Push-Away Problems: Videos & Practice Problems

In push away problems, conservation of momentum is crucial. When two objects, like a gun and bullet, push away from each other, their initial momentum is zero, leading to the equation: m_1 × v_1 × f + m_2 × v_2 × f = 0. Additionally, energy conservation relates elastic potential energy in springs to kinetic energy post-release, allowing calculations of stored energy and compression distance. Understanding these principles is essential for solving related physics problems effectively.

"Push-Away"Problems

In physics, understanding the conservation of momentum is crucial for solving problems involving interactions between objects. These interactions typically fall into two categories: collisions and push-away problems. While collisions involve objects moving towards each other and colliding, push-away problems occur when objects are initially together and then push away from each other, resulting in movement in opposite directions. This phenomenon is rooted in Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

Regardless of the type of interaction, the principle of conservation of momentum applies. Momentum, defined as the product of an object's mass and velocity, must remain constant in an isolated system. This means that the total momentum before an interaction must equal the total momentum after the interaction. The equation for conservation of momentum can be expressed as:

\[M_1 V_{1i} + M_2 V_{2i} = M_1 V_{1f} + M_2 V_{2f}\]

Where:

• M₁ and M₂ are the masses of the objects.
• V_1i and V_2i are the initial velocities of the objects.
• V_1f and V_2f are the final velocities of the objects.

In push-away problems, it is common for the initial velocities of the objects to be zero, simplifying the equation to:

\[0 = M_1 V_{1f} + M_2 V_{2f}\]

This indicates that the momentum gained by one object is equal in magnitude and opposite in direction to the momentum gained by the other object. For example, consider a scenario where a 4 kg sniper rifle fires a 5 g bullet. Before firing, both the rifle and bullet are at rest, resulting in an initial momentum of zero. After the bullet is fired at a velocity of 600 m/s, the recoil speed of the rifle can be calculated using the conservation of momentum principle.

By rearranging the momentum equation, we find:

\[V_{1f} = -\frac{M_2 V_{2f}}{M_1}\]

Substituting the known values:

\[V_{1f} = -\frac{0.005 \, \text{kg} \times 600 \, \text{m/s}}{4 \, \text{kg}} = -0.75 \, \text{m/s}\]

This negative sign indicates that the rifle recoils in the opposite direction to the bullet's motion. Thus, if the bullet moves to the right, the rifle recoils to the left at a speed of 0.75 m/s.

It is important to note that momentum is conserved only in an isolated system, where all forces acting on the objects are internal. In the example of the rifle and bullet, the forces exerted during firing are internal to the system. However, if an external force were applied, such as pushing the rifle while firing, momentum would not be conserved.

In summary, mastering the concept of conservation of momentum allows for the effective analysis of both collision and push-away problems, providing a foundational understanding of interactions in physics.

An 80-kg astronaut is stranded floating in space is 30 m away from his spaceship. He wants to return to his spaceship in 20 s. How fast must he throw his 2-kg space hammer, directly away from the spaceship, to accomplish this?

• A. 40 m/s
• B. 120 m/s
• C. 240 m/s
• D. 60 m/s

Answer: (D)

Sliding Football Player

In this scenario, we analyze the motion of a football player on frictionless ice who throws a football. Initially, both the player and the football are moving together at a velocity of 2 meters per second. When the player throws the football, it is propelled forward at an additional speed of 15 meters per second relative to the player. This results in the football's final velocity being the sum of its initial velocity and the throw speed, calculated as:

v_{2, final} = v_{initial} + v_{throw} = 2 \, \text{m/s} + 15 \, \text{m/s} = 17 \, \text{m/s}

To determine the final velocity of the player after the throw, we apply the principle of conservation of momentum. The momentum before the throw must equal the momentum after the throw. The equation can be expressed as:

m_{1} v_{1, initial} + m_{2} v_{2, initial} = m_{1} v_{1, final} + m_{2} v_{2, final}

Where:

• m_{1} = 70 \, \text{kg} (mass of the player)
• m_{2} = 0.45 \, \text{kg} (mass of the football)
• v_{1, initial} = 2 \, \text{m/s} (initial velocity of the player)
• v_{2, initial} = 2 \, \text{m/s} (initial velocity of the football)
• v_{2, final} = 17 \, \text{m/s} (final velocity of the football)

Substituting these values into the momentum equation gives:

70 \cdot 2 + 0.45 \cdot 2 = 70 \cdot v_{1, final} + 0.45 \cdot 17

Calculating the left side:

140 + 0.9 = 70 \cdot v_{1, final} + 7.65

Which simplifies to:

140.9 = 70 \cdot v_{1, final} + 7.65

Rearranging the equation to isolate v_{1, final} results in:

140.9 - 7.65 = 70 \cdot v_{1, final}

133.25 = 70 \cdot v_{1, final}

Dividing both sides by 70 yields:

v_{1, final} = \frac{133.25}{70} \approx 1.9 \, \text{m/s}

This calculation shows that after throwing the football, the player’s final velocity is approximately 1.9 meters per second. This indicates that the player has slowed down slightly due to the momentum transfer to the football, which, despite its lighter mass, is now moving at a significantly higher speed. The conservation of momentum principle illustrates how momentum is transferred between objects during interactions.

