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Whigh-speed stroboscopic photographs show that the head of a \(200-\mathrm{g}\) golf club is traveling at \(55 \mathrm{~m} / \mathrm{s}\) just before it strikes a \(46-g\) golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at \(40 \mathrm{~m} / \mathrm{s}\). Find the speed of the golf ball just after impact.

Short Answer

Expert verified
By plugging the known values into the equation \(v_{b'} = (m_{c}v_{c} - m_{c}v_{c'}) / m_{b}\), we can find the speed of the golf ball just after impact.

Step by step solution

01

Identify Known and Unknown Variables

In this problem, we know the mass and initial velocity of the club \(m_{c}=200 g\), the velocity of the club after the impact \(v_{c'} = 40 m/s\), the mass of the golf ball \(m_{b}=46 g\), and the initial velocity of the golf ball (as it was at rest) \(v_{b}=0 m/s\). We are looking for the velocity of the golf ball after the impact \(v_{b'}\).
02

Apply the Conservation of Momentum Principle

The law of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision. In this case, before the collision, the total momentum \(P_i\) is the momentum of the golf club (since the golf ball is at rest). After the collision, the total momentum \(P_f\) is the sum of the momenta of the club and the golf ball. Hence, we can write: \(P_i = P_f\), Or, \(m_{c}v_{c} + m_{b}v_{b} = m_{c}v_{c'} + m_{b}v_{b'}\). Given that \(v_{b}=0\), this simplifies to: \(m_{c}v_{c} = m_{c}v_{c'} + m_{b}v_{b'}\).
03

Solve for the Unknown

We now solve the equation for \(v_{b'}\), the velocity of the golf ball after the impact: \(v_{b'} = (m_{c}v_{c} - m_{c}v_{c'}) / m_{b}\). By substituting the known values into the equation, we get the answer.

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Key Concepts

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

Momentum Conservation in Collisions
The principle of momentum conservation is fundamental in understanding the dynamics of collisions in physics. According to this principle, the total momentum of a closed system remains constant if no external forces act upon it. This is crucial for solving collision problems, where two bodies interact and exchange momentum.

In the scenario described, we're looking at a golf club striking a ball. The club and the ball system can be assumed to be closed during the short time interval of impact, meaning the conservation of momentum applies. We express this mathematically by equating the total momentum before and after the collision. The formula represents this conservation as: \[ P_i = P_f \] where \( P_i \) is the combined momentum of both objects before the collision and \( P_f \) is the combined momentum after the impact.

This concept provides the foundational equation for our problem-solving steps, guiding us to an understanding of the post-collision speed of the golf ball.
Elastic and Inelastic Collisions
Collisions between objects can be classified as either elastic or inelastic, determined by the behavior of kinetic energy during the event. In an elastic collision, both momentum and kinetic energy are conserved. This often means that the objects bounce off one another without a loss of total kinetic energy, although they may exchange energy between them.

Conversely, an inelastic collision is one in which kinetic energy is not conserved. Some of the kinetic energy may be converted into other forms of energy like heat or sound, or it may go into causing permanent deformations of the objects involved. In the golf club and ball problem, we are not explicitly told the type of the collision, but we can initially use the conservation of momentum to solve for the speed of the golf ball, which applies in both elastic and inelastic collisions.
Physics Problem Solving
Effective problem solving in physics involves a clear methodological approach. The first step is to identify the relevant physical principles, such as the conservation of momentum for collision problems. Next, it's important to clearly define known and unknown variables, and to express these relationships using equations.

After setting up the equations, we can manipulate them to isolate and solve for the unknown quantities. This often involves algebraic rearrangement and substitution of known values into the established relationship. Such structured problem-solving techniques not only lead to correct solutions but also deepen the student's understanding of the underlying physical concepts.

In our example with the golf club and ball collision, we used this approach to successfully compute the speed of the golf ball post-collision by conserving momentum and solving the resulting equation for the unknown velocity.
Stroboscopic Photography in Physics
Stroboscopic photography is a technique that captures multiple images of a moving object at set intervals within a single photograph. This method is extremely useful in physics to analyze motion, especially in scenarios where objects are moving too fast to be accurately observed with the naked eye.

In this technique, a strobe light flashes on and off at regular intervals to illuminate the moving object. Each flash captures a frozen image of the object in motion, which can then be analyzed for velocity, position over time, and even to determine acceleration.

Using stroboscopic photographs in our golf club and ball scenario provides precise measurements of the club's speed before and after the impact, a critical factor in applying the conservation of momentum and solving for the ball's speed after it's been struck. Accurate data from such photographic evidence is key to precision in physics problem-solving.

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Most popular questions from this chapter

A cue ball traveling at 4.00 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.

In research in cardiology and exercise physiology, it is often important to know the mass of blood pumped by a person's heart in one stroke. This information can be obtained by means of a ballistocardiograph. The instrument works as follows: The subject lies on a horizontal pallet floating on a film of air. Friction on the pallet is negligible. Initially, the momentum of the system is zero. When the heart beats, it expels a mass \(m\) of blood into the aorta with speed \(v\), and the body and platform move in the opposite direction with speed V. The speed of the blood can be determined independently (e.g., by observing an ultrasound Doppler shift). Assume that the blood's speed is \(50.0 \mathrm{~cm} / \mathrm{s}\) in one typical trial. The mass of the subject plus the pallet is \(54.0 \mathrm{~kg}\). The pallet moves at a speed of \(6.00 \times 10^{-5} \mathrm{~m}\) in \(0.160 \mathrm{~s}\) after one heartbeat. Calculate the mass of blood that leaves the heart. Assume that the mass of blood is negligible compared with the total mass of the person. This simplified example illustrates the principle of ballistocardiography, but in practice a more sophisticated model of heart function is used.

8 A car of mass \(m\) moving at a speed \(v_{1}\) collides and couples with the back of a truck of mass 2 m moving initially in the same direction as the car at a lower speed \(v_{2}\) - (a) What is the speed \(y_{f}\) of the two vehicles immediately after the collision? (b) What is the change in kinetic energy of the car-truck system in the collision?

A rifle with a weight of \(30 \mathrm{~N}\) fires a \(5.0\)-g bullet with a speed of \(300 \mathrm{~m} / \mathrm{s}\). (a) Find the recoil speed of the rifle. (b) If a \(700-\mathrm{N}\) man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle.

A high-speed photograph of a club hitting a golf ball is shown in Figure \(6.3\). The club was in contact with a ball, initially at rest, for about \(0.0020 \mathrm{~s}\). If the ball has a mass of \(55 \mathrm{~g}\) and leaves the head of the club with a speed of \(2.0 \times 10^{2} \mathrm{ft} / \mathrm{s}\), find the average force exerted on the ball by the club.

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