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8\. A \(75.0\) -kg ice skater moving 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 without breaking a bone is \(4500 \mathrm{~N}\). If the impact time is \(0.100 \mathrm{~s}\), does a bone break?

Short Answer

Expert verified
Yes, a bone break is likely because the impact force, which is 7500N, exceeds the average tolerable force of 4500N.

Step by step solution

01

Identify what is given

We have the combined mass of skaters, \(M = 2*75.0 \mathrm{~kg} = 150.0 \mathrm{~kg}\). Their initial velocity, \(v_{i} = 10.0 \mathrm{~m/s}\), final velocity, \(v_{f} = 5.0 \mathrm{~m/s}\) and the duration of impact \(\Delta t = 0.100 \mathrm{~s}\).
02

Calculate acceleration

We calculate the acceleration by formula, \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v = v_{f} - v_{i}\). Substituting the values we get, \(\Delta v = 5.0 \mathrm{~m/s} - 10.0 \mathrm{~m/s} = -5.0 \mathrm{~m/s}\), which gives \(a = \frac{-5.0 \mathrm{~m/s}}{0.100 \mathrm{~s}} = -50.0 \mathrm{~m/s^2}\). The negative sign indicates that it's a deceleration, not an acceleration.
03

Calculate the force

We use Newton’s second law of motion to find the force which is \(F = m*a\). Substituting the values, we get \(F = 150.0 \mathrm{~kg} * -50.0 \mathrm{~m/s^2} = -7500 \mathrm{~N}\). The negative sign indicates that the force is exerted in the opposite direction in contrast to the direction of the motion. It causes the skaters to slow down.
04

Compare the force with the bone-breaking limitation

The calculated force \(7500 \mathrm{~N}\) is larger than the bone-breaking limit of \(4500 \mathrm{~N}\). Despite having a negative sign which only indicates direction, what's important here is its absolute value. Therefore, the bone will break at such impact force.

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Key Concepts

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

Impact Force
In the world of collision physics, impact force is a critical concept. It's the force exerted when two bodies collide, such as when ice skaters bump into each other. The magnitude of this force determines the outcome of the collision — whether objects stick together or bounce apart.
The impact force can be calculated using the formula derived from Newton’s second law, which is impacted by the mass of the objects and the change in velocity over the time duration of the impact. When the time of impact is shorter, the force is typically greater. This is why a sudden stop, like crashing into another skater, can be so forceful and damaging.
  • Short impact duration = higher force
  • Force depends on mass and velocity change
  • High impact force can cause injury
When the skaters collide in the textbook problem, the force was calculated at 7500 N, exceeding the threshold of 4500 N, indicating potential injury.
Newton's Second Law
Newton's second law is central to understanding motion and collisions. This law states that the force exerted on an object is equal to the rate of change of its momentum. In simpler terms: \[ F = m imes a \]Where \( F \) denotes force, \( m \) represents mass, and \( a \) signifies acceleration. This relationship forms the backbone of calculating forces in motion problems.
When applying this to the skater collision, you first determine the deceleration, then use it along with the combined mass of the skaters to find the force exerted during their collision.
  • Connects force, mass, and acceleration
  • Vital for calculating impact force in collisions
  • Helps determine safe limits in real-world impacts
Momentum Conservation
Momentum conservation is a key principle in physics, especially in collisions. It states that in the absence of external forces, the total momentum of a system remains constant. For colliding bodies like the ice skaters in the exercise, this means that the total initial momentum before collision equals the total final momentum after collision.
The formula for momentum \((p)\) is:\[ p = m imes v \]Where \( m \) is mass, and \( v \) is velocity. In our scenario, the stationary skater and the moving skater, after collision, share a combined momentum. This principle allowed us to solve for the final velocity after their interaction.
  • Total momentum before = total momentum after
  • Helps predict post-collision motion
  • Useful in calculating unknown velocities
Deceleration
Deceleration happens when an object slows down, and it is the negative form of acceleration. It's crucial in understanding collisions as it indicates the rate at which an object's velocity changes.
Using the skaters' initial and final velocities, the deceleration is calculated by:\[ a = \frac{\Delta v}{\Delta t} \]Where \( \Delta v \) is the change in velocity and \( \Delta t \) is the impact duration. A deceleration suggests the object is losing speed, as seen in our example where the deceleration is \(-50.0 \mathrm{~m/s^2}\).
  • Negative value indicates slowing down
  • Calculated as velocity change over time
  • Important for determining impact forces
By understanding deceleration, we gauge how quickly the skaters stop moving post-collision, providing insights into the force exerted.

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Most popular questions from this chapter

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ecp Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that exerted onthe truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false; Newton's third law tells us that both objects are acted upon by forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at \(8.00 \mathrm{~m} / \mathrm{s}\) and that they undergo a perfectly inelastic headon collision. Each driver has mass \(80.0 \mathrm{~kg} .\) Including the masses of the drivers, the total masses of the vehicles are \(800 \mathrm{~kg}\) for the car and \(4000 \mathrm{~kg}\) for the truck. If the collision time is \(0.120 \mathrm{~s}\), what force does the seat belt exert on each driver?

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