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Chapter 9: Problem 36

A \(0.25 \mathrm{~kg}\) puck is initially stationary on an ice surface with negligible friction. At time \(t=0\), a horizontal force begins to move the puck. The force is given by \(\vec{F}=\left(12.0-3.00 t^{2}\right) \hat{\mathrm{i}}\), with \(\vec{F}\) in newtons and \(t\) in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between \(t=0.500 \mathrm{~s}\) and \(t=1.25 \mathrm{~s} ?\) (b) What is the change in momentum of the puck between \(t=0\) and the instant at which \(F=0\) ?

Short Answer

Expert verified
(a) The impulse is 7.17 Ns. (b) The change in momentum is 16 Ns.

Step by step solution

01

Determine when the force becomes zero

The given force is \( \vec{F} = (12.0 - 3.00t^2) \hat{\mathrm{i}} \). To find when the force becomes zero, set the force equation to zero and solve for \( t \):\[12.0 - 3.00t^2 = 0 \3.00t^2 = 12.0 \t^2 = 4.0 \t = 2 \text{ s}\]The force is zero at \( t = 2 \text{ s} \).
02

Calculate Impulse from t=0.5s to t=1.25s

The impulse \( J \) is the integral of the force over the given time interval. This can be calculated as:\[J = \int_{0.5}^{1.25} (12.0 - 3.00t^2) \, dt\]Solving the integral:\[J = \left[ 12t - 3\frac{t^3}{3} \right]_{0.5}^{1.25} \J = \left[ 12(1.25) - (1.25)^3 \right] - \left[ 12(0.5) - (0.5)^3 \right] \J = \left[ 15 - 1.953125 \right] - \left[ 6 - 0.125 \right] \J = 13.046875 - 5.875 \J = 7.171875 \, \mathrm{Ns}\]
03

Find the change in momentum from t=0s to t=2s

Since the change in momentum \( \Delta p \) equals the impulse over a time period, calculate from \( t = 0 \) to \( t = 2 \), where force becomes zero:\[\Delta p = \int_{0}^{2} (12.0 - 3.00t^2) \, dt\]Evaluating the integral:\[\Delta p = \left[ 12t - 3\frac{t^3}{3} \right]_{0}^{2} \\Delta p = \left[ 12(2) - (2)^3 \right] - \left[ 12(0) - (0)^3 \right] \\Delta p = \left[ 24 - 8 \right] - [0] \\Delta p = 16 \, \mathrm{Ns}\]

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Key Concepts

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

Momentum
Momentum is an important concept in physics that describes the quantity of motion an object possesses. It is a vector quantity, meaning it has both magnitude and direction. The formula for momentum is given by \[ p = m imes v \]where \( p \) is the momentum, \( m \) is the mass of the object, and \( v \) is its velocity. This ensures that more massive objects and objects moving at higher speeds have greater momentum.
  • Measured in kilogram meters per second (kg m/s).
  • Directly proportional to an object's mass and velocity.
  • A crucial aspect in collision analysis and conservation laws.
Considering the impulse-momentum theorem, impulse (change in momentum) can show us how forces affect an object's motion. For the given exercise, the impulse applied to the puck altered its momentum from being stationary to moving as a result of the applied force.
Force
Force is an interaction that changes the motion of an object when unopposed. It is measured in newtons (N) and is also a vector quantity, characterized by its magnitude and direction. In the given problem, the force acting on the puck is expressed with the equation\[ \vec{F} = (12.0 - 3.00t^2) \hat{\mathrm{i}} \]where the size of the force changes over time. This specific force is a time-dependent force, implying that it varies as time \( t \) progresses.
  • The equation suggests the force decreases with increase in \( t^2 \).
  • The direction of the force is along the \( \hat{\mathrm{i}} \) unit vector.
  • The point when the force is zero will dramatically affect the motion of the puck.
Understanding how forces work and transform motion is foundational in dynamics and analysis of motion of objects.
Newton's Second Law
Newton's Second Law connects force, mass, and acceleration. It states that the force on an object is equal to the mass of that object multiplied by its acceleration:\[ \vec{F} = m \times \vec{a} \]This law is fundamental for solving mechanics problems where motion is involved. It signifies how the motion of an object changes when subjected to an applied force.For the puck exercise:
  • Initial force accelerates the puck from rest.
  • Acceleration can be derived using \( a = \frac{F}{m} \).
  • Movement continues until \( F \) becomes zero.
This principle helps in understanding the precisely changing velocity (or speed) of the puck when the specified force acts on it over time.
Integral Calculus
Integral calculus helps calculate quantities where accumulation is involved, like finding areas under curves or, as in physics, the total result of a variable force over time. In the exercise:The impulse \( J \) and change in momentum \( \Delta p \) are determined using integrals:\[ J = \int_{t_1}^{t_2} F(t) \, dt \]\[ \Delta p = \int_{t_1}^{t_2} F(t) \, dt \]These expressions give the total impulse and momentum change when forces act over a time interval.
  • Provides a precise method for computing impulse over the variable force.
  • Involves computing definite integrals for polynomial functions of \( t \).
  • Integration bounds are dictated by time points of interest, from when force starts to when it vanishes.
By using integral calculus, you accurately predict how much an object's momentum changes under the effect of time-varying forces.

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Most popular questions from this chapter

A force in the negative direction of an \(x\) axis is applied for \(27 \mathrm{~ms}\) to a \(0.40 \mathrm{~kg}\) ball initially moving at \(14 \mathrm{~m} / \mathrm{s}\) in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude \(32.4 \mathrm{~N} \cdot \mathrm{s}\). What are the ball's (a) speed and (b) direction of travel just after the force is applied? What are (c) the average magnitude of the force and (d) the direction of the impulse on the ball?

A \(0.15 \mathrm{~kg}\) ball hits a wall with a velocity of \((5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(6.50\) \(\mathrm{m} / \mathrm{s}) \hat{\mathrm{j}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). It rebounds from the wall with a velocity of \((2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(-3.20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). What are (a) the change in the ball's momentum, (b) the impulse on the ball, and (c) the impulse on the wall?

A \(6100 \mathrm{~kg}\) rocket is set for vertical firing from the ground. If the exhaust speed is \(1200 \mathrm{~m} / \mathrm{s}\), how much gas must ejected each second if the thrust (a) is to equal the magnitude the gravitational force on the rocket and (b) is to give the rocket initial upward acceleration of \(21 \mathrm{~m} / \mathrm{s}^{2}\) ?

A stone is dropped at \(t=0\). A second stone, with twice the mass of the first, is dropped from the same point at \(t=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stones at \(t=300 \mathrm{~ms} ?\) (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the twostone system moving at that time?

Particle \(A\) and particle \(B\) are held together with a compressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of \(A\) is \(2.00\) times the mass of \(B\), and the energy stored in the spring was \(60 \mathrm{~J}\). Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle \(A\) and (b) particle \(B\) ?
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