Push-Away Problems with Energy

In problems involving objects pushing away from each other, both momentum and energy conservation principles are essential for solving various aspects of the scenario. For instance, consider two blocks that are compressed against a spring. When released, they move in opposite directions, and we can analyze their motion using conservation laws.

Initially, both blocks are at rest, which means their initial momentum is zero. According to the momentum conservation equation:

\[m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}\]

Since the initial velocities are zero, the equation simplifies to:

\[0 = m_1 v_{1f} + m_2 v_{2f}\]

This indicates that the momentum gained by one block is equal in magnitude and opposite in direction to the momentum gained by the other block. For example, if a 3 kg block (Block 1) moves at 10 m/s to the left, the 4 kg block (Block 2) will have a final velocity that can be calculated as:

\[v_{2f} = -\frac{m_1 v_{1f}}{m_2}\]

Substituting the values gives:

\[v_{2f} = -\frac{3 \times (-10)}{4} = 7.5 \text{ m/s}\]

Next, to find the energy stored in the spring, we apply the principle of energy conservation. The initial energy in the system is the elastic potential energy stored in the spring, which converts into kinetic energy of the blocks once they are released. The energy conservation equation can be expressed as:

\[U_{initial} + K_{initial} = U_{final} + K_{final}\]

Given that the initial kinetic energy is zero (the blocks are at rest), and the final potential energy is zero (the spring is decompressed), we have:

\[U_{elastic} = K_{1f} + K_{2f}\]

The kinetic energy for each block is given by:

\[K = \frac{1}{2} m v^2\]

Thus, the total kinetic energy becomes:

\[U_{elastic} = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2\]

Substituting the known values results in:

\[U_{elastic} = \frac{1}{2} \times 3 \times 10^2 + \frac{1}{2} \times 4 \times (7.5)^2 = 262.5 \text{ joules}\]

Finally, to determine how much the spring was compressed before launching the blocks, we relate the elastic potential energy to the spring constant using the formula:

\[U_{elastic} = \frac{1}{2} k x^2\]

Rearranging for the compression distance \(x\) gives:

\[x = \sqrt{\frac{2 U_{elastic}}{k}}\]

Substituting the values, where \(k = 800 \text{ N/m}\), results in:

\[x = \sqrt{\frac{2 \times 262.5}{800}} \approx 0.81 \text{ meters}\]

In summary, by applying both momentum and energy conservation principles, we can effectively analyze and solve problems involving objects that push away from each other, determining their velocities, the energy stored in springs, and the compression distances involved.

Here’s what students ask on this topic:

What is a push away problem in physics?

A push away problem in physics involves two objects that start together and then push away from each other, moving in opposite directions. Common examples include the recoil of a gun when a bullet is fired, ice skaters pushing off each other, or astronauts in space. These problems are typically solved using the conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it. The initial momentum is usually zero, leading to the equation: $m_1\times v_1+m_2\times v_2=0$.

How do you solve a push away problem using conservation of momentum?

To solve a push away problem using conservation of momentum, follow these steps: 1) Draw a diagram of the initial and final states of the system. 2) Write the conservation of momentum equation: $m_1{v}_1+m_2{v}_2=0$. 3) Substitute the known values for masses and velocities. 4) Solve for the unknown velocity. For example, if a 4 kg gun fires a 0.005 kg bullet at 600 m/s, the recoil speed of the gun can be found using: $-m_1{v}_1=m_2{v}_2$.

What role does energy conservation play in push away problems?

Energy conservation is crucial in push away problems, especially when dealing with springs or other potential energy sources. The total mechanical energy (kinetic + potential) of the system remains constant if no external work is done. For example, if two blocks are pushed by a compressed spring, the elastic potential energy stored in the spring is converted into the kinetic energy of the blocks. The energy conservation equation is: $U_{\mathrm{elastic}}=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$.

How do you calculate the recoil speed of a gun in a push away problem?

To calculate the recoil speed of a gun in a push away problem, use the conservation of momentum. Assume the initial momentum is zero. The equation is: $m_1{v}_1+m_2{v}_2=0$. For a 4 kg gun firing a 0.005 kg bullet at 600 m/s, the recoil speed $v_1$ is found by solving: $-m_1{v}_1=m_2{v}_2$. Plugging in the values: $v_1=-\frac{0.005\times 600}{4}=-0.75 m/s$.

How do you determine the energy stored in a spring in a push away problem?

To determine the energy stored in a spring in a push away problem, use the conservation of energy. The elastic potential energy stored in the spring is converted into the kinetic energy of the objects. The equation is: $U_{\mathrm{elastic}}=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$. For example, if a 3 kg block moves at 10 m/s and a 4 kg block moves at 7.5 m/s, the stored energy is: $U_{\mathrm{elastic}}=\frac{1}{2}(3\times 10^2+4\times 7.5^2)=262.5 J$.

What is the importance of defining an isolated system in push away problems?

Defining an isolated system in push away problems is crucial because it ensures that the conservation of momentum applies. An isolated system means that all forces acting within the system are internal, and no external forces influence the system's momentum. For example, in a gun and bullet scenario, the forces between the gun and bullet are internal. If an external force, like a hand pushing the gun, is introduced, the system is no longer isolated, and momentum conservation does not hold. Therefore, correctly defining the system helps in accurately applying the conservation laws to solve the problem.