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Chapter 7: Problem 86

A golf ball of mass \(45.0 \mathrm{~g}\) moving at a speed of \(120 . \mathrm{km} / \mathrm{h}\) collides head on with a French TGV high-speed train of mass \(3.8 \cdot 10^{5} \mathrm{~kg}\) that is traveling at \(300 . \mathrm{km} / \mathrm{h}\). Assuming that the collision is elastic, what is the speed of the golf ball after the collision? (Do not try to conduct this experiment!)

Short Answer

Expert verified
Answer: To find the final velocity of the golf ball after the collision, follow the step-by-step solution provided, which involves converting units to SI units, using conservation of momentum and kinetic energy, solving for v1', and converting the final velocity back to km/h if needed. By solving the equations with the given values, you will find the final velocity of the golf ball after the collision.

Step by step solution

01

Convert units to SI units

Before doing any calculations, convert the given mass and velocities into SI units. The mass of golf ball is given in grams (g), so convert it into kilograms (kg). The velocities are given in kilometers per hour (km/h), so convert them into meters per second (m/s). Conversion factors: 1 kg = 1000 g 1 km/h = 1000 m/3600 s Mass of golf ball (m1) = 45.0 g = 45.0/1000 kg = 0.045 kg Velocity of golf ball (v1) = 120 km/h = 120 * (1000/3600) m/s = 33.3 m/s Mass of train (m2) = 3.8 * 10^5 kg (already in SI units) Velocity of train (v2) = 300 km/h = 300 * (1000/3600) m/s = 83.3 m/s
02

Use conservation of momentum

For an elastic collision, the momentum before and after the collision is conserved. The total momentum before collision (P_initial) is the sum of the momentum of golf ball and the train's momentum. The total momentum after the collision (P_final) can be expressed in terms of their final velocities (v1' and v2'). P_initial = P_final m1*v1 + m2*v2 = m1*v1' + m2*v2' Since we're interested in finding the final velocity of the golf ball (v1'), we can rearrange the equation to isolate v1' on one side: v1' = (m1*v1 + m2*v2 - m2*v2')/m1
03

Use conservation of kinetic energy

For an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. We can express the initial and final kinetic energy in terms of the initial and final velocities: KE_initial = KE_final 0.5*m1*v1^2 + 0.5*m2*v2^2 = 0.5*m1*(v1')^2 + 0.5*m2*(v2')^2 Our goal is to find v1', so we need to eliminate v2' from this equation. To do that, let's first solve for v2' from the momentum equation: v2' = (m1*v1 + m2*v2 - m1*v1')/m2 Now, substitute this expression for v2' into the kinetic energy equation. This will give us an equation only in terms of v1': 0.5*m1*v1^2 + 0.5*m2*v2^2 = 0.5*m1*(v1')^2 + 0.5*m2*((m1*v1 + m2*v2 - m1*v1')/m2)^2
04

Solve for the final velocity of the golf ball (v1')

Now, we have a single equation containing only v1'. Solve this equation for v1'. This might take a few steps involving algebraic manipulation, expanding, and simplifying. Note that dividing the whole equation by 0.5 will make it easier to deal with. Then expand and simplify the equation until you isolate v1' on one side of the equation. Once you have an expression for v1' in terms of known values, plug in the values for m1, m2, v1, and v2, and solve for v1'. This will give you the final velocity of the golf ball after the collision.
05

Convert the final velocity back to km/h (optional)

Once you have found the final velocity of the golf ball in meters per second (m/s), you can convert the velocity back to kilometers per hour (km/h) by using the conversion factor: v1' (km/h) = v1' (m/s) * (3600/1000) This gives you the final velocity of the golf ball in the original units.

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Key Concepts

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

Conservation of Momentum
In physics, the principle of conservation of momentum plays a crucial role in analyzing collisions, such as the case of the golf ball and the train. Momentum is defined as the product of an object's mass and velocity. It's a vector quantity, having both magnitude and direction.

The conservation of momentum states that the total momentum of a closed system remains constant if no external forces act upon it. For an elastic collision, which is a type of collision where both kinetic energy and momentum are conserved, this principle is particularly important.
  • The total momentum before a collision is equal to the total momentum after the collision.
  • In mathematical terms, this can be written as: \[ m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \] where \( m_1 \) and \( m_2 \) are the masses of the colliding objects, and \( v_1, v_2, v_1', \) and \( v_2' \) represent their initial and final velocities.

