Question

A particle \(\left(M_{1}=1.00 \mathrm{~kg}\right)\) moving at \(30.0^{\circ}\) downward from the horizontal with \(v_{1}=2.50 \mathrm{~m} / \mathrm{s}\) hits a second particle \(\left(M_{2}=2.00 \mathrm{~kg}\right)\) which is at rest. After the collision, the speed of \(M_{1}\) is reduced to \(0.500 \mathrm{~m} / \mathrm{s},\) and it is moving to the left and at an angle of \(32.0^{\circ}\) downward withrespect to the horizontal. You cannot assume that the collision is elastic. What is the speed of \(M_{2}\) after the collision?

Short answer

Use the given initial conditions and the conservation of momentum principle to find your answer. Initial conditions: Particle 1: mass = 1.00 kg, speed = 2.50 m/s, direction = 30.0° Particle 2: mass = 2.00 kg, at rest Final condition of particle 1: speed = 0.500 m/s, direction = 32.0° Final speed of particle 2:

Step-by-step solution

Write down the principle of linear momentum conservation

According to the principle of linear momentum conservation, the total momentum before and after the collision is the same. This means that the sum of the momentum of the two particles before the collision will be equal to the momentum of the two particles after the collision. We will write this principle separately for the x and y components of the momentum.

Write the momentum conservation equation in the x-direction

For the x-direction, the conservation of momentum equation is: $$m_1v_{1x} + m_2v_{2x} = m_{1}v'_{1x} + m_2v'_2x$$, where \(m_1, m_2\) are the masses of the particles, \(v_{1x}, v_{2x}\) are their initial velocities in the x-direction, and \(v'_{1x}, v'_{2x}\) are their final velocities in the x-direction.

Write the momentum conservation equation in the y-direction

For the y-direction, the conservation of momentum equation is: $$ m_1v_{1y} + m_2v_{2y} = m_{1}v'_{1y} + m_2v'_2y$$, where \(m_1, m_2\) are the masses of the particles, \(v_{1y}, v_{2y}\) are their initial velocities in the y-direction, and \(v'_{1y}, v'_{2y}\) are their final velocities in the y-direction.

Determine the known values

From the given information, we have: \(m_1 = 1.00\,\text{kg}\), \(v_1 = 2.50\,\text{m/s}\), \(v_1'\) = 0.500\,\text{m/s}$, \(m_2 = 2.00\,\text{kg}\), \(v_2 = 0\,\text{m/s}\) (since the second particle is at rest) We also have the angles for both particles: \(\theta_1 = 30.0^\circ\) (particle 1 before the collision) \(\theta'_1 = 32.0^\circ\) (particle 1 after the collision)

Compute the initial and final velocity components

Compute the components of the velocities for the two particles before and after the collision: \(v_{1x} = v_1 \cos\theta_1 = 2.50\,\text{m/s} \cdot \cos30^\circ\), \(v_{1y} = v_1 \sin\theta_1 = 2.50\,\text{m/s} \cdot \sin30^\circ\), \(v'_{1x} = v_1' \cos\theta'_1 = 0.500\,\text{m/s} \cdot \cos32^\circ\), \(v'_{1y} = v_1' \sin\theta'_1 = 0.500\,\text{m/s} \cdot \sin32^\circ\)

Replace the known values in the momentum conservation equations

Substitute the known values into the momentum conservation equations for the x and y components: For x-direction: $$1.00\,\text{kg} \cdot (2.50\,\text{m/s} \cdot \cos30^\circ) = 1.00\,\text{kg} \cdot (0.500\,\text{m/s} \cdot \cos32^\circ) + 2.00\,\text{kg} \cdot v'_2x$$ For y-direction: $$1.00\,\text{kg} \cdot (2.50\,\text{m/s} \cdot \sin30^\circ) = 1.00\,\text{kg} \cdot (0.500\,\text{m/s} \cdot \sin32^\circ) + 2.00\,\text{kg} \cdot v'_2y$$

