Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

During an ice-skating extravaganza, Robin Hood on Ice, a 50.0 -kg archer is standing still on ice skates. Assume that the friction between the ice skates and the ice is negligible. The archer shoots a \(0.100-\mathrm{kg}\) arrow horizontally at a speed of \(95.0 \mathrm{~m} / \mathrm{s}\). At what speed does the archer recoil?

Short Answer

Expert verified
Answer: The recoil speed of the archer is approximately -0.19 m/s, moving in the opposite direction to which the arrow was shot.

Step by step solution

01

Write down the given quantities.

We're given the following information: - Mass of archer (M1) = 50.0 kg - Mass of arrow (M2) = 0.100 kg - Initial speed of archer (U1) = 0 m/s (he's initially standing still) - Initial speed of arrow (U2) = 0 m/s (the arrow is not moving before being shot) - Final speed of arrow (V2) = 95.0 m/s (the speed at which the arrow is shot) - Final speed of archer (V1) = ? (this is what we need to find)
02

Apply the law of conservation of momentum.

The momentum before the arrow is shot must be equal to the momentum after the arrow is shot. Therefore, we can write the equation as: \( M1*U1 + M2*U2 = M1*V1 + M2*V2 \) Since \(U1\) and \(U2\) are both 0 (both archer and arrow are initially at rest), the equation becomes: \( M1*0 + M2*0 = M1*V1 + M2*V2 \)
03

Solve for the final speed of the archer (V1).

Now, we need to solve for \(V1\). We already have the values for \(M1\), \(M2\), and \(V2\). Plug those values into the equation: \( 50.0*0 + 0.100*0 = 50.0*V1 + 0.100*95.0 \) Simplify the equation: \( 0 = 50.0*V1 + 9.5 \) Now, we solve for \(V1\): \( V1 = \frac{-9.5}{50.0} \) \( V1 = -0.19 \mathrm{~m} / \mathrm{s} \)
04

Interpret the result.

The final speed of the archer is approximately -0.19 m/s. The negative sign indicates that the archer is recoiling in the opposite direction to which the arrow was shot. This makes sense, as we would expect the archer to move backward after shooting the arrow due to the conservation of momentum.

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 Vaia!

Key Concepts

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

Momentum in Collisions
Momentum is a key concept to understand when studying collisions in physics. It is defined as the product of an object's mass and its velocity, represented by the equation: \( p = mv \), where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity. During a collision or any interactive event between two objects, such as an archer shooting an arrow on ice, momentum plays a critical role in determining the outcomes of the motion involved.

When two objects interact, their combined momentum before the interaction must equal the combined momentum afterward if no external forces are acting. This fact is crucial for solving problems regarding collisions, like the one involving Robin Hood on ice, ensuring that you take into account the masses and velocities of all participating objects.
Law of Conservation of Momentum
The law of conservation of momentum states that, in a closed system where no external forces are present, the total momentum remains constant. This principle is a cornerstone in physics, especially in mechanics. The equation \( M1 * U1 + M2 * U2 = M1 * V1 + M2 * V2 \) depicts this law, where \( U \) and \( V \) denote initial and final velocities of objects with masses \( M1 \) and \( M2 \) respectively.

Understanding this law is essential for predicting the results of collision events. It allows us to calculate unknown variables such as the recoil velocity of an archer who has just shot an arrow. As seen in the Robin Hood example, prior to releasing the arrow, both entities' momenta are zero since they are at rest. However, once the arrow is released, the archer must recoil to conserve the system's total momentum.
Recoil Velocity
Recoil velocity is the speed at which an object moves backward after it exerts a force on another object. According to Newton's third law of motion, every action has an equal and opposite reaction. When an archer shoots an arrow, the force applied on the arrow also results in a reactive force on the archer, causing recoil.

This recoil velocity can be calculated using the conservation of momentum principle, where the initial momentum (before firing the arrow) is equal to the final momentum (after firing). As shown in the textbook exercise, after firing an arrow, the archer's recoil velocity is calculated to be negative, indicating the direction is opposite to the arrow's motion. This calculation is indispensable for predicting how an event will unfold in a frictionless environment.
Physics Problem Solving
Problem solving in physics often involves a systematic approach to applying laws and principles to find a solution. For the Robin Hood on ice problem, we follow specific steps to reach a conclusion: identifying known values, employing the relevant physics law (in this case, the conservation of momentum), manipulation of equations to solve for the unknowns, and finally, interpreting the results.

The process highlights the importance of understanding both the qualitative and quantitative aspects of the physical concepts. Moreover, developing a strong foundation in these problem-solving techniques is beneficial not only in academics but also in real-world scenarios that involve similar principles.

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

An alpha particle (mass \(=4.00 \mathrm{u}\) ) has a head-on, elastic collision with a nucleus (mass \(=166 \mathrm{u}\) ) that is initially at rest. What percentage of the kinetic energy of the alpha particle is transferred to the nucleus in the collision?

Assume the nucleus of a radon atom, \({ }^{222} \mathrm{Rn}\), has a mass of \(3.68 \cdot 10^{-25} \mathrm{~kg}\). This radioactive nucleus decays by emitting an alpha particle with an energy of \(8.79 \cdot 10^{-13} \mathrm{~J}\). The mass of an alpha particle is \(6.64 \cdot 10^{-27} \mathrm{~kg}\). Assuming that the radon nucleus was initially at rest, what is the velocity of the nucleus that remains after the decay?

In the movie Superman, Lois Lane falls from a building and is caught by the diving superhero. Assuming that Lois, with a mass of \(50.0 \mathrm{~kg},\) is falling at a terminal velocity of \(60.0 \mathrm{~m} / \mathrm{s}\), how much average force is exerted on her if it takes \(0.100 \mathrm{~s}\) to slow her to a stop? If Lois can withstand a maximum acceleration of \(7.00 g^{\prime}\) s, what minimum time should it take Superman to stop her after he begins to slow her down?

For a totally elastic collision between two objects, which of the following statements is (are) true? a) The total mechanical energy is conserved. b) The total kinetic energy is conserved. c) The total momentum is conserved. d) The momentum of each object is conserved. e) The kinetic energy of each object is conserved.

A 170.-g hockey puck moving in the positive \(x\) -direction at \(30.0 \mathrm{~m} / \mathrm{s}\) is struck by a stick at time \(t=2.00 \mathrm{~s}\) and moves in the opposite direction at \(25.0 \mathrm{~m} / \mathrm{s}\). If the puck is in contact with the stick for \(0.200 \mathrm{~s}\), plot the momentum and the position of the puck, and the force acting on it as a function of time, from 0 to \(5.00 \mathrm{~s}\). Be sure to label the coordinate axes with reasonable numbers.

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.

Sign-up for free