/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 BIO A \(75.0-\mathrm{kg}\) ice s... [FREE SOLUTION] | 91Ó°ÊÓ

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BIO A \(75.0-\mathrm{kg}\) ice skater mowing at \(10.0 \mathrm{~m} / \mathrm{s}\) crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at \(5.00 \mathrm{~m} / \mathrm{s}\). Suppose the average force a skater can experience withour breaking a bone is \(4500 \mathrm{~N}\). If the impact time is \(0.100 \mathrm{~s}\), does a bone break?

Short Answer

Expert verified
No, a bone does not break during the collision. The force experienced during the collision is less than the breaking limit.

Step by step solution

01

Calculate initial momentum

The initial momentum is calculated by multiplying the mass of the skater by their speed before the collision, using the formula \( p_{i} = mv \), where m is the mass and v is the velocity. Here, \( p_{i} = 75.0 kg \times 10.0 m/s = 750 kg \cdot m/s \)
02

Calculate final momentum

The final momentum after the collision is calculated in the same way. However, note that the total mass now includes two skaters, and the speed is their common speed after the collision. We use the formula \( p_{f} = Mv \), where M is the total mass and v is the velocity. Here, \( p_{f} = 2 \times 75.0 kg \times 5.00 m/s = 750 kg \cdot m/s \)
03

Calculate the change in momentum (impulse)

The impulse is calculated as the change in momentum, \( J = \Delta p = p_{f} - p_{i} \). For this problem, the impulse \( J = 750 kg \cdot m/s - 750 kg \cdot m/s = 0 kg \cdot m/s \).
04

Calculate the force

Since impulse is also equal to force multiplied by the impact time, we can use it to find the force during the collision. As the impulse equals zero, the force during the collision is also zero. F = J / t, where F is the force, J is the impulse, and t is the time. In this case, \( F = 0 kg \cdot m/s / 0.100 s = 0 N \).
05

Compare the force with the limit

It's now clear that the force experienced during the collision (0 N) is less than the average force a skater can experience without breaking a bone (4500 N).

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Key Concepts

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

Impulse-Momentum Theorem
The Impulse-Momentum Theorem is a fundamental principle in physics that relates the change in momentum of an object to the impulse applied to it. Momentum is the product of mass and velocity, expressed as \( p = mv \). When momentum changes, either due to a change in mass or velocity, it happens because of an applied impulse.
Impulse itself is defined as the product of force and the time period over which the force is applied, given by \( J = F \times t \). The relationship between impulse and momentum is given by the equation \( J = \Delta p \), where \( \Delta p \) is the change in momentum.
In the skater example, we see that the initial momentum was calculated, but after the collision the total momentum remained constant at 750 kg·m/s. Hence, no net impulse was applied, keeping everything in a state of equilibrium.
Collision Physics
Collision Physics delves into the interactions that objects experience when they collide. In such scenarios, forces are applied to the colliding bodies, causing changes in motion. While the momentum of individual objects might change during a collision, the total system momentum remains conserved, as seen in this ice skater example.
There are two primary types of collisions: elastic and inelastic. Elastic collisions conserve both momentum and kinetic energy; inelastic collisions only conserve momentum but not kinetic energy. The skaters experienced an inelastic collision, which is evident by their decrease in speed and shared velocity after impact.
  • The total mass post-collision was the sum of both skaters.
  • The velocity post-collision is the average of their combined momenta.
The final momentum matched the initial, exemplifying the core idea of conserved momentum, demonstrating that the overall kinetic energy wasn't preserved.
Impact Force Calculations
In this context, calculating the impact force is crucial to determine whether it can cause bodily harm. Impact force depends on the change in momentum and the time duration of the impact.
The calculation involves using the formula \( F = \frac{J}{t} \), where \( J \) is impulse and \( t \) is the time of impact. A longer impact time, given changes in momentum are constant, results in decreased force, reducing potential for damage.
For the skaters, the impulse was zero, indicating no net force during collision: \( F = \frac{0 \, \text{kg}\cdot\text{m/s}}{0.100 \, \text{s}} = 0 \, \text{N} \). This force is significantly less than the 4500 N threshold for bone fractures, confirming that neither skater experienced a force powerful enough to cause injury.
Key takeaways are:
  • Force decreases with longer impact time for a given impulse.
  • Even small force values can affect results in scenarios with brief impact times if impulse isn't zero.

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Most popular questions from this chapter

A 2.0-g particle moving at 8.0 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?

M An bullet of mass \(w=8.00 \mathrm{~g}\) is fired into a block of mass \(M=250 \mathrm{~g}\) that is initially at rest at the edge of a table of height \(h=1.00 \mathrm{~m}\) (Fig. P6.40). The bullet remains in the block, and after the impact the block lands \(d=2.00 \mathrm{~m}\) from the bottom of the table. Determine the initial speed of the bullet.

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.

S Show that the kinetic energy of a particle of mass \(m\) is related to the magnitude of the momentum \(p\) of that particle by \(K E=p^{2} / 2 \mathrm{~m}\). (Note: This expression is invalid for particles traveling at speeds near that of light.)

(a) A car traveling due east strikes a car traveling due north at an intersection, and the two move together as a unit. A property owner on the southeast corner of the intersection claims that his fence was torn down in the collision. Should he be awarded damages by the insurance company? Defend your answer. (b) Let the eastward-moving car have a mass of 1 300 kg and a speed of 30.0 km/h and the northward-moving car a mass of 1 100 kg and a speed of 20.0 km/h. Find the velocity after the collision. Are the results consistent with your answer to part (a)?

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