/*! 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 31 An ice hockey forward with mass ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An ice hockey forward with mass \(70.0 \mathrm{~kg}\) is skating due north with a speed of \(5.5 \mathrm{~m} / \mathrm{s}\). As the forward approaches the net for a slap shot, a defensive player (muss \(110 \mathrm{~kg}\) ) skates toward him in order to apply a body-check. The defensive player is traveling south at \(4.0 \mathrm{~m} / \mathrm{s}\) just before they collide. If the two players become intertwined and move together after they collide, in what direction and at what speed do they move after the collision? Friction between the two players and the ice can be neglected.

Short Answer

Expert verified
To get the short answer, solve the equation from step 3 for \(V\). The value of \(V\) gives the speed of the two players after the collision and its sign indicates the direction of motion. For the final answer, state the magnitude and the direction of \(V\).

Step by step solution

01

Understand Conservation of Momentum

The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. This can be represented by the formula \(m_1v_1 + m_2v_2 = (m_1 + m_2)V \), where m is mass, v is velocity and V is final velocity after collision. In this exercise, the two hockey players form the isolated system. Their total momentum before the collision is the sum of their individual momenta. After the collision, they become intertwined and move together as a single mass, thus their total momentum remains the same but is now represented as the single, combined mass times the final velocity \(V\).
02

Establish the Directions

We need to establish a direction sign convention before proceeding with the calculations. In this exercise, motion due north can be taken as positive and due south as negative. This means that the forward's velocity \(v_1\) is +5.5 m/s while the defensive player's velocity \(v_2\) is -4.0 m/s.
03

Apply the Conservation of Momentum

Substitute the masses and velocities into the conservation of momentum equation. Let's denote the forward's mass as \(m_1\) = 70.0 kg and the defensive player's mass as \(m_2\) = 110.0 kg. After substituting, we get the equation \(70.0*5.5 + 110.0*(-4.0) = (70.0+110.0) * V \) . This equation can be solved to find \(V\), the speed of both players after the collision.
04

Solve for the Final Velocity

By solving the equation in step 3, we get the value of \(V\). The sign of \(V\) will determine the direction of motion. If \(V\) turns out to be positive, then the two players move due north after the collision. If \(V\) is negative, then the two players move due south after the collision. Let's solve for \(V\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Collisions in Physics
Collisions occur when two or more objects come into contact and exert forces on each other. In physics, understanding collisions involves examining how these objects interact and change motion as a result. Collisions can be classified in various ways:
  • Elastic collisions: Both kinetic energy and momentum are conserved. The objects bounce off each other without losing energy to deformation or heat.
  • Inelastic collisions: Momentum is conserved, but kinetic energy is not. Typically, objects may stick together post-collision, as in the exercise scenario with the two hockey players.
Each type of collision follows the principle of conservation of momentum. This exercise focuses on an inelastic collision, where two hockey players collide and move together as one mass. This helps to explore how momentum is shared and transformed within a system after impact. Understanding these principles helps in predicting outcomes in a wide range of practical scenarios, from sports to vehicle safety.
Momentum
Momentum is a fundamental concept in physics that describes the quantity of motion an object possesses. It is a vector quantity, meaning it has both magnitude and direction, and is calculated as the product of an object's mass and velocity, given by the formula:\[ p = m imes v \]Here, \( p \) represents momentum, \( m \) is mass, and \( v \) is velocity. Momentum plays a crucial role in understanding how objects interact with each other, especially during collisions.In the presented exercise, each hockey player has a distinct momentum based on their mass and velocity:
  • The forward has a mass of 70.0 kg and is moving north at 5.5 m/s, giving him a momentum of \( p_1 = 70.0 \, \text{kg} \times 5.5 \, \text{m/s} \).
  • The defensive player, with a mass of 110 kg, moves south at 4.0 m/s, resulting in a momentum of \( p_2 = 110.0 \, \text{kg} \times (-4.0) \, \text{m/s} \).
Since they move together post-collision, understanding their initial momenta helps to determine their shared momentum and resultant velocity, considering all vector directions.
Principle of Conservation
The principle of conservation is a fundamental law in physics, which states that certain properties of isolated systems remain constant over time. One of the most significant laws under this principle is the conservation of momentum.In the context of the exercise, the principle of conservation of momentum can be understood as:- The total momentum before the collision must equal the total momentum after the collision, provided no external forces act on the system.Formally, it is expressed as:\[ m_1v_1 + m_2v_2 = (m_1 + m_2) \times V \]Where:
  • \( m_1, v_1 \): Mass and velocity of the first player (the forward).
  • \( m_2, v_2 \): Mass and velocity of the second player (the defensive player).
  • \( V \): The final velocity of both players moving together.
The exercise demonstrates this principle by calculating the final shared velocity after the collision while maintaining the system's total momentum. By substituting known values into the momentum equation, the result provides insight into both the speed and direction of the post-collision movement, showcasing how momentum is conserved and redistributed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two sticky spheres are suspended from light ropes of length \(L\) that are attached to the ceiling at a common point. Sphere \(A\) has mass \(2 m\) and is hanging at rest with its rope vertical. Sphere \(B\) has mass \(m\) and is held so that its rope makes an angle with the vertical that puts \(B\) a vertical height \(H\) above \(A\). Sphere \(B\) is released from rest and swings down, collides with sphere \(A,\) and sticks to it. In terms of \(H,\) what is the maximum height above the original position of \(A\) reached by the combined spheres after their collision?

\(A 12.0 \mathrm{~kg}\) shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of \(150 \mathrm{~m} / \mathrm{s}\). At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was relcased in the explosion?

When cars are equipped with flexible bumpers, they will bounce off cach other during low-specd collisions, thus causing less damage, In one such accident, a \(1750 \mathrm{~kg}\) car traveling to the right at \(1.50 \mathrm{~m} / \mathrm{s}\) collides with a \(1450 \mathrm{~kg}\) car going to the left at \(1.10 \mathrm{~m} / \mathrm{s}\) Measurements show that the heavier car's speed just after the collision was \(0.250 \mathrm{~m} / \mathrm{s}\) in its original direction. Ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic eneryy of the two-car system during this collision.

Two carts of cqual mass are on a horizontal, frictionless air track. Initially cart \(A\) is moving toward stationary cart \(B\) with a speed of \(v_{A}\). The carts undergo an inelastic collision, and after the collision the total kinetic energy of the two carts is one-half their initial total kinctic energy before the collision. What is the speed of cach cart after the collision?

You (mass 55kg) are riding a frictionless skateboard (mass 5.0 kg) in a straight line at a speed of \(4.5 \mathrm{~m} / \mathrm{s}\). \(\Lambda\) friend standing on a balcuny ahove you drops a \(2.5 \mathrm{~kg}\) sack of flont straight down into your arms. (a) What is your new speed while you hold the suck? (b) since the sack was dropped vertically, how can it affect your horizontal thotion? Explain. (c) Now you try to rid yourscif of the extra weight hy throwing the sack straight up. What will be your speed while the sack is in the air? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.