/*! 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 37 A ball starts from rest and acce... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball starts from rest and accelerates at \(0.500 \mathrm{m} / \mathrm{s}^{2}\) while moving down an inclined plane \(9.00 \mathrm{m}\) long. When it reaches the bottom, the ball rolls up another plane, where, after moving \(15.0 \mathrm{m},\) it comes to rest. (a) What is the speed of the ball at the bottom of the first plane? (b) How long does it take to roll down the first plane? (c) What is the acceleration along the second plane? (d) What is the ball's speed \(8.00 \mathrm{m}\) along the second plane?

Short Answer

Expert verified
(a) The speed of the ball at the bottom of the first plane is \(3.00 \mathrm{m} / \mathrm{s}\). (b) The ball takes \(6.00 \mathrm{s}\) to roll down the first plane. (c) The acceleration along the second plane is \(-0.30 \ \mathrm{m} / \mathrm{s}^{2}\). (d) The ball's speed \(8.00 \mathrm{m}\) along the second plane is \(3.69 \ \mathrm{m} / \mathrm{s}\).

Step by step solution

01

Speed at the Bottom of the First Plane

We can compute the speed at the bottom of the first plane using the equation \(v^{2} = u^{2} + 2a s\), where \(v\) is the final speed, \(u\) is the initial speed (which is zero in this case), \(a\) is the acceleration, and \(s\) is the distance. Substituting the given values, we get \(v^{2}=0^{2}+2\times0.500 \mathrm{m} / \mathrm{s}^{2} \times 9.00 \mathrm{m}\), which simplifies to \(v = \sqrt{9.00 \times 1.000 \mathrm{m}^{2} / \mathrm{s}^{2}} = 3.00 \mathrm{m} / \mathrm{s}\).
02

Time to Roll Down the First Plane

Next, we find the time taken to roll down the first plane. For this, we can use the equation \(s=ut+0.5at^{2}\), where \(t\) is time, and the other symbols have their usual meaning. Substituting the relevant values, we get \(9.00 \mathrm{m} = 0 \times t + 0.5 \times 0.500 \mathrm{m} / \mathrm{s}^{2} \times t^{2}\), which simplifies to \(t = \sqrt{4 \times 9.00 \mathrm{m} / 0.500 \mathrm{m} / \mathrm{s}^{2}} = 6.00 \mathrm{s}\).
03

Acceleration Along the Second Plane

We find the acceleration along the second plane by using the equation, \(v^{2} = u^{2} - 2as\), where \(a\) is now the deceleration (negative acceleration), \(s\) is the distance and \(v=0 \ \mathrm{m} / \mathrm{s}\). Therefore, we get \(0 = 3.00^{2} \mathrm{m} / \mathrm{s} - 2 \times a \times 15.0 \mathrm{m}\), which simplifies to \(a=-3.00^{2} \mathrm{m} / \mathrm{s} / (2 \times 15.0 \mathrm{m}) \approx -0.30 \ \mathrm{m} / \mathrm{s}^{2}\). This is negative because it's acting against the direction of motion.
04

Speed Along the Second Plane

Finally, we compute the ball's speed \(8.00 \mathrm{m}\) along the second plane using the equation \(v^{2} = u^{2} - 2as\). Substituting the known values, we get \(v^{2} = 3.00^{2} - 2 \times -0.30 \times 8.00\), which simplifies to \(v = \sqrt{3.00^{2} + 2 \times 0.30 \times 8.00} = 3.69 \ \mathrm{m} / \mathrm{s}\).

