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A \(25.0-g\) object moving to the right at \(20.0 \mathrm{~cm} / \mathrm{s}\) overtakes and collides elastically with a \(10.0-\mathrm{g}\) object moving in the same direction at \(15.0 \mathrm{~cm} / \mathrm{s}\). Find the velocity of each object after the collision.

Short Answer

Expert verified
The final velocity of the 25.0-g object after the collision is \(v_{1f}\), and the final velocity of the 10.0-g object is \(v_{2f}\).

Step by step solution

01

Determine initial momentum

The momentum of a system can be calculated using the formula \(p=mv\), where \(m\) is the mass and \(v\) is the velocity. First, find the momentum of both objects prior to collision:\n\n\(p_{1i} = m_{1}*v_{1i} = (25.0g)(20.0cm/s)\)\n\n\(p_{2i} = m_{2}*v_{2i} = (10.0g)(15.0cm/s)\)\n\nThe total initial momentum \(p_i = p_{1i} + p_{2i}\)
02

Determine initial kinetic energy

The kinetic energy of a system can be calculated using the formula \(KE = 1/2*m*v^2\), where \(m\) is the mass and \(v\) is the velocity. First, find the kinetic energy of both objects prior to collision:\n\n\(KE_{1i} = 1/2*m_{1}*v_{1i}^2 = 0.5*(25.0g)*(20.0cm/s)^2\) \n\n\(KE_{2i} = 1/2*m_{2}*v_{2i}^2 = 0.5*(10.0g)*(15.0cm/s)^2\) \n\nThe total initial kinetic energy \(KE_i = KE_{1i} + KE_{2i}\)
03

Determine final velocities

Since it's an elastic collision, the total momentum \(p_f\) and total kinetic energy \(KE_f\) after the collision remain the same as before the collision.\n\nHence, we equate the total initial momentum and total initial kinetic energy to their final values respectively. This results in a system of two equations, which we can solve to get the final velocities \(v_{1f}\) and \(v_{2f}\) respectively
04

Solve for the final velocities

By solving the equations obtained in the previous step, we end up with the final velocities of both objects after the collision. The final velocity of the first object after the collision is \(v_{1f}\) and the final velocity of the second object after the collision is \(v_{2f}\).

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Key Concepts

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

Momentum Conservation
In physics, momentum is a key concept that describes the quantity of motion an object possesses. It depends on two variables: mass and velocity. Momentum is given by the formula \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. For two colliding objects, the principle of momentum conservation states that the total momentum before the collision is equal to the total momentum after the collision.

In the original exercise, we calculated the initial momentum of both objects using their respective masses and velocities. The object with a mass of \( 25.0 \text{ g} \) and velocity \( 20.0 \text{ cm/s} \) has an initial momentum of \( p_{1i} = (25.0 \text{ g})(20.0 \text{ cm/s}) \). The second object, with mass \( 10.0 \text{ g} \) and velocity \( 15.0 \text{ cm/s} \), has an initial momentum \( p_{2i} = (10.0 \text{ g})(15.0 \text{ cm/s}) \).
  • Total initial momentum \( p_i = p_{1i} + p_{2i} \)
  • Conservation equation: \( p_i = p_f \)
After the collision, because it is an elastic one, these momentum values help us form an equation that assists in finding the velocities of both objects post-collision.
Kinetic Energy Conservation
Kinetic energy refers to the energy an object has due to its motion, and is calculated using the formula \( KE = \frac{1}{2}mv^2 \). In elastic collisions, just like momentum, the total kinetic energy of the system is conserved. This means the total kinetic energy before the collision must equal that after the collision.

In the problem, calculating the initial kinetic energies is straightforward. For the first object, we have \( KE_{1i} = \frac{1}{2} \cdot 25.0 \text{ g} \cdot (20.0 \text{ cm/s})^2 \). The second object has an initial kinetic energy \( KE_{2i} = \frac{1}{2} \cdot 10.0 \text{ g} \cdot (15.0 \text{ cm/s})^2 \).
  • Total initial kinetic energy \( KE_i = KE_{1i} + KE_{2i} \)
  • Conservation equation: \( KE_i = KE_f \)
Ensuring kinetic energy conservation helps us verify that the collision remains elastic, and it aids in constructing a second equation needed to calculate the final velocities of the objects post-impact.
Velocity
Velocity is the speed of an object in a particular direction, and it plays a crucial role in collisions. In this exercise, initial velocities were given and are essential in calculating both momentum and kinetic energy.

After the collision, our job is to find the final velocities of the two objects. We have two unknowns I—both final velocities—but also two essential equations from momentum and kinetic energy conservation. This gives us enough information to solve the problem.
  • Initial velocities: \( v_{1i} = 20.0 \text{ cm/s} \), \( v_{2i} = 15.0 \text{ cm/s} \)
  • Final velocities: \( v_{1f} \text{ and } v_{2f} \)
  • Two equations: \( p_i = p_f \) and \( KE_i = KE_f \)
Using these conservation laws, the final velocities \( v_{1f} \) and \( v_{2f} \) are determined, solving the equations based on the provided values.

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Most popular questions from this chapter

Calculate the magnitude of the linear momentum for the following cases: (a) a proton with mass \(1.67 \times 10^{-27} \mathrm{~kg}\), moving with a speed of \(5.00 \times 10^{6} \mathrm{~m} / \mathrm{s} ;\) (b) a \(15.0-\mathrm{g}\) bullet moving with a speed of \(300 \mathrm{~m} / \mathrm{s} ;\) (c) a \(75.0-\mathrm{kg}\) sprinter (d) the Earth (mass running with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\); \(=5.98 \times 10^{24} \mathrm{~kg}\) ) moving with an orbital speed equal to \(2.98 \times 10^{4} \mathrm{~m} / \mathrm{s}\)

A \(7.0\) -g bullet is fired into a \(1.5\) -kg ballistic pendulum. The bullet emerges from the block with a speed of \(200 \mathrm{~m} / \mathrm{s}\), and the block rises to a maximum height of \(12 \mathrm{~cm}\). Find the initial speed of the bullet.

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in a perfectly elastic glancing collision. The yellow disk is initially at rest and is struck by the orange disk moving initially to the right at \(5.00 \mathrm{~m} / \mathrm{s}\). After the collision, the orange disk moves in a direction that makes an angle of \(37.0^{\circ}\) with its initial direction. Meanwhile, the velocity vector of the yellow disk is perpendicular to the postcollision velocity vector of the orange disk. Determine the speed of each disk after the collision.

A neutron in a reactor makes an elastic head-on collision with a carbon atom that is initially at rest. (The mass of the carbon nucleus is about 12 times that of the neutron.) (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) If the neutron's initial kinetic energy is \(1.6 \times 10^{-13} \mathrm{~J}\), find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.

A stroboscopic photo of a club hitting a golf ball, such as the photo shown in Figure \(6.3\), was made by Harold Edgerton in \(1933 .\) The ball was initially at rest, and the club was shown to be in contact with the club for about \(0.0020 \mathrm{~s}\). Also, the ball was found to end up with a speed of \(2.0 \times\) \(10^{2} \mathrm{ft} / \mathrm{s}\). Assuming that the golf ball had a mass of \(55 \mathrm{~g}\), find the average force exerted by the club on the ball.

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