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Chapter 15: Problem 3

A hockey puck is traveling to the left with a velocity of \(v_{1}=10 \mathrm{~m} / \mathrm{s}\) when it is struck by a hockey stick and given a velocity of \(v_{2}=20 \mathrm{~m} / \mathrm{s}\) as shown. Determine the magnitude of the net impulse exerted by the hockey stick on the puck. The puck has a mass of \(0.2 \mathrm{~kg}\).

Short Answer

Expert verified
The magnitude of the impulse exerted on the puck by the hockey stick is 6 kg*m/s.

Step by step solution

01

Understanding Impulse and calculating Momentum

The Impulse can be found using the Impulse-Momentum Theorem, which states that the change in momentum of an object equals the impulse applied to it. This can be written as: \[ I= \Delta p=m(v_2-v_1)\] Where \(I\) is the impulse, \(\Delta p\) is the change in momentum, \(m\) is the mass of the object and \(v_2, v_1\) are the final and initial velocities respectively. Here, the momentum is the product of mass and velocity. First calculate the initial momentum which is the product of mass and the initial velocity \(v_1\) and the final momentum: the product of mass and final velocity \(v_2\).
02

Calculating Initial and Final Momentums

The initial momentum (P1) is given by the mass of the puck times its initial velocity. In this case, the initial velocity is -10 m/s (going to the left) and the mass is 0.2 kg. So, we multiply: \[ P_1 = m* v_1 = 0.2 kg * -10 m/s = -2 kg*m/s \] The negative sign indicates that the momentum was to the left. Similarly, the final momentum (P2) is given by the mass of the puck times its final velocity: \[ P_2 = m* v_2 = 0.2 kg * 20 m/s = 4 kg*m/s \]
03

Calculate Impulse

Now that we have both initial and final momenta, we can substitute these into the Impulse-Momentum theorem: \[ I = \Delta p = P_2 - P_1 = 4 kg*m/s - (-2 kg*m/s) = 6 kg*m/s \] Hence, the magnitude of the impulse exerted on the puck is 6 kg*m/s.

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Key Concepts

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

Momentum Calculation
Momentum is a crucial concept in physics, especially when exploring the movement and forces acting on an object. It is defined as the product of an object's mass and its velocity. 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 the velocity of the object.
In the context of the original exercise involving a hockey puck, we calculate the puck's momentum twice: once for its initial state before being hit, and once for its state after being hit by the hockey stick.

The initial momentum is calculated using the puck's mass and its initial velocity to the left, resulting in a value of \(-2 \text{ kg} \cdot \text{m/s}\). The negative sign indicates direction. After the hockey stick applies an impulse, the puck has a final momentum determined using the same formula, yielding \(4 \text{ kg} \cdot \text{m/s}\). Each calculation considers both magnitude and direction.
Impulse Calculation
Impulse is the change in momentum of an object when a force is applied over a period of time. The Impulse-Momentum Theorem is expressed as:\[ I = \Delta p \]where:
  • I is the impulse exerted on the object,
  • \(\Delta p\) represents the change in momentum.
This theorem allows us to calculate the impulse by considering the initial and final momentum of the object. In simple terms, impulse is the effect of applying a force to an object for a period of time, resulting in a change in the object's velocity.

From the exercise, after finding the initial and final momenta, the change in momentum \(\Delta p\) can be calculated by subtracting the initial momentum from the final momentum, i.e.,\(\Delta p = 4 - (-2) = 6 \text{ kg} \cdot \text{m/s}\). This magnitude represents the impulse delivered to the hockey puck by the hockey stick, prompting its change in velocity.
Physics Problem Solving
Solving physics problems systematically involves understanding the concepts and using the appropriate formulas. The exercise shows a clear method leveraging the Impulse-Momentum Theorem to find the impulse exerted on the puck.

Here are steps to approach such a problem:
  • Identify relevant concepts: Know what physical principles apply. In this case, momentum and impulse.
  • Use the correct formulas: Employ the momentum formula and the Impulse-Momentum Theorem provided.
  • Account for direction: Recognize that velocity and momentum are vector quantities, with direction being a crucial factor.
  • Calculate each required value: Step through determining initial and final momenta before calculating the net impulse.
  • Verify calculations: Ensure the consistency and correctness of numerical results.
By following these structured steps, solving physics problems becomes more manageable, helping you to grasp the underlying concepts effectively.

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Most popular questions from this chapter

Water is flowing from the 150 -mm-diameter fire hydrant with a velocity \(v_{B}=15 \mathrm{~m} / \mathrm{s}\). Determine the horizontal and vertical components of force and the moment developed at the base joint \(A,\) if the static (gauge) pressure at \(A\) is \(50 \mathrm{kPa}\). The diameter of the fire hydrant at \(A\) is \(200 \mathrm{~mm} \cdot \rho_{w}=1 \mathrm{Mg} / \mathrm{m}^{3}\)

A toboggan and rider, having a total mass of \(150 \mathrm{~kg}\), enter horizontally tangent to a \(90^{\circ}\) circular curve with a velocity of \(v_{A}=70 \mathrm{~km} / \mathrm{h}\). If the track is flat and banked at an angle of \(60^{\circ},\) determine the speed \(v_{B}\) and the angle \(\theta\) of "descent," measured from the horizontal in a vertical \(x-z\) plane, at which the toboggan exists at \(B\). Neglect friction in the calculation.

Each of the two stages \(A\) and \(B\) of the rocket has a mass of \(2 \mathrm{Mg}\) when their fuel tanks are empty. They each carry \(500 \mathrm{~kg}\) of fuel and are capable of consuming it at a rate of \(50 \mathrm{~kg} / \mathrm{s}\) and eject it with a constant velocity of \(2500 \mathrm{~m} / \mathrm{s}\), measured with respect to the rocket. The rocket is launched vertically from rest by first igniting stage \(B\). Then stage \(A\) is ignited immediately after all the fuel in \(B\) is consumed and \(A\) has separated from \(B\). Determine the maximum velocity of stage \(A\). Neglect drag resistance and the variation of the rocket's weight with altitude.

A ball is thrown onto a rough floor at an angle \(\theta\). If it rebounds at an angle \(\phi\) and the coefficient of kinetic friction is \(\mu,\) determine the coefficient of restitution \(e\). Neglect the size of the ball. 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 motor, \(M\), pulls on the cable with a force \(F=\left(10 t^{2}+300\right) \mathrm{N},\) where \(t\) is in seconds. If the \(100 \mathrm{~kg}\) crate is originally at rest at \(t=0,\) determine its speed when \(t=4 \mathrm{~s}\). Neglect the mass of the cable and pulleys. Hint: First find the time needed to begin lifting the crate.
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