Problem 21 - College Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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:\[ \Delta p = 0.25 \, \text{kg} \times v_f - 0 = 0.25 v_f \, \text{kg}\cdot\text{m/s} \]

Step 4: 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 \(v_f\):\[ v_f = \frac{3.2 \, \text{N}\cdot\text{s}}{0.25 \, \text{kg}} = 12.8 \, \text{m/s} \].

Key Concepts

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.