}\) . (a) Which car has the greater kinetic energy? What is the ratio of the kinetic energy of the Nash to that of the Packard? (b) Which car has the greater magnitude of momentum? What is the ratio of the magnitude of momentum of the Nash to that of the Packard? (c) Let \(F_{\mathrm{N}}\) be the net force required to stop the Nash in time \(t,\) and let \(F_{\mathrm{P}}\) be the net force required to stop the Packard in the same time. Which is larger: \(F_{\mathrm{N}}\) or \(F_{\mathrm{P}} ?\) What is the ratio \(F_{\mathrm{N}} / F_{\mathrm{P}}\) of these two forces? (d) Now let \(F_{\mathrm{N}}\) be the net force required to stop the Nash in a distance \(d,\) and let \(F_{\mathrm{P}}\) be the net force required to stop the Packard in the same distance. Which is larger: \(F_{\mathrm{N}}\) or \(F_{\mathrm{P}} ?\) What is the ratio \(F_{\mathrm{N}} / F_{\mathrm{P}}\)?
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