Completely Inelastic Collisions: Videos & Practice Problems

In a completely inelastic collision, two objects collide and stick together, moving with a common final velocity. For example, a 1 kg block colliding with a 9 kg block at 20 m/s results in both moving together at 2 m/s after the collision. This outcome is derived from the conservation of momentum, expressed as m_1 × v_1i + m_2 × v_2i = (m1 + m2)v_f. This illustrates how increased mass leads to decreased speed, maintaining momentum conservation.

Completely Inelastic Collilsions

In physics, a completely inelastic collision is a specific type of collision where two objects collide and stick together, moving as a single entity after the impact. This scenario is characterized by the conservation of momentum, which states that the total momentum before the collision equals the total momentum after the collision.

To illustrate this concept, consider a 1-kilogram block colliding with a 9-kilogram block that is initially at rest. The initial speed of the 1-kilogram block is 20 meters per second. In a completely inelastic collision, both blocks will move together with the same final velocity after the collision.

The conservation of momentum can be expressed with the formula:

$$m_1 v_{1\text{initial}} + m_2 v_{2\text{initial}} = (m_1 + m_2) v_{\text{final}}$$

Here, \(m_1\) is the mass of the first block (1 kg), \(v_{1\text{initial}}\) is its initial velocity (20 m/s), \(m_2\) is the mass of the second block (9 kg), and \(v_{2\text{initial}}\) is its initial velocity (0 m/s). The final velocity \(v_{\text{final}}\) is what we want to determine.

Substituting the known values into the equation gives:

$$1 \times 20 + 9 \times 0 = (1 + 9) v_{\text{final}}$$

This simplifies to:

$$20 = 10 v_{\text{final}}$$

Solving for \(v_{\text{final}}\) yields:

$$v_{\text{final}} = \frac{20}{10} = 2 \text{ m/s}$$

Thus, after the collision, the combined mass of the two blocks (10 kg) moves at a final speed of 2 m/s. This demonstrates how, in a completely inelastic collision, the total momentum is conserved even though the speed of the system decreases due to the increase in mass. The initial momentum of the system is transformed into the motion of the combined mass, illustrating the fundamental principles of momentum conservation in physics.

A 70-kg hockey player skates on frictionless ice at 5m/s. An opposing player of mass 110kg collides head-on with him at 4m/s. The two players become entangled after the collision and move together. Determine their final velocity.

• A. 0.5 m/s in the direction of the opposing player
• B. 0.5 m/s in the direction of the 70 kg player
• C. 4.39 m/s in the direction of the opposing player
• D. 4.39 m/s in the direction of the 70 kg player

Answer: (A)

Bullet Fired Into Block

In this scenario, we analyze a bullet embedding itself into a block of wood, illustrating a completely inelastic collision. The bullet has a mass of 0.01 kg, while the block of wood has a mass of 5 kg. After the collision, both the bullet and the block move together with a final velocity of 0.6 m/s. Our goal is to determine the initial speed of the bullet.

To solve this, we apply the principle of conservation of momentum, which states that the total momentum before the collision equals the total momentum after the collision. The equation can be expressed as:

$$m_1 v_{1\text{initial}} + m_2 v_{2\text{initial}} = (m_1 + m_2) v_{\text{final}}$$

Here, \(m_1\) is the mass of the bullet, \(m_2\) is the mass of the block, \(v_{1\text{initial}}\) is the initial velocity of the bullet, and \(v_{2\text{initial}}\) is the initial velocity of the block, which is 0 since it is stationary. The final velocity \(v_{\text{final}}\) is given as 0.6 m/s.

Substituting the known values into the equation, we have:

$$0.01 v_{1\text{initial}} + 5 \cdot 0 = (0.01 + 5) \cdot 0.6$$

This simplifies to:

$$0.01 v_{1\text{initial}} = 5.01 \cdot 0.6$$

Calculating the right side gives:

$$0.01 v_{1\text{initial}} = 3.006$$

Dividing both sides by 0.01 yields:

$$v_{1\text{initial}} = 301 \text{ m/s}$$

This result indicates that the bullet was traveling at a speed of 301 m/s before the collision.

