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Chapter 10: Problem 71

Box of Marbles \(\mathrm{A}\) box is put on a scale that is marked in units of mass and adjusted to read zero when the box is empty. A stream of marbles is then poured into the box from a height \(h\) above its bottom at a rate of \(R\) (marbles per second). Each marble has mass \(m .(\mathrm{a})\) If the collisions between the marbles and the box are completely inelastic, find the scale reading at time \(t\) after the marbles begin to fill the box. (b) Determine a numerical answer when \(R=100 \mathrm{~s}^{-1}, h=7.60 \mathrm{~m}, m=4.50 \mathrm{~g}\), and \(t=10.0 \mathrm{~s}\)

Short Answer

Expert verified
49.635 N

Step by step solution

01

- Understanding Inelastic Collisions

Inelastic collisions mean that when the marbles hit the bottom of the box, they do not bounce back. Instead, they stick to the bottom of the box and transfer their entire momentum to it.
02

- Calculating the Force Due to Falling Marbles

Each marble has a mass of \( m \) and the gravitational force acting on each marble is \( F_g = mg \). When the marbles fall from a height \( h \), their velocity just before hitting the bottom of the box is given by \( v = \sqrt{2gh} \).
03

- Finding the Change in Momentum

The change in momentum of each marble when it hits the box is \( m \sqrt{2gh} \). Since the collisions are completely inelastic, the momentum change can be written as force times time, which is \( F = \frac{dp}{dt} \).
04

- Computing the Force from Collision Rate

The rate at which the marbles hit the bottom of the box is given by \( R \) (marbles per second). So the total force due to the impacts is \( F_{collision} = R m \sqrt{2gh} \).
05

- Summing Forces on the Box

The scale reading will reflect the sum of the gravitational force due to the mass of marbles in the box and the force due to the collisions. The gravitational force is \( F_g = t R mg \) (since \( t R \) marbles have fallen into the box at time \( t \)). The total scale reading will thus be the sum of both forces \[ F_{total} = t R mg + R m \sqrt{2gh} \].
06

- Substituting the Given Values

Now substitute the values: \( R = 100 \) marbles/s, \( h = 7.60 \) m, \( m = 4.50 \) g (\( 4.50 \) g = 0.0045 kg), \( t = 10.0 \) s, and \( g = 9.81 \) m/s\textsuperscript{2}. Calculate each term separately and then sum them.
07

- Calculation of Gravitational Force Component

First, calculate the gravitational force due to the mass of marbles: \[ t R mg = 10.0 \times 100 \times 0.0045 \times 9.81 = 44.145 \text{ N} \]
08

- Calculation of Collision Force Component

Next, calculate the force due to collisions: \[ R m \sqrt{2gh} = 100 \times 0.0045 \times \sqrt{2 \times 9.81 \times 7.60} = 100 \times 0.0045 \times 12.2 = 5.49 \text{ N} \]
09

- Getting the Total Scale Reading

Finally, sum both forces to get the total scale reading at time \( t = 10.0 \) s: \[ F_{total} = 44.145 + 5.49 = 49.635 \text{ N} \]

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Key Concepts

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

momentum transfer
In physics, momentum transfer happens when an object, like a marble, changes its motion due to a collision. If a marble with mass \(m\) and velocity \(v\) hits the bottom of the box and stops, it transfers its momentum to the box. Momentum is calculated using \(p = mv\). For inelastic collisions, like in this problem, the marbles don't bounce back, so they stick and transfer all their momentum to the box.

When marbles fall from height \(h\), gaining speed due to gravity, they hit the box with velocity \(v = \sqrt{2gh}\). Therefore, the momentum transfer per marble is \(m \sqrt{2gh}\). Since the marbles hit the box at a rate of \(R\) marbles per second, the force exerted due to this momentum transfer is calculated as \( F = Rm\sqrt{2gh}\).
gravitational force
Gravitational force is the force exerted by the Earth's gravity on an object. It is calculated using \(F_g = mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity (9.81 m/s\textsuperscript{2}). In this exercise, as marbles fall into the box, they add to the total mass in the box, which increases the gravitational force on the scale.

At any time \(t\), the number of marbles in the box can be calculated as \(Rt\) (marbles per second * time in seconds). Therefore, the total gravitational force due to the marbles in the box at time \(t\) is \(t R m g\). This force contributes to the overall reading on the scale.
collision force
Collision force comes from marbles hitting and sticking to the bottom of the box, which happens during inelastic collisions. Each marble impacts the box with a force derived from its momentum change over time. When marbles fall and hit the box, their velocity before impact is given by \(v = \sqrt{2gh}\).

The rate at which marbles hit the box is \(R\) marbles per second. Hence, the force due to continuous collisions is because each marble's momentum change is \(m \sqrt{2gh}\).

Mathematically, the collision force can be expressed as \( Rm\sqrt{2gh}\). This force combines with the gravitational force to produce the total force measured by the scale.

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Most popular questions from this chapter

Two Snowy Peaks Two snowy peaks are \(850 \mathrm{~m}\) and \(750 \mathrm{~m}\) above the valley between them. A ski run extends down from the top of the higher peak and then back up to the top of the lower one, with a total length of \(3.2 \mathrm{~km}\) and an average slope of \(30^{\circ}\) (Fig. 10-49). (a) A skier starts from rest at the top of the higher peak. At what speed will he arrive at the top of the lower peak if he coasts without using ski poles? Ignore friction. (b) Approximately what coefficient of kinetic friction between snow and skis would make him stop just at the top of the lower peak?

Billiard Ball A billiard ball moving at a speed of \(2.2 \mathrm{~m} / \mathrm{s}\) strikes an identical stationary ball a glancing blow. After the collision, one ball is found to be moving at a speed of \(1.1 \mathrm{~m} / \mathrm{s}\) in a direction making a \(60^{\circ}\) angle with the original line of motion. (a) Find the velocity of the other ball. (b) Can the collision be inelastic, given these data?

Spring Attached to Wall A \(1.0 \mathrm{~kg}\) block at rest on a horizontal frictionless surface is connected to an unstretched spring \((k=\) \(200 \mathrm{~N} / \mathrm{m})\) whose other end is fixed (Fig. \(10-58\) ). A \(2.0 \mathrm{~kg}\) block moving at \(4.0 \mathrm{~m} / \mathrm{s}\) collides with the \(1.0 \mathrm{~kg}\) block. If the two blocks stick together after the one- dimensional collision, what maximum compression of the spring occurs when the blocks momentarily stop?

Momentum and Energy? Is it possible for a system of interacting objects to conserve momentum and also mechanical energy (kinetic plus potential)? Discuss and defend your answer, then give an example that illustrates the case you are trying to make.

Block Slides Down an Incline In Fig. \(10-48\), a block is moved down an incline a distance of \(5.0 \mathrm{~m}\) from point \(A\) to point \(B\) by a force \(\vec{F}\) that is parallel to the incline and has magnitude \(2.0 \mathrm{~N}\). The magnitude of the frictional force acting on the block is \(10 \mathrm{~N}\). If the kinetic energy of the block increases by \(35 \mathrm{~J}\) between \(A\) and \(B\), how much work is done on the block by the gravitational force as the block moves from \(A\) to \(B\) ?
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