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A pool player imparts an impulse of \(3.2 \mathrm{~N} \cdot \mathrm{s}\) to a stationary \(0.25-\mathrm{kg}\) cue ball with a cue stick. What is the speed of the ball just after impact?

Short Answer

Expert verified
The speed of the ball just after impact is 12.8 m/s.

Step by step solution

01

Understand Impulse and Momentum

Impulse is a quantity that describes the effect of a force acting over time. It is defined as the product of force and the time during which the force acts. The given impulse, expressed in momentum units, is used to change the momentum of the object. Here, the impulse is given as 3.2 N·s.
02

Apply Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse acting on an object is equal to the change in its momentum. Mathematically, it can be expressed as:\[ J = \Delta p \]where \( J \) is the impulse and \Delta p is the change in momentum.
03

Calculate the Change in Momentum

Change in momentum (\( \Delta p \)) can be calculated using the formula:\[ \Delta p = mv_f - mv_i \]where \(m\) is the mass of the object, \(v_f\) is the final velocity, and \(v_i\) is the initial velocity. Since the cue ball is initially stationary, \(v_i = 0\). Thus:\[ \Delta p = 0.25 \, \text{kg} \times v_f - 0 = 0.25 v_f \, \text{kg}\cdot\text{m/s} \]
04

Solve for Final Velocity

Since we know the impulse \( J = 3.2 \, \text{N}\cdot\text{s} \), substitute into the equation from Step 3:\[ 3.2 \, \text{N}\cdot\text{s} = 0.25 \, \text{kg} \times v_f \]Rearrange to solve for \(v_f\):\[ v_f = \frac{3.2 \, \text{N}\cdot\text{s}}{0.25 \, \text{kg}} = 12.8 \, \text{m/s} \].

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Key Concepts

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

Understanding Impulse
Impulse is an important concept in physics, especially when you want to understand how forces affect the motion of objects. In simple terms, impulse is the effect of a force applied over a period of time. For instance, when you hit a stationary cue ball with a cue stick, you are applying a force for a certain duration. This creates an impulse. The unit of impulse is newton-seconds (N·s) and it’s calculated as the product of the force ( F ) and the time ( t ) the force is applied:

- Impulse ( J ) = Force ( F ) × Time ( t )
When a player strikes a cue ball, for example, the impulse transfers to the ball, changing its motion from being stationary to moving with a certain velocity. Thus, the impulse is intimately tied to the change in the ball's momentum.
Momentum and Its Change
Momentum can be thought of as the 'quantity of motion' an object possesses. It is a product of the object’s mass and velocity. The greater the momentum, the harder it is to stop the object.

- Momentum ( p ) = Mass ( m ) × Velocity ( v )
In the context of our pool exercise, the pool player imparts an impulse that changes the momentum of the cue ball, initially at rest. According to the impulse-momentum theorem, the change in the momentum of the ball ( Δp ) is equivalent to the impulse imparted:

- Impulse ( J ) = Change in Momentum ( Δp ).
The momentum change can be written as: - Δp = mv_f - mv_i where m is mass, v_f is final velocity, and v_i is initial velocity. Since the cue ball starts from rest, v_i = 0 , simplifying our calculations to Δp = mv_f .
Final Velocity Calculation
Calculating the final velocity of the cue ball after it receives an impulse involves using the relationship between impulse and momentum change. From the previous discussions, the final velocity (v_f) is obtained by rearranging the impulse-momentum equation for v_f:

- J = 0.25 \, \text{kg} \times v_fGiven:- Impulse (J) = 3.2 N·s- Mass (m) = 0.25 kg
You plug these values into the equation and solve for the final velocity:

\[v_f = \frac{3.2 \, \text{N}\cdot\text{s}}{0.25 \, \text{kg}} = 12.8 \, \text{m/s}\]
So, the speed of the cue ball just after the impact is 12.8 meters per second. This demonstrates how a seemingly simple action, such as striking a ball, involves calculated physics principles to determine outcomes like speed.

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Most popular questions from this chapter

A volleyball is traveling toward you. (a) Which action will require a greater force on the volleyball, your catching the ball or your hitting the ball back? Why? (b) A 0.45 -kg volleyball travels with a horizontal velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) over the net. You jump up and hit the ball back with a horizontal velocity of \(7.0 \mathrm{~m} / \mathrm{s}\). If the contact time is \(0.040 \mathrm{~s}\), what was the average force on the ball?

In a pairs figure-skating competition, a \(65-\mathrm{kg}\) man and his \(45-\mathrm{kg}\) female partner stand facing each other on skates on the ice. If they push apart and the woman has a velocity of \(1.5 \mathrm{~m} / \mathrm{s}\) eastward, what is the velocity of her partner? (Neglect friction.)

A car with a mass of \(1500 \mathrm{~kg}\) is rolling on a level road at \(30.0 \mathrm{~m} / \mathrm{s}\). It receives an impulse with a magnitude of \(2000 \mathrm{~N} \cdot \mathrm{s}\) and its speed is reduced as much as possible by an impulse of this size. (a) Was this impulse caused by (1) the driver hitting the accelerator, (2) the driver putting on the brakes, or (3) the driver turning the steering wheel? (b) What was the car's speed after the impulse was applied?

In nuclear reactors, subatomic particles called neutrons are slowed down by allowing them to collide with the atoms of a moderator material, such as carbon atoms, which are 12 times as massive as neutrons. (a) In a head-on elastic collision with a carbon atom, what percentage of a neutron's energy is lost? (b) If the neutron has an initial speed of \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\), what will be its speed after collision?

A 15000 -N automobile travels at a speed of \(45 \mathrm{~km} / \mathrm{h}\) northward along a street, and a 7500 -N sports car travels at a speed of \(60 \mathrm{~km} / \mathrm{h}\) eastward along an intersecting street. (a) If neither driver brakes and the cars collide at the intersection and lock bumpers, what will the velocity of the cars be immediately after the collision? (b) What percentage of the initial kinetic energy will be lost in the collision?

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