Solve for the final velocity components \(v'_2x\) and \(v'_2y\)

Solve the x and y momentum conservation equations for \(v'_2x\) and \(v'_2y\), respectively: \(v'_2x = \frac{1.00\,\text{kg} \cdot (2.50\,\text{m/s} \cdot \cos30^\circ) - 1.00\,\text{kg} \cdot (0.500\,\text{m/s} \cdot \cos32^\circ)}{2.00\,\text{kg}}\) \(v'_2y = \frac{1.00\,\text{kg} \cdot (2.50\,\text{m/s} \cdot \sin30^\circ) - 1.00\,\text{kg} \cdot (0.500\,\text{m/s} \cdot \sin32^\circ)}{2.00\,\text{kg}}\)

Find the magnitude of the final velocity, \(v'_2\)

The magnitude of the final velocity, \(v'_2\), can be found using the Pythagorean theorem: $$v'_2 = \sqrt{(v'_2x)^2 + (v'_2y)^2}$$ Substitute the values of \(v'_2x\) and \(v'_2y\) and solve for \(v'_2\)- this gives us the final speed of particle 2 after the collision.

Key concepts

Momentum Conservation Equations

When we talk about collisions in physics, one of the most critical ideas is the conservation of linear momentum. The principle of linear momentum conservation states that if no external forces act on a system, the total momentum of the system remains constant. For a two-particle system, this means that the momentum before and after a collision is the same. To apply this principle, we especially look at the components of the momentum in the horizontal (x) and vertical (y) direction.

Let's use an example. Suppose two particles collide. Particle 1 has mass m1 and initial velocity v1, and Particle 2 has mass m2 and is initially at rest. We can write two separate equations for the x and y components of momentum:

• For the x-direction: \(m_1v_{1x} + m_2v_{2x} = m_{1}v'_{1x} + m_2v'_2x\)
• For the y-direction: \(m_1v_{1y} + m_2v_{2y} = m_{1}v'_{1y} + m_2v'_2y\)

Here, the primed variables (v'1x, v'1y, v'2x, v'2y) represent velocities after the collision. These equations are foundational for determining the velocities of the particles after any collision, provided that no external forces interfere.

Inelastic Collision

When analyzing collisions, we often categorize them as elastic or inelastic based on whether kinetic energy is conserved during the collision. An inelastic collision is one where the kinetic energy is not conserved. Instead, some of it is transformed into other forms of energy, such as heat or sound. However, even in inelastic collisions, linear momentum is conserved.

In the case we're looking at, the problem specifies that we 'cannot assume that the collision is elastic.' This means that we should not expect the kinetic energies before and after the collision to be equal. In inelastic collisions, we often find that objects stick together or deform, which is why kinetic energy isn't conserved. But, for our momentum calculations, that doesn't matter – conservation of momentum still applies and is key to solving the problem at hand.

Velocity Components

The concept of velocity components is crucial when dealing with collisions that occur at an angle. In two dimensions, any velocity vector can be broken down into horizontal (x) and vertical (y) components. For instance, if an object is moving at an angle to the horizontal, its velocity in the x-direction is obtained by multiplying its speed by the cosine of the angle, and its velocity in the y-direction is found by multiplying its speed by the sine of the angle.

In our textbook exercise, particle 1 has initial and final velocities at angles to the horizontal. To work with these velocities in momentum conservation equations, we must break them down into components:

• \(v_{1x} = v_1 \times \text{cos}(\theta_1)\)
• \(v_{1y} = v_1 \times \text{sin}(\theta_1)\)
• \(v'_{1x} = v'_1 \times \text{cos}(\theta'_1)\)
• \(v'_{1y} = v'_1 \times \text{sin}(\theta'_1)\)

This breakdown allows us to handle the x and y directions separately in our momentum conservation equations. Knowing how to calculate these components is critical for understanding the motion of objects after a collision.