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Chapter 6: Problem 30

A ball of density \(\rho\) is dropped onto a horizontal solid surface. It bounces elastically from the surface and returns to its original position in time \(t_{1}\). Next, the ball is released from the same height, but this time it strikes the surface of a liquid of density \(\rho_{L}(>\rho\) but less than \(2 \rho)\), and takes \(t\) : second to come back to its original height (1) \(t_{2}>t_{1}\) (2) \(t_{1}>t_{2}\) (3) the motion of the ball is not simple harmonic (4) If \(\rho=\rho_{L}\), then the speed of the ball inside the liquid will be independent of its depth

Short Answer

Expert verified
(1) \( t_2 > t_1 \); the liquid introduces resistance and buoyancy, increasing return time.

Step by step solution

01

Understand the Initial Situation

In the first scenario, the ball is dropped onto a solid surface, bounces back elastically, and returns to the original height in time \( t_1 \). Since the bounce is elastic and there is no loss of energy, the time taken to return is only dictated by gravity and the initial height from which the ball was dropped.
02

Understand the Liquid Scenario

In the second scenario, the ball is dropped into a liquid. Here, the density of the liquid is \( \rho_L \), which is greater than that of the ball (\( \rho \)) but less than twice the density (\( \rho < \rho_L < 2\rho \)). Upon entering the liquid, the ball experiences a buoyant force causing it to slow down, altering the time \( t_2 \) taken to return to the original position.
03

Compare Forces in Both Scenarios

When the ball drops onto the solid, the only force at play upon bouncing is the elastic force, which conserves energy. In the liquid scenario, the ball encounters resistance and buoyancy affecting its upward motion. The net force is less than gravity alone due to these effects, which increases the time taken to reach the original height.
04

Evaluate Timing Differences

Given the presence of resistance and buoyancy in the liquid, \( t_2 \) must be greater than \( t_1 \), since \( t_2 \) includes both the slowing down from these forces and the fact that the ball doesn't bounce elastically off the liquid but is subjected to viscous drag and buoyancy.
05

Evaluate Other Options

The motion of the ball cannot be simple harmonic as constant forces are not acting uniformly throughout the movement unlike a spring-mass system. At \( \rho = \rho_L \), while the speeds are influenced by density, they are not truly independent of depth due to gravitational effects.

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Key Concepts

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

Elastic Collision
When the ball bounces off a solid surface, it demonstrates what we call an **elastic collision**. In an elastic collision, no kinetic energy is lost. This means the ball returns to its original height because the energy it starts with is the same as when it returns. The time, denoted as \( t_1 \), for the ball to return after bouncing is determined by the following factors:
  • The height from which it was dropped
  • Gravitational force acting on it
These collisions are idealized situations as most real-world collisions are slightly inelastic, meaning some energy is lost due to factors like air resistance or heat. However, in this exercise, we imagine a perfect scenario. On a solid surface, the perfect elastic nature makes it straightforward for the ball to return in time \( t_1 \).
Density
Density is a critical concept to understand the behavior of the ball, especially when it strikes a liquid. By definition, **density** is mass per unit volume, given by the formula: \[ \rho = \frac{m}{V} \] where \( \rho \) is density, \( m \) is mass, and \( V \) is volume. In this scenario, the liquid the ball strikes has a greater density (\( \rho_L \)) than the ball's own density (\( \rho \)). However, \( \rho_L \) must be less than twice the density of the ball, i.e., \( \rho < \rho_L < 2\rho \). This difference in density between the object and the fluid creates a **buoyant force**, which will act against the gravitational force, slowing down the ball's upward motion. The ball's motion through the liquid is altered here, leading to a longer time \( t_2 \) to return to its original height.
Fluid Mechanics
**Fluid mechanics** explores how fluids (liquids and gases) move under various forces. When the ball enters a liquid, it is subject to several forces:
  • **Gravity** pulling it downward
  • **Buoyant force** pushing it up, which is due to the displaced liquid's weight
  • **Viscous drag** which opposes the ball's motion through the liquid
These dynamics are influenced by the fluid's properties, such as density and viscosity. While gravity always acts downward, the buoyant force depends on the density difference, which is why the ball slows down compared to its elastic bounce off a solid surface. Fluid mechanics helps us understand why \( t_2 \) is greater than \( t_1 \), as the ball faces resistance from the liquid in addition to gravity.
Projectile Motion
The principles of **projectile motion** are at play in both scenarios of the bounced ball. Projectile motion involves objects in two-dimensional motion influenced by gravity alone, assuming no air resistance.In the first scenario, after bouncing elastically, the ball exhibits classic projectile motion:
  • Initial and final heights are the same
  • Time of ascent equals the time of descent
When submerged in a liquid, however, the situation changes due to additional forces at play, which means the motion ceases to be simple projectile motion. Instead, the forces of buoyancy and viscous drag intervene. As a result, the ball spends more time (\( t_2 \)) in the liquid because these forces slow it down compared to its trajectory along the solid surface, which only gravity influences.