Next, we examine the kinetic energy before and after the collision to determine how much energy is lost during the inelastic collision. The kinetic energy (KE) is calculated using the formula:

$$KE = \frac{1}{2} m v^2$$

For the final kinetic energy, since both objects move together after the collision, we use the combined mass:

$$KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v_{\text{final}}^2 = \frac{1}{2} (0.01 + 5) (0.6)^2$$

Calculating this gives:

$$KE_{\text{final}} = \frac{1}{2} \cdot 5.01 \cdot 0.36 = 0.902 \text{ joules}$$

For the initial kinetic energy, only the bullet contributes since the block is stationary:

$$KE_{\text{initial}} = \frac{1}{2} m_1 v_{1\text{initial}}^2 = \frac{1}{2} \cdot 0.01 \cdot (301)^2$$

Calculating this yields:

$$KE_{\text{initial}} = \frac{1}{2} \cdot 0.01 \cdot 90601 = 453.005 \text{ joules}$$

To find the change in kinetic energy (ΔKE), we subtract the final kinetic energy from the initial kinetic energy:

$$\Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 0.902 - 453.005 = -452.103 \text{ joules}$$

This negative value indicates a significant loss of kinetic energy, which is transformed into other forms of energy, such as heat and sound, during the collision. Thus, the system loses almost all of its kinetic energy as the bullet embeds itself into the wood.

2D Collision

In this scenario, we analyze a completely inelastic collision between a car and a truck, where both vehicles lock together after the collision. The car, with a mass of 1,000 kg, is traveling east at a velocity of 20 m/s, while the truck, with a mass of 3,000 kg, is moving north at 10 m/s. The goal is to determine the final velocity and direction of the combined mass after the collision.

To begin, we establish a coordinate system with the x-axis representing east and the y-axis representing north. The total mass after the collision is the sum of the two vehicles, which is 4,000 kg (1,000 kg + 3,000 kg). The final velocity vector, denoted as \( \mathbf{v}_{\text{final}} \), will have both x and y components, \( \mathbf{v}_{\text{final}_x} \) and \( \mathbf{v}_{\text{final}_y} \).

Using the principle of conservation of momentum, we can write separate equations for the x and y components:

For the x-axis:

\[ m_a v_{a_x} + m_b v_{b_x} = (m_a + m_b) v_{\text{final}_x} \] Substituting the known values: \[ (1,000 \, \text{kg} \cdot 20 \, \text{m/s}) + (3,000 \, \text{kg} \cdot 0) = 4,000 \, \text{kg} \cdot v_{\text{final}_x \] This simplifies to: \[ 20,000 = 4,000 v_{\text{final}_x} \] Thus, we find: \[ v_{\text{final}_x} = 5 \, \text{m/s} \]

For the y-axis:

\[ m_a v_{a_y} + m_b v_{b_y} = (m_a + m_b) v_{\text{final}_y} \] Substituting the known values: \[ (1,000 \, \text{kg} \cdot 0) + (3,000 \, \text{kg} \cdot 10 \, \text{m/s}) = 4,000 \, \text{kg} \cdot v_{\text{final}_y} \] This simplifies to: \[ 30,000 = 4,000 v_{\text{final}_y} \] Thus, we find: \[ v_{\text{final}_y} = 7.5 \, \text{m/s} \]

Now that we have both components, we can calculate the magnitude of the final velocity:

\[ v_{\text{final}} = \sqrt{v_{\text{final}_x}^2 + v_{\text{final}_y}^2} = \sqrt{(5 \, \text{m/s})^2 + (7.5 \, \text{m/s})^2} = \sqrt{25 + 56.25} = \sqrt{81.25} \approx 9.01 \, \text{m/s} \]

To find the direction of the final velocity, we use the inverse tangent function:

\[ \theta = \tan^{-1}\left(\frac{v_{\text{final}_y}}{v_{\text{final}_x}}\right) = \tan^{-1}\left(\frac{7.5}{5}\right) \approx 56.3^\circ \]

In conclusion, after the collision, the combined mass of the car and truck moves with a final velocity of approximately 9.01 m/s at an angle of 56.3 degrees with respect to the x-axis (east).

