/*! 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 2 An elevator platform is going up... [FREE SOLUTION] | 91Ó°ÊÓ

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An elevator platform is going up at a speed \(20 \mathrm{~ms}^{-1}\) at during its upward motion a small ball of \(50 \mathrm{~g}\) mass falling in downward direction strikes the platform elastically a speed \(5 \mathrm{~ms}^{-1}\). Find the speed (in \(\mathrm{ms}^{-1}\) ) with which the ball rebounds.

Short Answer

Expert verified
The ball rebounds with a speed of 5 m/s upward.

Step by step solution

01

Understand the Problem

We have a ball falling and striking an upward-moving platform elastically. You must find the rebound speed of the ball. The platform moves upward at 20 m/s, and the ball moves downward at 5 m/s before collision.
02

Define Velocities Relative to the Platform

Use relative motion principles. Because the platform moves at 20 m/s upwards, convert the ball's velocity to a platform-relative frame of reference. The ball initially approaches the platform at a relative speed of 5 m/s + 20 m/s = 25 m/s downwards.
03

Calculate the Relative Rebound Velocity

In an elastic collision, the relative speed of separation equals the relative speed of approach. Therefore, the relative rebound velocity of the ball in the upward direction is 25 m/s.
04

Convert Back to Ground Reference Frame

Now convert the rebound velocity back to the ground reference frame. Since the platform is moving upwards at 20 m/s, the velocity of the ball relative to an observer on the ground will be 25 m/s (relative rebound speed) - 20 m/s (platform speed) = 5 m/s upward.

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Key Concepts

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

Relative Motion
Relative motion is an essential concept when dealing with problems involving moving frames of reference, such as the elevator platform and the ball in this exercise. To make sense of the movements, it's important to understand how velocities change when considered relative to different objects.

In this example, the platform moves upwards at 20 m/s, while the ball moves downward at 5 m/s. By using relative motion, we view the ball's movement as it appears from the platform, which is also moving. This involves calculating the ball's velocity relative to the platform's velocity. The formula used is:
  • Relative velocity = velocity of ball - velocity of platform
  • In this case: 5 m/s down + 20 m/s up = 25 m/s down relative to the platform.
So, to the observer on the platform, the ball seems to approach at 25 m/s. This simplification helps determine the rebound and impact speeds more easily.
Rebound Velocity
Rebound velocity is the speed and direction with which the ball moves after bouncing back from the platform. Understanding rebound velocity is crucial, especially in elastic collisions, where no kinetic energy is lost.

In this exercise, once the ball strikes the platform, it rebounds elastically. Since the relative motion principle determined that the ball approaches the platform at 25 m/s, and because of the nature of elastic collisions (where the speed of separation is equal to the speed of approach), the ball will rebound upwards at 25 m/s. But remember, this is in the platform's reference frame.

Calculating the ground frame velocity involves adjusting the rebound speed by the platform's speed. So:
  • Rebounded velocity = relative rebound speed - platform speed
  • Therefore, 25 m/s - 20 m/s = 5 m/s upwards in the ground frame.
This means that the ball essentially reverses its direction, bouncing back with a 5 m/s speed upwards according to an observer on the ground.
Conservation of Momentum
The principle of conservation of momentum is pivotal in understanding collisions, including elastic ones. It states that, within a closed system not influenced by external forces, the total momentum before collision equals the total momentum after the collision.

In our scenario, we focused on finding rebound velocity primarily through relative motion, but conservation of momentum also ensures that the platform-ball system adheres to this equilibrium. For elastic collisions, both momentum and kinetic energy are conserved. This means:
  • Momentum before = Momentum after
  • Total kinetic energy before = Total kinetic energy after
Although the mass of the platform is not considered here due to the small size of the ball, this principle underlies every calculation we performed. In exercises where mass and velocity vary significantly, conservation of momentum is used to find unknowns such as the velocities or masses involved post-collision, ensuring consistency and equilibrium in all calculations.

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Most popular questions from this chapter

Two equal spheres \(B\) and \(C\), each of mass \(m\), are in contact on a smooth horizontal table. A third sphere \(A\) of same size as that of \(B\) or \(C\) but mass \(m / 2\) impinges symmetrically on them with a velocity \(u\) and is itself brought to rest. The coefficient of restitution between the two spheres \(A\) and \(B\) (or between \(A\) and \(C\) ) is (1) \(1 / 3\) (2) \(1 / 4\) (3) \(2 / 3\) (4) \(3 / 4\)

A small steel ball \(A\) is suspended by an inextensible thread of length \(l=1.5 \mathrm{~m}\) from O. Another identical ball is thrown vertically downwards such that its surface remains just in contact with thread during downward motion and collides elastically with the suspended ball. If the suspended ball just completes vertical circle after collision, calculate the velocity (in \(\mathrm{cm} / \mathrm{s}\) ) of the falling ball just before collision \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

A ball impinges directly on another ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by the impact, the value of coefficient of restitution is (1) \(\frac{1}{2}\) (2) \(\frac{1}{\sqrt{3}}\) (3) \(\frac{1}{\sqrt{2}}\) (4) \(\frac{\sqrt{3}}{2}\)

A \(70 \mathrm{~g}\) ball \(B\) dropped from a height \(h_{0}=9 \mathrm{~m}\) reaches a height \(h_{2}=0.25 \mathrm{~m}\) after bouncing twice from identical \(210 \mathrm{~g}\) plates. Plate \(A\) rests directly on hard ground, while plate \(C\) rests on a foam-rubber mat. Find the height \(h_{1}\) (in \(\mathrm{m}\) ) of the ball's first bounce.

Velocity of a particle of mass \(2 \mathrm{~kg}\) changes from \(\vec{v}_{1}=(-2 \hat{i}-2 \hat{j}) \mathrm{m} / \mathrm{s}\) to \(\vec{v}_{2}=(\hat{i}-\hat{j}) \mathrm{m} / \mathrm{s}\) after colliding with a plane surface. (1) The angle made by the plane surface with the positive \(x\)-axis is \(90^{\circ}+\tan ^{-1}\left(\frac{1}{3}\right)\) (2) The angle made by the plane surface with the positive \(x\)-axis is \(\tan ^{-1}\left(\frac{1}{3}\right)\) (3) The direction of change in momentum makes an angle \(\tan ^{-1}\left(\frac{1}{3}\right)\) with the positive \(x\)-axis. (4) The direction of change in momentum makes an angle \(90^{\circ}+\tan ^{-1}\left(\frac{1}{3}\right)\) with the plane surface.

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