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Most popular questions from this chapter

An empty balloon weighs \(W_{1}\). If air equal in weight to \(W\) is pumped into the balloon, the weight of the balloon becomes \(W_{2}\). Suppose that the density of air inside and outside the balloon is the same. Then (1) \(W_{2}=W_{1}+W\) (2) \(W_{2}=\sqrt{W_{1} W}\) (3) \(W_{2}=W_{1}\) (4) \(W_{2}=W_{1}-W\)

For Problems \(18-20\) A liquid having density \(6000 \mathrm{~kg} / \mathrm{m}^{3}\) stands to a height of \(4 \mathrm{~m}\) in a sealed tank as shown in the figure. The tank contains compressed air at a gauge pressure of \(3 \mathrm{~atm}\). The horizontal outlet pipe has a cross-sectional area of \(6 \mathrm{~cm}^{2}\) and \(3 \mathrm{~cm}^{2}\) at larger and smaller sections. Atmospheric pressure \(=1 \mathrm{~atm}, g=10 \mathrm{~m} / \mathrm{s}^{2}\), \(1 \mathrm{~atm}=10^{5} \mathrm{~N} / \mathrm{m}^{2}\). Assume that depth of water in the tank remains constant due to \(s\) very large base and air pressure above it remains constant. Based on the above information, answer the following questions. The discharge rate from the outlet is (1) \(4.02 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) (2) \(6 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}\) (3) \(7.56 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) (4) \(4.02 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}\)

A solid block of volume \(V=10^{-3} \mathrm{~m}^{3}\) and density \(d=800\) \(\mathrm{kg} / \mathrm{m}^{3}\) is tied to one end of a string, the other end of which is tied to the bottom of the vessel. The vessel contains two immiscible liquids of densities \(\rho_{1}=1000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\rho_{2}=\) \(1500 \mathrm{~kg} / \mathrm{m}^{3}\). The solid block is immersed with \(2 / 5\) th of its volume in the liquid of higher density and \(3 / 5\) th in the liquid of lower density. The vessel is placed in an elevator which is moving up with an acceleration of \(a=g / 2\). Find the tension in the string (in N). [g=10 m/s \(^{2}\) ].

The cubical container \(A B C D E F G H\) which is completely filled with an ideal (non-viscous and incompressible) fuid, moves in a gravity free space with a acceleration of \(a=a_{0}(\hat{i}-\hat{j}+\hat{k})\) where \(a_{0}\) is a positive constant. Then the only point in the container where pressure can be zero is (1) \(B\) (2) \(C\) (3) \(E\) (4) \(H\)

A block of silver of mass \(4 \mathrm{~kg}\) hanging from a string is immersed in a liquid of relative density \(0.72\). If relative density of silver is 10 , then tension in the string will be (1) \(37.12 \mathrm{~N}\) (2) \(42 \mathrm{~N}\) (3) \(73 \mathrm{~N}\) (4) \(21 \mathrm{~N}\)
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