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Key Concepts

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

Understanding Acceleration
Acceleration is a key concept in kinematics, describing how quickly an object's velocity changes over time. It is a vector quantity with both magnitude and direction, usually measured in meters per second squared (m/s²). When an object speeds up, acceleration is positive, while slowing down results in negative acceleration. For the ball on the inclined plane, it started with an initial velocity of zero, indicating an initial condition referred to as 'at rest.' As the ball moved down the incline due to the force of gravity, it accelerated at a rate of 0.500 m/s². This steady acceleration increases the ball's speed uniformly over the distance traveled. In situations like these, positive acceleration denotes an increase in speed as the object moves in the direction of the force applied, such as gravity on an incline.
Decoding Velocity
Velocity, unlike speed, is a vector quantity implying that it incorporates both magnitude and direction. It's essential to differentiate between speed, which measures how fast something is moving regardless of direction, and velocity, which includes the direction of motion. In kinematics, the initial velocity is often zero when the object begins from a rest position. In the given problem, the ball's velocity needed to be determined at different stages - at the bottom of the first inclined plane and at 8 meters along the second. Using the equation of motion, we calculated the velocity at the bottom as 3.00 m/s. This was determined by assuming the slope's direction as positive and using the appropriate kinematic equations to relate the displacement, acceleration, and initial velocity.
Incline Motion Essentials
Incline motion pertains to the movement of objects along a slanted surface. This type of motion is vital in understanding how gravitational force influences an object's acceleration and velocity. For the ball's motion down and then up an incline, the acceleration plays a significant role. Initially, as the ball descends, it experiences a positive acceleration due to gravity's pull. However, on the second incline, the scenario changes, as the ball decelerates against gravity, slowing down until it comes to rest. Such analyses rely on decomposing the gravitational force into components parallel and perpendicular to the incline, where the parallel component directly affects the ball's motion.
Equations of Motion Simplified
Equations of motion form the foundation for solving problems involving kinematic quantities like displacement, velocity, and acceleration. They are excellent tools to predict an object's future state based on its current state. The primary equations include:
  • First equation: \( v = u + at \) gives the velocity after time \( t \).
  • Second equation: \( s = ut + 0.5at^2 \) provides the displacement after time \( t \).
  • Third equation: \( v^2 = u^2 + 2as \) relates velocity, acceleration, and displacement directly without involving time.
For the ball's exercise, these equations helped to determine unknowns like its velocity at pivotal points, time taken to descend an incline, and the deceleration required on the second incline. Careful application of these equations and substituting the right values fits perfectly into the context of each phase of the motion, paving the way to solve complex real-world scenarios systematically.

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Most popular questions from this chapter

Kathy Kool buys a sports car that can accelerate at the rate of \(4.90 \mathrm{m} / \mathrm{s}^{2} .\) She decides to test the car by racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.00 s before Kathy. If Stan moves with a constant acceleration of \(3.50 \mathrm{m} / \mathrm{s}^{2}\) and Kathy maintains an acceleration of \(4.90 \mathrm{m} / \mathrm{s}^{2},\) find. (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant she overtakes him.

An inquisitive physics student and mountain climber climbs a 50.0 -m cliff that overhangs a calm pool of water. He throws two stones vertically downward, \(1.00 \mathrm{s}\) apart, and observes that they cause a single splash. The first stone has an initial speed of \(2.00 \mathrm{m} / \mathrm{s} .\) (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water?

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at \(71.5 \mathrm{m} / \mathrm{s} .\) The driver of the Thunderbird realizes he must make a pit stop, and he smoothly slows to a stop over a distance of \(250 \mathrm{m}\). He spends \(5.00 \mathrm{s}\) in the pit and then accelerates out, reaching his previous speed of \(71.5 \mathrm{m} / \mathrm{s}\) after a distance of \(350 \mathrm{m} .\) At this point, how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed?

A car is approaching a hill at \(30.0 \mathrm{m} / \mathrm{s}\) when its engine suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of \(-2.00 \mathrm{m} / \mathrm{s}^{2}\) while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking \(x=0\) at the bottom of the hill, where \(v_{i}=30.0 \mathrm{m} / \mathrm{s} .\) (b) Determine the maximum distance the car rolls up the hill.

A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of \(80.0 \mathrm{m} / \mathrm{s}\) at ground level. Its engines then fire and it accelerates upward at \(4.00 \mathrm{m} / \mathrm{s}^{2}\) until it reaches an altitude of \(1000 \mathrm{m} .\) At that point its engines fail and the rocket goes into free fall, with an acceleration of \(-9.80 \mathrm{m} / \mathrm{s}^{2} .\) (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (You will need to consider the motion while the engine is operating separate from the free-fall motion.)

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