/*! 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 36 . Two identical objects travelin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

. Two identical objects traveling in opposite directions with the same speed \(V\) make a head-on collision. Find the speed of each object after the collision if (a) they stick together and (b) if the collision is perfectly elastic.

Short Answer

Expert verified
(a) Speed is 0; (b) Speeds are \( V \) and \(-V \).

Step by step solution

01

Understanding the Problem

We are given two objects that collide head-on. Initially, both objects travel in opposite directions with the same speed \( V \). We need to analyze two scenarios: (a) when the objects stick together after collision, and (b) when the collision is perfectly elastic.
02

Conservation of Momentum for Inelastic Collision

In an inelastic collision where the two objects stick together, the conservation of momentum principle applies. Initial momentum = final momentum. If the mass of each object is \( m \), then the total initial momentum is \( m \times V + m \times (-V) = 0 \). After collision, the combined mass \( 2m \) moves with speed \( V_f \), so the new momentum is \( 2m \times V_f \). Thus, \( 2m \times V_f = 0 \), implying \( V_f = 0 \).
03

Conservation of Momentum for Elastic Collision

For a perfectly elastic collision, both momentum and kinetic energy are conserved. The total initial momentum is \( 0 \) (as in Step 2), so the total final momentum must also be zero. This means the objects will have equal and opposite speeds after collision.
04

Conservation of Kinetic Energy for Elastic Collision

For a perfectly elastic collision, the initial and final kinetic energies are equal. Initially, total kinetic energy is \( \frac{1}{2}mV^2 + \frac{1}{2}mV^2 = mV^2 \). After collision, the kinetic energy remains \( mV^2 \), suggesting each object retains its speed but moves in opposite direction. Thus, final speeds are \( V \) and \(-V \).
05

Conclusion

(a) In the inelastic collision where they stick together, the speed after collision is 0. (b) In the perfectly elastic collision, each object maintains its speed \( V \) but moves in the opposite direction.

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.

Inelastic Collision
In an inelastic collision, the colliding objects stick together and move as a single mass post-collision. During such collisions, one key principle remains intact - the conservation of momentum. However, inelastic collisions do not conserve kinetic energy.

Let's delve deeper into how momentum applies here. When two identical objects, each with mass \( m \), collide head-on, their momentums cancel out if they travel with the same speed \( V \) but in opposite directions. This leads to an initial momentum of zero.

After a completely inelastic collision where the objects merge, the final momentum remains zero since they stick together and don't move. This simply means that all kinetic energy is transformed into other types of energy, like sound or thermal energy, and hence the objects come to a halt altogether.
Elastic Collision
In an elastic collision, unlike an inelastic collision, both momentum and kinetic energy are conserved. This type of collision is characterized by the two objects bouncing off each other without any loss of total mechanical energy.

For identical objects colliding head-on with the same speed, the principle of conservation of momentum dictates that if the initial total momentum was zero, the final total momentum must also be zero. They move away with equal magnitudes of velocity in opposite directions.

What makes elastic collisions special is the conservation of kinetic energy. Initially, each object holds energy \( \frac{1}{2}mV^2 \), and the total is \( mV^2 \). After impact, each retains their original speed thus ensuring the total kinetic energy remains \( mV^2 \).
Conservation of Kinetic Energy
Kinetic energy, a measure of energy due to motion, is only conserved in elastic collisions. It implies that the total kinetic energy before the collision equals the total kinetic energy after collision.

When analyzing a head-on elastic collision between two identical masses with equal speed but opposite velocities, the kinetic energy before and after remains \( mV^2 \). Each object retains its speed following the collision, demonstrating no kinetic energy is "lost" to other forms like thermal or sound energy.

This contrasts with inelastic collisions, where the final kinetic energy is typically less due to its conversion to other energy forms, aligning with the inelastic collision outcomes.
Head-On Collision
A head-on collision is a type of impact where two objects collide directly along their line of motion. This is often analyzed in terms of both momentum and kinetic energy conservation depending on whether the collision is elastic or inelastic.

In the case of two identical objects traveling at the same speed in opposite directions, a head-on collision can succinctly illustrate principles of momentum conservation. Such scenarios are ideal for understanding basic physics, as they provide clear outcomes like stopped objects post-collision for inelastic cases or objects reversing their velocities in elastic ones.

By breaking down these collisions into simpler conditions with equal masses and velocities, students can grasp the fundamental differences between inelastic and elastic collisions more easily.

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

\(\bullet\) The magnitude of the momentum of a cat is \(p .\) What would be the magnitude of the momentum (in terms of \(p )\) of a dog having three times the mass of the cat if it had (a) the same speed as the cat, and (b) the same kinetic energy as the cat?

\(\bullet\) Tennis, anyone? Tennis players sometimes leap into the air to return a volley. (a) If a 57 g tennis ball is traveling horizontally at 72 \(\mathrm{m} / \mathrm{s}\) (which does occur), and a 61 kg tennis player leaps vertically upward and hits the ball, causing it to travel at 45 \(\mathrm{m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{ms},\) what force does her racket exert on the ball? What force does the ball exert on the racket?

The structure of the atom. During \(1910-1911,\) Sir Ernest Rutherford performed a series of experiments to determine the structure of the atom. He aimed a beam of alpha particles (helium nuclei, of mass \(6.65 \times 10^{-27} \mathrm{kg}\) ) at an extremely thin sheet of gold foil. Most of the alphas went right through with little deflection, but a small percentage bounced directly back. These results told him that the atom must be mostly empty space with an extremely small nucleus. The alpha particles that bounced back must have made a head-on collision with this nucleus. A typical speed for the alpha particles before the collision was \(1.25 \times 10^{7} \mathrm{m} / \mathrm{s},\) and the gold atom has a mass of \(3.27 \times 10^{-25} \mathrm{kg}\) . Assuming (quite reasonably) elastic collisions, what would be the speed after the collision of a gold atom if an alpha particle makes a direct hit on the nucleus?

Two identical 1.50 \(\mathrm{kg}\) masses are pressed against opposite ends of a light spring of force constant \(1.75 \mathrm{N} / \mathrm{cm},\) compress- ing the spring by 20.0 \(\mathrm{cm}\) from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless horizontal lab table.

You (mass 55 \(\mathrm{kg}\) ) are riding your frictionless skateboard (mass 5.0 \(\mathrm{kg}\) ) in a straight line at a speed of 4.5 \(\mathrm{m} / \mathrm{s}\) when a friend standing on a balcony above you drops a 2.5 \(\mathrm{kg}\) sack of flour straight down into your arms. (a) What is your new speed, while holding the flour sack? (b) Since the sack was dropped vertically, how can it affect your horizontal motion? Explain. (c) Suppose you now try to rid yourself of the extra weight by throwing the flour 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.