Three-Body Collisions

In this problem, we analyze a scenario involving a bullet and two blocks, focusing on the conservation of momentum during two collisions. The bullet, with a mass of \( m_1 = 0.01 \, \text{kg} \), travels at an initial speed of \( v_{1a} = 430 \, \text{m/s} \) and collides with the first block, \( m_2 = 0.5 \, \text{kg} \), which is initially at rest. After passing through the first block, the bullet embeds itself into the second block, also with a mass of \( m_3 = 0.5 \, \text{kg} \).

To solve for the final velocity of the combined bullet and second block system after the second collision, we label the key events: point A (before the first collision), point B (after the first collision), and point C (after the second collision). The conservation of momentum principle is applied at each collision.

For the first collision (from A to B), the momentum equation is:

\[ m_1 v_{1a} + m_2 v_{2a} = m_1 v_{1b} + m_2 v_{2b} \]

Since the second block is at rest initially, \( v_{2a} = 0 \). Thus, the equation simplifies to:

\[ 0.01 \times 430 + 0.5 \times 0 = 0.01 v_{1b} + 0.5 v_{2b} \]

Here, \( v_{2b} \) is the velocity of the first block after the collision, which is given as \( 5.6 \, \text{m/s} \). Plugging in the values, we can solve for \( v_{1b} \), the bullet's velocity after exiting the first block.

Next, we calculate \( v_{1b} \) as follows:

\[ 4.3 = 0.01 v_{1b} + 2.8 \]

Rearranging gives:

\[ 1.5 = 0.01 v_{1b} \implies v_{1b} = 150 \, \text{m/s} \]

Now, we can analyze the second collision (from B to C). The momentum equation for this inelastic collision is:

\[ m_1 v_{1b} + m_3 v_{3b} = (m_1 + m_3) v_c \]

Since the second block is stationary before the collision, \( v_{3b} = 0 \). Thus, the equation simplifies to:

\[ 0.01 \times 150 + 0.5 \times 0 = (0.01 + 0.5) v_c \]

Calculating gives:

\[ 1.5 = 0.51 v_c \implies v_c = \frac{1.5}{0.51} \approx 2.94 \, \text{m/s} \]

In conclusion, the final velocity of the bullet and second block system after the second collision is approximately \( 2.94 \, \text{m/s} \). This problem illustrates the importance of organizing the events and applying the conservation of momentum correctly in multi-collision scenarios.

Here’s what students ask on this topic:

What is a completely inelastic collision?

A completely inelastic collision is a type of collision where two objects collide and stick together, moving with a common final velocity. This means that after the collision, the two objects move as a single combined mass. The defining characteristic of a completely inelastic collision is that the objects do not bounce off each other but instead remain attached. The final velocity of the combined mass can be determined using the conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

How do you calculate the final velocity in a completely inelastic collision?

To calculate the final velocity in a completely inelastic collision, you use the conservation of momentum principle. The equation is:

$m_1{v}_{1}i+m_2{v}_{2}i=\; (m_1+m_2)v_f$

Here, m1 and m2 are the masses of the two objects, v1i and v2i are their initial velocities, and vf is the final velocity of the combined mass. By solving for vf, you can find the final velocity after the collision.

What is the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved. The objects bounce off each other without any loss of kinetic energy. In contrast, in an inelastic collision, only momentum is conserved, and some kinetic energy is converted into other forms of energy, such as heat or sound. A completely inelastic collision is a special case where the colliding objects stick together and move with a common final velocity, resulting in the maximum possible loss of kinetic energy.

Why is kinetic energy not conserved in a completely inelastic collision?

Kinetic energy is not conserved in a completely inelastic collision because some of the kinetic energy is transformed into other forms of energy, such as heat, sound, or deformation energy. When the objects stick together, the internal forces during the collision cause energy dissipation. While momentum is conserved, the loss of kinetic energy is a characteristic feature of inelastic collisions, especially completely inelastic ones where the objects move together as a single mass after the collision.

Can you provide an example of a completely inelastic collision?

Sure! Consider a 1 kg block moving at 20 m/s that collides with a 9 kg block initially at rest. After the collision, the two blocks stick together and move as a single mass. Using the conservation of momentum:

$1\times 20+9\times 0=\; (1+9)v_f$

Simplifying, we get:

$20=10v_f$

Solving for vf, we find:

$v_f=2\text{m/s}$

So, the combined mass moves at 2 m/s after the collision.