/*! 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 65 A \(1-\) lb ball \(A\) is travel... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(1-\) lb ball \(A\) is traveling horizontally at \(20 \mathrm{ft} / \mathrm{s}\) when it strikes a 10 -lb block \(B\) that is at rest. If the coefficient of restitution between \(A\) and \(B\) is \(e=0.6,\) and the coefficient of kinetic friction between the plane and the block is \(\mu_{k}=0.4,\) determine the time for the block \(B\) to stop sliding.

Short Answer

Expert verified
Involve the above detailed steps of calculations by Applying the equation of motion and linear momentum conservation.

Step by step solution

01

Calculate the Velocity of block B after impact

We first need to calculate the final velocity of the block B after being struck by the ball A, using the formula \(v_B = v_A \cdot \frac{m_A \cdot (1+e)}{m_A + m_B}\), where \(m_A\) and \(m_B\) are the masses of the ball A and block B respectively, \(v_A\) is the initial velocity of the ball, and \(e\) is the coefficient of restitution. Substituting \(m_A = 1lb\), \(m_B = 10lb\), \(v_A = 20ft/s\), and \(e = 0.6\), we get \(v_B = 20 \cdot \frac{1 \cdot (1+0.6)}{1 + 10}\) ft/s.
02

Calculate the deceleration due to friction

Next, the negative acceleration (deceleration) due to friction must be calculated, using the formula \(a_f = g \cdot \mu_k\), where \(g\) is the acceleration due to gravity and \(\mu_k\) is the coefficient of kinetic friction. When \(g = 32.2 ft/s^2\) and \(\mu_k = 0.4\), we have \(a_f = 32.2 \cdot 0.4\) ft/s^2.
03

Calculate the time for block B to stop

Finally we calculate the time it takes for B to stop sliding using the formula \(t_b = \frac{v_B}{a_f}\). This gives us the time it takes for block B to come to a stop.

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Key Concepts

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

Understanding Kinetic Friction
Kinetic friction, often symbolized as \(\mu_k\), is the resistive force that occurs between surfaces in motion relative to one another. This frictional force acts in the opposite direction to the motion and is crucial for understanding why objects in motion eventually come to a stop when no additional forces are applied.

For instance, when a block slides over a surface, the roughness and characteristics of both the block and the surface contribute to the strength of the kinetic frictional force. This force can be calculated using the equation \(f_k = \mu_k \times N\), where \(f_k\) is the kinetic frictional force, \(N\) is the normal force, and \(\mu_k\) is the coefficient of kinetic friction.

In our exercise, the coefficient of kinetic friction between the block and the plane was given as \(\mu_k = 0.4\). This means that for every unit of normal force, the kinetic frictional force exerted is 0.4 units. This value is essential for calculating how quickly the block slows down and eventually stops after the impact.
Impulse and Momentum
The concepts of impulse and momentum are central to analyzing collisions and their outcomes. Momentum is a measure of an object's motion and is given by the product of its mass and velocity (momentum \(= m \times v\)). When two objects collide, such as a ball and a block, their momenta interact.

Impulse, on the other hand, refers to the change in momentum of an object when a force is applied over a period of time (impulse \(= F \times t\)). It is also equal to the product of the average force and the time interval over which this force acts.

The law of conservation of momentum states that in the absence of external forces, the total momentum of a system before and after a collision remains constant. When considering the coefficient of restitution, which tells us how 'elastic' a collision is, we can understand how the impact changes the velocity of the objects involved, hence altering their momenta. By calculating the impulse, we can predict the aftermath of such collisions on the movement of the objects.
Deceleration Due to Friction
When we talk about 'deceleration due to friction', we are referring to the negative acceleration that objects experience as they slide across a surface and are gradually brought to a halt due to kinetic friction. Deceleration (\(a_f\)) can be calculated using the formula \(a_f = g \times \mu_k\), where \(g\) is the gravitational acceleration and \(\mu_k\) is the coefficient of kinetic friction.

In the context of our exercise, after the collision, block B begins to slide and then decelerates due to the force of kinetic friction between the block and the surface. The deceleration is directly proportional to \(\mu_k\), illustrating that a higher coefficient would result in a quicker stop. Moreover, by knowing the initial velocity of the block and the deceleration, we can use the kinematic equations to find the time it takes for the block to come to a complete stop. This understanding of deceleration is essential not just in theoretical physics problems but also in practical applications like vehicle braking systems and conveyor belts.

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Most popular questions from this chapter

A ball is thrown onto a rough floor at an angle of \(\theta=45^{\circ} .\) If it rebounds at the same angle \(\phi=45^{\circ},\) determine the coefficient of kinetic friction between the floor and the ball. The coefficient of restitution is \(e=0.6 .\) Hint: Show that during impact, the average impulses in the \(x\) and \(y\) directions are related by \(I_{x}=\mu I_{y} .\) since the time of impact is the same, \(F_{x} \Delta t=\mu F_{y} \Delta t\) or \(F_{x}=\mu F_{y}\)

The 20 -g bullet is traveling at \(400 \mathrm{m} / \mathrm{s}\) when it becomes embedded in the 2 -kg stationary block. Determine the distance the block will slide before it stops. The coefficient of kinetic friction between the block and the plane is \(\mu_{k}=0.2\).

The \(5-\mathrm{Mg}\) bus \(B\) is traveling to the right at \(20 \mathrm{m} / \mathrm{s}\). Meanwhile a \(2-\mathrm{Mg}\) car \(A\) is traveling at \(15 \mathrm{m} / \mathrm{s}\) to the right. If the vehicles crash and become entangled, determine their common velocity just after the collision. Assume that the vehicles are free to roll during collision.

The 50 -kg crate is pulled by the constant force \(\mathbf{P}\). If the crate starts from rest and achieves a speed of \(10 \mathrm{m} / \mathrm{s}\) in \(5 \mathrm{s},\) determine the magnitude of \(\mathbf{P}\). The coefficient of kinetic friction between the crate and the ground is \(\mu_{k}=0.2\).

The automobile has a weight of 2700 lb and is traveling forward at \(4 \mathrm{ft} / \mathrm{s}\) when it crashes into the wall. If the impact occurs in 0.06 s, determine the average impulsive force acting on the car. Assume the brakes are not applied. If the coefficient of kinetic friction between the wheels and the pavement is \(\mu_{k}=0.3,\) calculate the impulsive force on the wall if the brakes were applied during the crash.The brakes are applied to all four wheels so that all the wheels slip.

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