Problem 55 - College Physics

Question

A 2.0-g particle moving at 8.0 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object. (a) Find the speed of each particle after the collision. (b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g. (c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b). In which case does the incident particle lose more kinetic energy?

Step-by-Step Solution

Answer

Speeds after collision part (a): 2.0-g particle: 5.33 m/s (away), 1.0-g particle: -5.33 m/s (away) ; part (b): 2.0-g particle: 3.33 m/s (away), 10.0-g particle: 13.33 m/s (away). The incident 2.0-g particle loses more kinetic energy in part (b).

Step 1: Conservation of Linear Momentum

Apply the conservation of linear momentum: \( m_1 v_1 = m_1 v_1' + m_2 v_2' \), where \( m_1 = 2 \, g \), \( v_1 = 8 \, m/s \) (initial velocities of particle 1), and \( v_1', v_2' \) (final velocities) are unknowns. With \( m_2 = 1 \, g \) and \( v_2 = 0 \, m/s \), the equation can be simplified to \( 16 = 2 v_1' + v_2' \).

Step 2: Conservation of Kinetic Energy

Apply conservation of kinetic energy: \( \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 {v_1'}^2 + \frac{1}{2} m_2 {v_2'}^2 \). This simplifies to \( 8 = v_1' + \frac{v_2'}{2} \).

Step 3: Solve the Equations

Solving Equations from Step 1 and Step 2 simultaneously gives \( v_1' = 16/3 \, m/s \) and \( v_2' = −16/3 \, m/s \) for part (a).

Step 4: Repeat with Different Mass

Repeat Steps 1 and 2 with \( m_2 = 10 \, g \) for part (b). Solving these equations gives \( v_1' = 10/3 \, m/s \) and \( v_2' = −40/3 \, m/s \).

Step 5: Calculate Final Kinetic Energy

Finally to solve part (c), calculate the final kinetic energies of the 2.0-g particle for both scenarios: \( KE_1' = \frac{1}{2} m_1 {v_1'}^2 \). This gives \( KE_1' = 88.89 \, mJ \) for part (a), and \( KE_1' = 11.11 \, mJ \) for part (b). Comparing these values allows to conclude that the incident particle loses more kinetic energy in part (b).

Key Concepts

Conservation of Linear Momentum

Imagine two skaters pushing off from one another on an ice rink: regardless of their size difference, they move apart without losing overall momentum. This illustrates the conservation of linear momentum, a fundamental principle in physics that tells us the total momentum in a closed system remains constant when it's not disturbed by external forces.
In our textbook problem, a moving particle collides with a stationary one. To find the speeds after the collision, we use the conservation of linear momentum. This principle is expressed mathematically as the total momentum before collision being equal to the total momentum after collision. It's what allows us to set up an equation like this when particle 1 with mass m₁ and velocity v₁ strikes a stationary particle 2 with mass m₂:
\[ m_1 v_1 = m_1 v_1' + m_2 v_2' \]
By knowing the masses and the initial velocity of the moving particle, we can use this formula to find the unknown final velocities v₁' and v₂'.

Conservation of Kinetic Energy

Another critical principle observed during an elastic collision is the conservation of kinetic energy. In such collisions, not only is momentum conserved, but the total kinetic energy before the impact is the same as after.
When two bodies collide and bounce off each other without any energy loss to heat, sound, or deformation, the kinetic energy – the energy due to motion – is preserved. For the collision described in our exercise, the equation reflecting this conservation looks like this:
\[ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 {v_1'}^2 + \frac{1}{2} m_2 {v_2'}^2 \]
Solving this equation along with the one for linear momentum gives us the final velocities that are consistent with both energy and momentum conservation. This is how we ensure the collision is indeed elastic as the task states.

Final Kinetic Energy

In analyzing collisions, the final kinetic energy of involved objects helps us understand how the energies redistribute after the event. For a perfectly elastic collision, we expect the kinetic energy of each object to adjust according to the conservation laws.
Using the velocities calculated from the principles of conservation, we can find the kinetic energy by plugging the final velocity into the kinetic energy formula:
\[ KE' = \frac{1}{2} m {v'}^2 \]
For our example, we calculate this for the moving 2.0-g particle post-collision, revealing the kinetic energy in both scenarios (a) and (b). The comparison demonstrates the differences in kinetic energy distribution depending on the masses and speeds involved.

Elastic Head-On Collision

A head-on collision is where the primary direction of motion is along the line joining the centers of the two colliding bodies. If the collision is elastic, as in our problem, both kinetic energy and momentum are conserved. In practical terms, this means that the two bodies will collide, exchange energy, and move apart without any energy being converted into other forms like heat or sound.
In the context of the textbook problem, we are dealing with a simplified scenario of a one-dimensional elastic collision. To determine the outcomes of such an event, two key calculations come into play—the conservation of linear momentum and the conservation of kinetic energy, which, when solved together, enable us to predict the final velocities of the particles. Our step-by-step solution calculates what happens when a moving particle hits another at rest, a fundamental concept useful across physics, from basic demonstrations to advanced applications in areas like billiards, vehicle safety design, and even astrophysics.