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Chapter 8: Problem 28

Imagine that a \(1.20-\mathrm{kg}\) hard-rubber ball traveling at \(10.0 \mathrm{~m} / \mathrm{s}\) bounces off a brick wall in an essentially elastic collision. Determine the change in the momentum of the ball. [Hint: What change in momentum will just stop the ball?]

Short Answer

Expert verified
The change in momentum of the ball is -24.0 kg m/s.

Step by step solution

01

Understanding Momentum

Momentum (p"]) is a measure of an object's motion and is calculated as the product of an object's mass (m"]) and velocity ("]). The formula for momentum is given as: \[ p = mv \].
02

Initial Momentum Calculation

To find the initial momentum of the ball, we use the formula for momentum: \( p = mv \). Substituting the given values, the initial momentum \( p_i \) of the ball is: \[ p_i = 1.20 \, \text{kg} \times 10.0 \, \text{m/s} = 12.0 \, \text{kg} \, \text{m/s} \].
03

Final Momentum Calculation

After an essentially elastic collision, the ball will have the same speed but in the opposite direction. Thus, the final velocity of the ball \( v_f \) is \(-10.0 \, \text{m/s} \). The final momentum \( p_f \) is: \[ p_f = 1.20 \, \text{kg} \times (-10.0 \, \text{m/s}) = -12.0 \, \text{kg} \, \text{m/s} \].
04

Change in Momentum

The change in momentum (\( \Delta p \)) is the difference between the final momentum and the initial momentum. It is given by: \[ \Delta p = p_f - p_i \]. Substituting the values, we get: \[ \Delta p = -12.0 \, \text{kg} \, \text{m/s} - 12.0 \, \text{kg} \, \text{m/s} = -24.0 \, \text{kg} \, \text{m/s} \].

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Key Concepts

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

Elastic Collision
An elastic collision is a type of collision where the total kinetic energy and momentum of the system remain conserved before and after the interaction. In such scenarios, objects bounce off each other without deformation or generation of heat.

In an elastic collision, the relative speed of separation after collision is equal to the relative speed of approach before collision. This means that the colliding objects swap velocities, provided their masses remain constant.
  • Imagine two objects colliding elastically - they spring apart similarly to how a basketball bounces on a hard court.
  • If an object collides elastically with a wall, like the rubber ball in our exercise, it rebounds with unchanged speed but opposite direction.
In our exercise, the ball strikes a brick wall and rebounds, showcasing the essence of an elastic collision.
Momentum Calculation
Momentum is a crucial concept in physics, capturing how much mass in motion a body possesses. The basic formula for momentum is: \[ p = mv \] where \( p \) stands for momentum, \( m \) is mass, and \( v \) is velocity.

Calculating momentum involves multiplying the object's mass by its velocity. This helps to determine how much force is needed to change the object's path or bring it to a stop.
  • In our example, the initial momentum of the ball is calculated by multiplying its mass (1.20 kg) with its initial velocity (10.0 m/s), yielding 12.0 kg m/s.
  • Similarly, the final momentum considers the same mass and a reversed velocity of -10.0 m/s, resulting in -12.0 kg m/s.
This calculation is fundamental to understanding how objects behave under motion.
Change in Momentum
The change in momentum, also known as impulse, occurs when an object's velocity changes due to an unbalanced force. This can result from a collision or another force acting upon the object.

In the exercise, the change in momentum is determined by finding the difference between the object's final and initial momentum:\[ \Delta p = p_f - p_i \]In our problem, the rubber ball's initial and final momenta are 12.0 kg m/s and -12.0 kg m/s, respectively. Thus, the change is:\[ \Delta p = -12.0 \, \text{kg m/s} - 12.0 \, \text{kg m/s} = -24.0 \, \text{kg m/s} \]This substantial change reflects the reversal of the ball's velocity upon impact with the wall.
Momentum Formula
The momentum formula \( p = mv \) provides a simple method to determine an object鈥檚 momentum by multiplying its mass \( m \) by its velocity \( v \).

This formula is fundamental for calculating how much 'push' an object carries and is essential in numerous physics problems where motion is analyzed.
  • Key points about the momentum formula include the fact that momentum is a vector quantity, meaning it has both magnitude and direction.
  • The direction of momentum is the same as the object's velocity direction.
Understanding this formula helps clarify how objects interact during collisions. In the exercise, it helped demonstrate how an object's reversal in direction results in a change of momentum.

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Most popular questions from this chapter

Imagine a space vehicle floating in the void. It fires a small thruster that delivers a forward force of \(2000 \mathrm{~N}\) for \(25.0 \mathrm{~s}\). Determine the resulting change in momentum of the craft. Do you need the mass of the ship?

A 1200 -kg gun mounted on wheels shoots an \(8.00\) -kg projectile with a muzzle velocity of \(600 \mathrm{~m} / \mathrm{s}\) at an angle of \(300^{\circ}\) above the horizontal. Find the horizontal recoil speed of the gun.

Two girls (masses \(m_{1}\) and \(m_{2}\) ) are on roller skates and stand at rest, close to each other and face to face. Girl-1 pushes squarely against girl-2 and sends her moving backward. Assuming the girls move freely on their skates, write an expression for the speed with which girl-1 moves. We take the two girls to comprise the system under consideration. The problem states that girl-2 moves "backward," so let that be the negative direction; therefore, the "forward" direction is positive. There is no resultant external force on the system (the push of one girl on the other is an internal force), and so momentum is conserved: 1 from which Girl-1 recoils with this speed. Notice that if \(m_{2} / m_{1}\) is very large, \(v_{1}\) is much larger than \(v_{2}\). The velocity of girl-1, \(\overrightarrow{\mathbf{v}}_{1}\), points in the positive forward direction. The velocity of girl-2, \(\overrightarrow{\mathbf{v}}_{2}\), points in the negative backward direction. If we put numbers into the equation, \(\mathrm{v}_{2}\) would have to be negative and \(u_{1}\) would come out positive.

A 15 -g bullet moving at \(300 \mathrm{~m} / \mathrm{s}\) passes through a \(2.0\) -cm-thick sheet of foam plastic and emerges with a speed of \(90 \mathrm{~m} / \mathrm{s}\). Assuming that the speed change takes place uniformly, what average force impeded the bullet's motion through the plastic? We can determine the change in momentum, and that suggests using the impulse equation to find the force \(F\) on the bullet as it takes a time \(\Delta t\) to pass through the plastic. Taking the initial direction of motion to be positive, $$ F \Delta t=m v_{f}-m v_{i} $$ We can find \(\Delta t\) by assuming uniform deceleration and using \(x=\) \(v_{a v} t\), where \(x=0.020 \mathrm{~m}\) and \(v_{a v}=\frac{1}{2}\left(v_{i}+v_{f}\right)=195 \mathrm{~m} / \mathrm{s}\). This gives \(\Delta t=1.026 \times 10^{-4} \mathrm{~s}\). Then \((F)\left(1.026 \times 10^{-4} \mathrm{~s}\right)=(0.015 \mathrm{~kg})(90 \mathrm{~m} / \mathrm{s})-(0.015 \mathrm{~kg})(300 \mathrm{~m} / \mathrm{s})\) which yields \(F=-3.1 \times 10^{4} \mathrm{~N}\) as the average retarding force. How could this problem have been solved using \(F=\) ma instead of the impulse equation? By using energy methods?

A billiard ball moving at a speed \(v_{1 i}\) strikes, head-on, another billiard ball that is at rest. Assuming the collision is completely elastic, show that $$ v_{1 i}=v_{1 f}+v_{2 f} $$
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