Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 18

A \(60-\mathrm{kg}\) skater, at rest on frictionless ice, tosses a 12 -kg snowball with velocity \(\vec{v}=53.0 \hat{\imath}+14.0 \hat{\jmath} \mathrm{m} / \mathrm{s},\) where the \(x\) - and \(y\) -axes are in the horizontal plane. Find the skater's subsequent velocity.

Short Answer

Expert verified
The skater's subsequent velocity is \(\vec{v} = -10.6\hat{\imath} - 2.8\hat{\jmath} m/s\)

Step by step solution

01

Define the Initial and Final Momenta

The total momentum before the action must equal the total momentum after the action according to the law of conservation of momentum, because the system is isolated (no external forces are acting). The skater's initial velocity was zero, therefore the initial momentum of the system was 0. When the skater throws the snowball, the final momentum must still equal to 0. Mathematically, we write as follows: Initial Momentum = Final Momentum ⇒ \(0 = m_1v_1 + m_2v_2\) where, \(m_1\) and \(v_1\) are the mass and velocity of the skater respectively and \(m_2\) and \(v_2\) are the mass and velocity of the snowball.
02

Decompose the Snowball's Velocity

The velocity of the snowball is given in the form of vector composed of \(\hat{\imath}\) component along x-axis and \(\hat{\jmath}\) component along y-axis. This implies that the snowball moves in two dimensions. The principle of conservation of momentum should be applied to each direction (x and y) separately. The snowball's velocity is given as \(\vec{v}=53.0 \hat{\imath}+14.0 \hat{\jmath} m/s\). Thus, \(v_{2x} = 53.0 m/s\) and \(v_{2y} = 14.0 m/s\).
03

Applying Conservation of Momentum for X-axis

We can plug the values of the masses of skater and snowball, and the x-component of snowball's velocity into the conservation of momentum equation. This gives: \(0 = m_1v_{1x} + m_2v_{2x}\) ⇒ \(0 = 60kg*v_{1x} + 12kg*53.0m/s\). Solving this equation, we find the x-component of skater's velocity: \(v_{1x} = - (12kg*53.0m/s) / 60kg = -10.6 m/s\). Here, the negative sign indicates the skater moves in the opposite direction of the x-axis, which is expected.
04

Applying Conservation of Momentum for Y-axis

Similarly, we apply the conservation of momentum equation for the y-axis, knowing that no movement has happened along this direction initially: \(0 = m_1v_{1y} + m_2*v_{2y}\) ⇒ \(0 = 60kg*v_{1y} + 12kg*14.0m/s\). Solving for \(v_{1y}\), we get \(v_{1y} = - (12kg*14.0m/s) / 60kg = - 2.8 m/s\). Again the negative sign indicates that the skater moves in the direction opposite to that of the snowball along the y-axis.
05

Finding the Final Answer

Now we simply put the two components of the velocity together to express the skater's velocity in the vector form. Hence, the skater's subsequent velocity is \(\vec{v} = -10.6\hat{\imath} - 2.8\hat{\jmath} m/s\) which indicates that the skater moved in the exact opposite direction of the snowball's throw.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider a system of three equal-mass particles moving in a plane; their positions are given by \(a_{i} \hat{\imath}+b_{i} \hat{\jmath},\) where \(a_{i}\) and \(b_{i}\) are functions of time with the units of position. Particle 1 has \(a_{1}=3 t^{2}+5\) and \(b_{1}=0 ;\) particle 2 has \(a_{2}=7 t+2\) and \(b_{2}=2 ;\) particle 3 has \(a_{3}=3 t\) and \(b_{3}=2 t+6 .\) Find the center-of-mass position, velocity, and acceleration of the system as functions of time.

Physicians perform needle biopsies to sample tissue from internal-organs. A spring-loaded gun shoots a hollow needle into the tissue; extracting the needle brings out the tissue core. A particular device uses 8.3 -mg needles that take 90 ms to stop in the tissue, which exerts a stopping force of \(41 \mathrm{mN}\). (a) Find the impulse imparted by the tissue. (b) How far into the tissue does the needle penetrate?

Two objects of unequal mass, one initially at rest, undergo a onedimensional elastic collision. For a given mass ratio, show that the fraction of the initial energy transferred to the initially stationary object doesn't depend on which object it is.

High-speed photos of a 220 - \(\mu \mathrm{g}\) flea jumping vertically show that the jump lasts \(1.2 \mathrm{ms}\) and involves an average vertical acceleration of \(100 g .\) What (a) average force and (b) impulse does the ground exert on the flea during its jump? (c) What's the change in the flea's momentum during its jump?

In a ballistic pendulum demonstration gone bad, a 0.52 -g pellet, fired horizontally with kinetic energy \(3.25 \mathrm{J},\) passes straight through a 400 -g Styrofoam pendulum block. If the pendulum rises a maximum height of \(0.50 \mathrm{mm}\), how much kinetic energy did the pellet have after emerging from the Styrofoam?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Dynamics

Read Explanation

Solid State Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Classical Mechanics

Read Explanation

Rotational Dynamics

Read Explanation
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