Knowing the initial velocities and masses allows us to calculate the final velocities of the objects after the collision. This is especially useful in solving problems involving collisions, to determine the aftermath of the interaction.
Conservation of Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion, and it's calculated using the formula \( KE = \frac{1}{2}mv^2 \). In an elastic collision, such as the one between the golf ball and the train, the total kinetic energy is conserved. This means it remains the same before and after the collision.

For conservation of kinetic energy in a collision, the sum of the kinetic energies of all bodies involved, before the collision, equals the sum after the collision. The equation for this is:
\[ 0.5m_1v_1^2 + 0.5m_2v_2^2 = 0.5m_1v_1'^2 + 0.5m_2v_2'^2 \]
  • In this expression, \( m_1, v_1, v_1' \) similarly refer to the mass and initial, final velocities of one object, while \( m_2, v_2, v_2' \) refer to those of the second object.
  • The conservation of kinetic energy helps determine the aftermath of an elastic collision by allowing one to solve the equations to calculate unknown variables, such as final velocities.
Understanding these concepts is key to solving problems related to elastic collisions effectively. They provide insight into how energy and motion are distributed and conserved during such interactions.
SI Unit Conversion
When dealing with physics problems, especially those involving calculations like in our collision scenario, it's essential to use a consistent set of units, typically the International System of Units (SI). This ensures accuracy and clarity in your results.
SI units are the standard for scientific calculations: mass is measured in kilograms (kg), velocity in meters per second (m/s), and so on.
  • For example, converting mass from grams to kilograms requires dividing by 1000: \( 45.0 \text{ g} = 0.045 \text{ kg} \).
  • Velocity, given initially in kilometers per hour, must be converted to meters per second using the conversion factor: \[ 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} \] This equates to multiplying the velocity by \( \frac{1000}{3600} \).

Adopting SI units from the beginning assists in applying physical laws correctly, enabling straightforward manipulation of equations to find the solution. Without it, calculations might be prone to errors due to inconsistent units, complicating the process and results.

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Most popular questions from this chapter

After several large firecrackers have been inserted into its holes, a bowling ball is projected into the air using a homemade launcher and explodes in midair. During the launch, the 7.00 -kg ball is shot into the air with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\) at a \(40.0^{\circ}\) angle; it explodes at the peak of its trajectory, breaking into three pieces of equal mass. One piece travels straight up with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Another piece travels straight back with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). What is the velocity of the third piece (speed and direction)?

Using momentum and force principles, explain why an air bag reduces injury in an automobile collision.

NASA has taken an increased interest in near Earth asteroids. These objects, popularized in recent blockbuster movies, can pass very close to Earth on a cosmic scale, sometimes as close as 1 million miles. Most are small-less than \(500 \mathrm{~m}\) across \(-\) and while an impact with one of the smaller ones could be dangerous, experts believe that it may not be catastrophic to the human race. One possible defense system against near Earth asteroids involves hitting an incoming asteroid with a rocket to divert its course. Assume a relatively small asteroid with a mass of \(2.10 \cdot 10^{10} \mathrm{~kg}\) is traveling toward the Earth at a modest speed of \(12.0 \mathrm{~km} / \mathrm{s}\). a) How fast would a large rocket with a mass of \(8.00 \cdot 10^{4} \mathrm{~kg}\) have to be moving when it hit the asteroid head on in order to stop the asteroid? b) An alternative approach would be to divert the asteroid from its path by a small amount to cause it to miss Earth. How fast would the rocket of part (a) have to be traveling at impact to divert the asteroid's path by \(1.00^{\circ}\) ? In this case, assume that the rocket hits the asteroid while traveling along a line perpendicular to the asteroid's path.

Here is a popular lecture demonstration that you can perform at home. Place a golf ball on top of a basketball, and drop the pair from rest so they fall to the ground. (For reasons that should become clear once you solve this problem, do not attempt to do this experiment inside, but outdoors instead!) With a little practice, you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). a) If the balls are released from a height where the bottom of the basketball is at \(0.701 \mathrm{~m}\) above the ground, what is the absolute value of the basketball's momentum just before it hits the ground? b) What is the absolute value of the momentum of the golf ball at this instant? c) Treat the collision of the basketball with the floor and the collision of the golf ball with the basketball as totally elastic collisions in one dimension. What is the absolute magnitude of the momentum of the golf ball after these collisions? d) Now comes the interesting question: How high, measured from the ground, will the golf ball bounce up after its collision with the basketball?

A soccer ball with mass \(0.265 \mathrm{~kg}\) is initially at rest and is kicked at an angle of \(20.8^{\circ}\) with respect to the horizontal. The soccer ball travels a horizontal distance of \(52.8 \mathrm{~m}\) after it is kicked. What is the impulse received by the soccer ball during the kick? Assume there is no air resistance.
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