Collisions & Motion (Momentum & Energy): Videos & Practice Problems

In problems involving collisions and subsequent motion, apply conservation of momentum during the collision and conservation of energy for motion. For a completely inelastic collision, where two objects stick together, use the equation ${\frac{m}{1}}^2=\frac{m}{g}y$. The final height can be calculated as $y=\frac{v^2}{2g}$, emphasizing the importance of energy conservation post-collision.

Collision Problems with Motion/Energy

In solving problems that involve both collisions and subsequent motion, it is essential to apply the principles of conservation of momentum and conservation of energy effectively. These problems typically consist of a collision phase followed by a motion phase where the objects may change speeds or heights. For instance, consider a scenario where a crate collides with another crate, and they stick together, subsequently moving up an inclined plane.

To tackle such problems, it is beneficial to clearly label points of interest throughout the process. For example, designate point A as the moment just before the collision, point B as immediately after the collision when the crates are moving together, and point C as the point where they reach their maximum height on the incline. The goal is to determine the height (y) at point C.

Start by writing the conservation of momentum equation for the collision phase. This can be expressed as:

$$ m_1 v_{1A} + m_2 v_{2A} = (m_1 + m_2) v_B $$

Here, \( m_1 \) and \( m_2 \) are the masses of the crates, and \( v_{1A} \) and \( v_{2A} \) are their initial velocities. Since the second crate is at rest before the collision, its initial velocity is zero.

Next, for the motion phase, apply the conservation of energy principle. The energy conservation equation can be written as:

$$ \frac{1}{2} m v_B^2 = m g y_C $$

In this equation, \( g \) represents the acceleration due to gravity, and \( y_C \) is the height reached at point C. Notably, the mass \( m \) cancels out, simplifying the equation to:

$$ y_C = \frac{v_B^2}{2g} $$

To find \( v_B \), substitute the values obtained from the momentum equation into the energy equation. After calculating \( v_B \) using the momentum equation, you can plug this value back into the energy equation to find the height \( y_C \).

It is important to note that while energy is not conserved during the collision (due to inelasticity), it is conserved in the motion phase after the collision, provided there are no non-conservative forces acting on the system, such as friction. This principle allows for the calculation of the height reached after the collision.

In summary, when faced with problems involving collisions followed by motion, systematically apply conservation laws, clearly label your points of interest, and ensure to differentiate between the phases of the problem. This structured approach will help in accurately solving for the desired variables, such as height or velocity, in various scenarios, including those involving springs or pendulum-like motions.

A 300g bullet is fired horizontally into a 10-kg wooden block initially at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.6. The bullet remains embedded in the block, which then slides 35 m along the surface before stopping. What was the initial speed of the bullet?

• A. 492.5 m/s
• B. 20.3 m/s
• C. 677 m//s
• D. 697 m/s

Answer: (D)

Bullet-Block with Height

In this scenario, we analyze the motion of a bullet fired into a wooden block, which subsequently rises after the bullet passes through it. The goal is to determine the maximum height the block reaches after the collision. This problem involves both momentum conservation during the collision and energy conservation as the block rises.

We define three key points in the motion: point A (before the collision), point B (immediately after the collision), and point C (at the maximum height). The bullet, with a mass of \( m_1 = 0.006 \, \text{kg} \) (6 grams), is fired upwards at an initial velocity of \( v_{1a} = 800 \, \text{m/s} \). The block has a mass of \( m_2 = 1.2 \, \text{kg} \) and is initially at rest, so \( v_{2a} = 0 \, \text{m/s} \).

