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Chapter 9: Problem 11

In a slow-pitch softball game, a \(0.200-\mathrm{kg}\) softball crosses the plate at \(15.0 \mathrm{m} / \mathrm{s}\) at an angle of \(45.0^{\circ}\) below the horizontal. The batter hits the ball toward center field, giving it a velocity of \(40.0 \mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal. (a) Determine the impulse delivered to the ball. (b) If the force on the ball increases linearly for \(4.00 \mathrm{ms}\), holds constant for \(20.0 \mathrm{ms},\) and then decreases to zero linearly in another \(4.00 \mathrm{ms},\) what is the maximum force on the ball?

Short Answer

Expert verified
To determine the impulse, first find the initial and final momentum in both the horizontal and vertical directions and subtract these to find the change in momentum (impulse). The impulse can then be calculated as the square root of the sum of the squares of the changes in the horizontal and vertical momentum (following Pythagoras' Theorem). The maximum force is then determined as twice the impulse divided by the sum of the time at which the force is maximum and when it starts and ends, as the shape under the force-time graph is a trapezoid.

Step by step solution

01

Calculate the initial and final momentum

The initial momentum \(p_{i}\) and the final momentum \(p_{f}\) in the horizontal and vertical components separately, using the equations: \[ p_{i_{x}} = m * V_{i_{x}} = 0.200 kg *\(-15 m/s * cos(45)\) \] \[ p_{i_{y}} = m * V_{i_{y}} = 0.200 kg *\(-15 m/s * sin(45)\) \] \[p_{f_{x}} = m * V_{f_{x}} = 0.200 kg *\(-40 m/s * cos(30)\) \] \[ p_{f_{y}} = m * V_{f_{y}} = 0.200 kg *\(-40 m/s * sin(30)\) \]
02

Determine the impulse

The impulse \( J \) delivered to the ball can be calculated as: \[ J = 螖p = p_{f} - p_{i} \] Doing this for each component: \[ J_{x} = 螖p_{x} = p_{f_{x}} - p_{i_{x}} \] and \[ J_{y} = 螖p_{y} = p_{f_{y}} - p_{i_{y}} \] Then, find the magnitude of the impulse vector \( J \) using Pythagoras theorem: \[ J = sqrt(J_{x}^{2} + J_{y}^{2}) \]
03

Find the maximum force

Since the force increases linearly for 4 ms, stays constant for 20 ms, and then decreases linearly for another 4 ms, the shape under the force-time graph is a trapezoid. The area of this trapezoid (the impulse) is calculated as \[ J = (1/2)(F_{max} + F_{0})(T_{max} + T_{0}) \] The force at the start and end is zero (\(F_{0}\)) and the time of the maximum force (\(T_{max}\)) is 20 ms while \(T_{0}\) is 4 ms. Simplifying the above equation for \(F_{max}\), we get \[ F_{max} = 2J /(T_{max} + T_{0})-F_{0} \] Thus, we can determine the maximum force on the ball.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics, representing the quantity of motion an object has. It is a vector quantity with both magnitude and direction, and is calculated by multiplying the mass of an object by its velocity. In the given problem, we're looking at momentum in terms of its components along the horizontal (x) and vertical (y) axes before and after a softball is hit by a batter.

Understanding momentum is crucial for analyzing collisions and interactions between objects. If we know the mass and velocity of an object, we can compute its momentum using the equation:
\[ p = m \times v \]Where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. For two-dimensional motion, we separately calculate the x and y components of the momentum, considering angles and direction.
Force-Time Graph
A force-time graph is a powerful tool in physics that visually represents the variation of force exerted on an object over time. The area under the curve on this graph is of particular interest because it quantifies impulse, which is equal to the change in momentum of the object.

In our given exercise, the graph would consist of three linear sections representing the increase, constancy, and decrease of the force applied to the softball over specific time intervals. The trapezoidal shape under the force-time curve corresponds to the impulse delivered to the softball. To put it simply, determining the area of the trapezoid on the force-time graph provides the magnitude of impulse.
\[ J = \text{Area under the force-time graph} \]By knowing impulse, we can backtrack to find the maximum force when the shape of the force-time curve and the durations of force application are known.
Kinematics in Two Dimensions
Kinematics in two dimensions is an aspect of physics that describes the motion of objects in planes using variables such as displacement, velocity, acceleration, and time. This type of motion is more complex because it involves separate calculations for the horizontal and vertical components.

In analyzing the softball's trajectory post-hit, we are dealing with kinematics in two dimensions, as it moves in a curved path with both horizontal and vertical motions. The initial and final velocities given at angles require us to use trigonometric functions to compute individual components. This is essential for finding out how much the softball鈥檚 momentum changes, which tells us the impulse.
For example, the horizontal velocity component is found using \( v_x = v \cos(\theta) \) and the vertical component with \( v_y = v \sin(\theta) \), where \( v \) is the velocity and \( \theta \) is the angle of motion relative to the horizontal.

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Most popular questions from this chapter

A 10.0 -g bullet is fired into a stationary block of wood \((m=\) \(5.00 \mathrm{kg}) .\) The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is \(0.600 \mathrm{m} / \mathrm{s} .\) What was the original speed of the bullet?

A neutron in a nuclear reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) If the initial kinetic energy of the neutron is \(1.60 \times 10^{-13} \mathrm{J},\) find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is nearly 12.0 times the mass of the neutron.)

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton's third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at \(8.00 \mathrm{m} / \mathrm{s}\) and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the drivers, the total vehicle masses are \(800 \mathrm{kg}\) for the car and 4000 kg for the truck. If the collision time is \(0.120 \mathrm{s},\) what force does the seatbelt exert on each driver?

Two blocks of masses \(M\) and \(3 M\) are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig. P9.4). A cord initially holding the blocks together is burned; after this, the block of mass \(3 M\) moves to the right with a speed of \(2.00 \mathrm{m} / \mathrm{s}\) (a) What is the speed of the block of mass \(M ?\) (b) Find the original elastic potential energy in the spring if \(M=0.350 \mathrm{kg}\)

A glider of mass \(m\) is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant \(k\) compressed by a distance \(x\). The glider is released from rest. (a) Show that the glider attains a speed of \(v=x(k / m)^{1 / 2} .\) (b) Does a glider of large or of small mass attain a greater speed? (c) Show that the impulse imparted to the glider is given by the expression \(x(k m)^{1 / 2} .\) (d) Is a greater impulse injected into a large or a small mass? (e) Is more work done on a large or a small mass?
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