Problem 10 - College Physics

Question

\(\bullet\) On a frictionless air track, a 0.150 \(\mathrm{kg}\) glider moving at 1.20 \(\mathrm{m} / \mathrm{s}\) to the right collides with and sticks to a stationary 0.250 \(\mathrm{kg}\) glider. (a) What is the net momentum of this two- glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the glid- ers after the collision. (d) Is kinetic energy conserved during the collision?

Step-by-Step Solution

Answer

(a) 0.180 kg·m/s; (b) 0.180 kg·m/s; (c) 0.450 m/s; (d) No, kinetic energy is not conserved.

Step 1: Understand the System

We have two gliders: Glider 1 with mass \( m_1 = 0.150 \, \mathrm{kg} \) moving at velocity \( v_1 = 1.20 \, \mathrm{m/s} \), and Glider 2 with mass \( m_2 = 0.250 \, \mathrm{kg} \) which is stationary (velocity \( v_2 = 0 \)).

Step 2: Calculate Initial Momentum

The net momentum of the system before the collision is given by: \( p_{ ext{initial}} = m_1 v_1 + m_2 v_2 \). Substituting the given values: \( p_{ ext{initial}} = 0.150 \, \mathrm{kg} \times 1.20 \, \mathrm{m/s} + 0.250 \, \mathrm{kg} \times 0 \, \mathrm{m/s} = 0.180 \, \mathrm{kg \cdot m/s} \).

Step 3: Conservation of Momentum

According to the law of conservation of momentum, the net momentum before the collision is equal to the net momentum after the collision. Thus, \( p_{ ext{final}} = p_{ ext{initial}} = 0.180 \, \mathrm{kg \cdot m/s} \).

Step 4: Calculate Final Velocity After Collision

Since the gliders stick together after the collision, they move with a common velocity \( v_f \). The equation for final momentum is \( p_{ ext{final}} = (m_1 + m_2) \times v_f \). Setting this equal to the initial momentum, we have, \( 0.180 \, \mathrm{kg \cdot m/s} = (0.150 \, \mathrm{kg} + 0.250 \, \mathrm{kg}) \times v_f \). Solving for \( v_f \), we get \( v_f = \frac{0.180}{0.400} = 0.450 \, \mathrm{m/s} \).

Step 5: Check Kinetic Energy Conservation

Calculate the initial kinetic energy \( KE_i = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} \times 0.150 \, \mathrm{kg} \times (1.20 \, \mathrm{m/s})^2 + 0 = 0.108 \, \mathrm{J} \). Calculate the final kinetic energy \( KE_f = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \times 0.400 \, \mathrm{kg} \times (0.450 \, \mathrm{m/s})^2 = 0.0405 \, \mathrm{J} \). Since \( KE_f eq KE_i \), kinetic energy is not conserved.

Key Concepts

Inelastic Collision

Inelastic collisions are a fascinating topic in physics, especially when demonstrating the conservation of momentum. In an inelastic collision, two objects collide and stick together, moving as a single unit afterwards. This process is common in real-world scenarios where energy is dissipated through deformation, sound, and heat.
During our exercise, the collision between two gliders exemplifies a perfectly inelastic collision. Before the impact, we observe glider 1 moving with a velocity of 1.20 m/s, while glider 2 remains stationary. After they collide, they stick together, showcasing the essence of inelasticity, where the two masses combine and proceed with the same velocity.
It's important to note that although the total momentum of the system is conserved in such collisions, kinetic energy is not. The decrease in kinetic energy typically transforms into other forms, unlike in elastic collisions where kinetic energy remains constant.

• In inelastic collisions, objects stick together.
• Momentum is conserved, but kinetic energy is not.
• They are common in real-world interactions.

Momentum Calculation

Calculating momentum is crucial in understanding how objects interact in collisions. Momentum, a vector quantity, is defined as the product of an object's mass and its velocity. The law of conservation of momentum states that the total momentum in an isolated system remains constant if no external forces act upon it.

For our two-glider system, the initial total momentum is derived from the moving glider alone since the second glider is at rest. We calculate this using the formula: \[ p_{\text{initial}} = m_1 imes v_1 + m_2 imes v_2 = 0.150 \, \mathrm{kg} \times 1.20 \, \mathrm{m/s} = 0.180 \, \mathrm{kg\cdot m/s} \]
After the collision, the total momentum remains the same, according to the conservation law, but the mass term reflects both gliders moving together: \[ p_{\text{final}} = (m_1 + m_2) \times v_f \] Here, solving for the final velocity \( v_f \) tells us how fast the combined gliders will move post-collision. Since momentum is conserved, this equation results in us being able to calculate the shared velocity of the gliders.

• Momentum is mass times velocity: \( p = mv \).
• Conservation of momentum applies if no external forces exist.
• Initial and final total momenta in an isolated system are equal.

Kinetic Energy Conservation

While momentum conservation is a staple in physics scenarios like collisions, kinetic energy behaves differently, especially in inelastic collisions. Kinetic energy is defined as \[ KE = \frac{1}{2}mv^2 \] and it represents the energy an object possesses due to its motion.
In the exercise we're exploring, the kinetic energy before and after the collision differs significantly. Initial kinetic energy solely depends on the moving glider's speed, computed as \[ KE_i = \frac{1}{2} \times 0.150 \, \mathrm{kg} \times (1.20 \, \mathrm{m/s})^2 = 0.108 \, \mathrm{J} \].
After the collision, both gliders now have this new shared velocity. Consequently, the final kinetic energy drops to \[ KE_f = \frac{1}{2} \times 0.400 \, \mathrm{kg} \times (0.450 \, \mathrm{m/s})^2 = 0.0405 \, \mathrm{J} \].
The reduction in kinetic energy confirms that it is not conserved in inelastic collisions, as some energy is transformed into other non-mechanical forms during the collision.

• Kinetic energy in motion is \( KE = \frac{1}{2}mv^2 \).
• Inelastic collisions involve energy transformation into other forms.
• Kinetic energy is not conserved in inelastic collisions, unlike momentum.