To analyze the collision, we apply the principle of conservation of momentum, which states:

$$ m_1 v_{1a} + m_2 v_{2a} = m_1 v_{1b} + m_2 v_{2b} $$

Here, \( v_{1b} \) is the velocity of the bullet after the collision, which is given as \( 150 \, \text{m/s} \). The equation simplifies to:

$$ 0.006 \times 800 + 0 = 0.006 \times 150 + 1.2 v_{2b} $$

Solving this equation allows us to find \( v_{2b} \), the velocity of the block immediately after the collision. After calculations, we find:

$$ v_{2b} = 3.25 \, \text{m/s} $$

Next, we analyze the motion from point B to point C using the conservation of energy principle. The kinetic energy at point B is converted into gravitational potential energy at point C. The relevant equation is:

$$ \frac{1}{2} m_2 v_{2b}^2 = m_2 g y_{max} $$

Where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \). Since \( m_2 \) appears on both sides of the equation, it cancels out, leading to:

$$ \frac{1}{2} v_{2b}^2 = g y_{max} $$

Substituting \( v_{2b} = 3.25 \, \text{m/s} \) into the equation gives:

$$ \frac{1}{2} (3.25)^2 = 9.8 y_{max} $$

Solving for \( y_{max} \) yields:

$$ y_{max} = \frac{(3.25)^2}{2 \times 9.8} \approx 0.54 \, \text{m} $$

This result indicates that the block rises to a maximum height of approximately \( 0.54 \, \text{m} \) after the bullet passes through it. The analysis demonstrates the interplay between momentum and energy conservation in a vertical motion scenario.

Here’s what students ask on this topic:

How do you solve problems involving both collisions and subsequent motion?

To solve problems involving both collisions and subsequent motion, you need to apply the conservation of momentum during the collision and the conservation of energy for the motion. First, identify the points of interest: before the collision (point A), after the collision (point B), and the final state of motion (point C). Use the conservation of momentum equation for the collision: $m_1{v}_1{A}_{}+m_2{v}_2{A}_{}=m_1{v}_1{B}_{}+m_2{v}_2{B}_{}$. Then, use the conservation of energy equation for the motion: $K_B+U_B=K_C+U_C$. This approach helps you systematically solve for unknowns like final height or speed.

What is the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved. The objects collide and bounce off each other without any loss of kinetic energy. The equation for an elastic collision is: $m_1{v}_1{i}_{}+m_2{v}_2{i}_{}=m_1{v}_1{f}_{}+m_2{v}_2{f}_{}$. In an inelastic collision, momentum is conserved, but kinetic energy is not. The objects may stick together, and some kinetic energy is converted into other forms of energy, such as heat or sound. The equation for a completely inelastic collision is: $m_1{v}_1{i}_{}+m_2{v}_2{i}_{}=\; (m_1+m_2)v_{\mathrm{f}}$.

How do you calculate the final height in a collision problem involving an inclined plane?

To calculate the final height in a collision problem involving an inclined plane, first use the conservation of momentum to find the velocity after the collision: $m_1{v}_1{A}_{}+m_2{v}_2{A}_{}=\; (m_1+m_2)v_{\mathrm{B}}$. Then, apply the conservation of energy to find the height: $\frac{v^2}{2}$. The final height $y$ is given by $y=\frac{v^2}{2}$, where $v$ is the velocity after the collision and $g$ is the acceleration due to gravity.

Why is energy not conserved in a completely inelastic collision?

In a completely inelastic collision, energy is not conserved because some of the kinetic energy is converted into other forms of energy, such as heat, sound, or deformation energy. While momentum is conserved, the loss of kinetic energy occurs due to the internal forces acting during the collision. The equation for a completely inelastic collision is: $m_1{v}_1{i}_{}+m_2{v}_2{i}_{}=\; (m_1+m_2)v_{\mathrm{f}}$. The kinetic energy before and after the collision is different, indicating that some energy has been transformed into other forms.

How do you use conservation of momentum and energy in a problem involving a collision and an inclined plane?

To use conservation of momentum and energy in a problem involving a collision and an inclined plane, follow these steps: 1) Apply the conservation of momentum to the collision to find the velocity after the collision: $m_1{v}_1{A}_{}+m_2{v}_2{A}_{}=\; (m_1+m_2)v_{\mathrm{B}}$. 2) Use the conservation of energy to find the height the objects reach on the inclined plane: $K_B+U_B=K_C+U_C$. The final height $y$ can be calculated using $y=\frac{v^2}{2}$, where $v$ is the velocity after the collision and $g$ is the acceleration due to gravity.