/*! 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 3 A man standing on the edge of th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 3

A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with a speed of \(20 \mathrm{~m} / \mathrm{s}\). Two seconds later, an identical stone is thrown vertically downwards with the same speed of \(20 \mathrm{~m} / \mathrm{s}\). Then (1) the relative velocity between the two stones remains constant till one hits the ground (2) both will have the same kinetic energy, when they hit the ground (3) the time interval between their hitting the ground is \(2 \mathrm{~s}\) (4) if the collision on the ground is perfectly elastic, both will rise to the same height above the ground

Short Answer

Expert verified
(1) True, (2) True, (3) True, (4) True: All statements are correct.

Step by step solution

01

Analyze Given Situation

Two stones are thrown from the same height: one upwards and one downwards, both with a speed of 20 m/s, but with a 2-second delay between throws. We need to examine their relative velocities, kinetic energies upon impact, time interval between impacts, and heights reached if collisions were perfectly elastic.
02

Determine Relative Velocity

The relative velocity of two objects, where both have speeds of 20 m/s in opposite directions (one upwards, one downwards), can be calculated as: \(20 + 20 = 40\) m/s. Since relative velocity depends solely on the initial speeds and directions (and not on gravitational effects after they start moving), it will remain constant until one hits the ground.
03

Analyze Kinetic Energy When Hitting the Ground

Since both stones are thrown with the same speed, and they both eventually undergo the same change in potential energy (falling from the same height), according to the conservation of mechanical energy, they will have identical speeds and thus identical kinetic energies when hitting the ground.
04

Time Interval Between Hitting the Ground

The stone thrown upwards will slow down, stop, and then fall back, taking longer to reach the ground than the stone thrown 2 seconds later with a downward velocity. The time to reach the maximum height (upwards stop point) is 2 seconds for the upwards thrown stone. The delay means both stones undergo a simultaneous free fall with equal duration. Thus, the time between their ground impacts will be the initial 2 seconds delay from the throws.
05

Height Reached After Elastic Collision

In a perfectly elastic collision, both stones will rebound to the same height they were released from initially. Despite different paths initially (upwards vs. downwards), full elasticity implies a return to the original potential energy state, so each stone reaches the same height from where it was thrown.

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 91影视!

Key Concepts

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

Relative Velocity
Relative velocity is an important concept when analyzing positions and motions of objects relative to each other. In this scenario, two stones are involved, one thrown upwards and another downwards at the same speed but with a delay. The key idea behind relative velocity is to measure how fast one object is moving in relation to another.
This can be calculated by considering their individual speeds and directions. Here, since the stones are moving in opposite directions, the relative velocity is the sum of their speeds.
  • Stone A moves upwards at 20 m/s while Stone B moves downwards at the same speed.
  • Their relative velocity is therefore 20 m/s + 20 m/s = 40 m/s.
This value remains constant prior to any stone hitting the ground, because both stones maintain their initial speeds until affected by external forces.
Kinetic Energy
Kinetic energy is defined as the energy an object possesses due to its motion. The kinetic energy of an object can be calculated using the formula: \[KE = \frac{1}{2}mv^2\] where \(m\) is the mass of the object and \(v\) is its velocity.
In the exercise, both stones are identical and thrown with the same speed of 20 m/s, meaning their kinetic energy when they hit the ground will be identical too.
This is due to the principle of conservation of energy, which states that in the absence of external forces, the total energy in a system remains constant.
  • Both stones start and end at the same heights and gain or lose potential energy in the same manner.
  • Thus, each stone's potential energy transforms equally to kinetic energy upon impact.
Elastic Collision
Elastic collision is a type of collision where there is no net loss in kinetic energy. When objects undergo perfectly elastic collisions, both kinetic energy and momentum are conserved. In the context of the stones being thrown:
  • If the collision with the ground is perfectly elastic, the stones will rebound to the same height from which they were initially released.
  • This is because the system is closed with conservation of kinetic energy, resulting in both stones rising back to their original energies and positions.
This implies that any initial kinetic energy and potential energy stored due to gravity will be entirely returned post-collision, allowing each stone to achieve its starting point above the ground.
Free Fall
Free fall describes the motion of objects under the sole influence of gravity. Such objects experience acceleration due to gravity (\( g \approx 9.81 \, \text{m/s}^2\) on Earth's surface) without any resistance from air or other forces. In the exercise, after being thrown, each stone enters a state of free fall:
  • The stone thrown upwards slows as it moves against gravity until it stops momentarily, then accelerates downward.
  • The stone thrown downwards immediately starts accelerating downwards.
Since gravity acts equally on both stones, they experience the same acceleration during their descent, with only initial velocity affecting the time it takes to reach the ground.

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

A ball of mass \(m\) is projected with a speed \(v\) into the barrel of a spring gun of mass \(M\) initially at rest lying on a frictionless surface. The mass sticks in the barrel at the point of maximum compression in the spring. The fraction of kinetic energy of the ball stored in the spring is (1) \(\frac{m}{M}\) (2) \(\frac{M}{m+M}\) (3) \(\frac{m}{M+M}\) (4) none of these

A man stands at one end of the open truck which can run on frictionless horizontal rails. Initially, the man and the truck are at rest. Man now walks to the other end and stops. Then which of the following is true? (1) The truck moves opposite to direction of motion of the man even after the man ceases to walk. (2) The centre of mass of the man and the truck remains at the same point throughout the man's walk. (3) The kinetic energy of the man and the truck are exactly equal throughout the man's walk. (4) The truck does not move at all during the man's walk.

A gun of mass \(M\), fires a shell of mass \(m\) horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height ' \(h\) '. The recoil velocity of the gun is (1) \(\left(\frac{2 m^{2} g h}{M(m+M)}\right)^{\frac{1}{2}}\) (2) \(\left(\frac{2 m^{2} g h}{M(m-M)}\right)^{\frac{1}{2}}\) (3) \(\left(\frac{2 m^{2} g h}{2 M(m-M)}\right)^{\frac{1}{2}}\) (4) \(\left(\frac{2 m^{2} g h}{2 M(m+M)}\right)^{\frac{1}{2}}\)

A car of mass \(m\) is initially at rest on the boat of mass \(M\) tied to the wall of dock through a massless, inextensible string. The caraccelerates from rest to velocity \(v_{0}\) in times \(t_{0}\). At \(t=t_{0}\) the car applies brake and comes to rest relative to the boat in negligible time. Neglect friction between the boat and water; the time ' \(t\) ' at which boat will strike the wall is (1) \(\frac{L(M+m)}{m v_{0}}\) (2) \(t_{0}+\frac{L(M+m)}{m v_{0}}\) (3) \(t_{0}+\frac{L(M+m)}{M v_{0}}\) (4) none of these

Two blocks of masses \(m\) and \(M\) are interconnected by an ideal spring of stiffness \(k\). If \(m\) is pushed with a velocity \(v_{0}\) (1) Velocity of centre of the system \((M+m) ; v_{C}=\frac{m v_{0}}{M+m}\) (2) Kinetic energy of the system ( \(M+m\) ) with respect to \(\mathrm{CM}\) is equal to \(\frac{M m v_{0}^{2}}{2(M+m)}\) (3) The maximum work done by the spring is equal to \(-\frac{M m v_{0}^{2}}{2(M+m)}\) (4) Maximum compression of spring $$ x_{\max }=v_{0} \sqrt{\frac{M m}{(M+m) k}} $$
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fields in Physics

Read Explanation

Mechanics and Materials

Read Explanation

Energy Physics

Read Explanation

Nuclear Physics

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.