An air-track glider undergoes a perfectly inelastic collision with an identical glider that is initially at rest. what fraction

Question

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An air-track glider undergoes a perfectly inelastic collision with an identical glider that is initially at rest. what fraction

of the first glider's initial kinetic energy is transformed into thermal energy in this collision?

Physics

2 answers:

Keith_Richards [3.2K] · Answer 1 · 2026-03-24T00:40:22+03:00

In this collision, half of the initial kinetic energy of the first glider is converted into thermal energy.

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Further explanation

According to Newton's second law of motion , the net force exerted on an object is directly related to its mass and acceleration.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Mass of the Object ( kg )

a = Acceleration ( m )

Now let's solve the problem!

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Given:

mass of first glider = m₁ = m

mass of second glider = m₂ = m

initial speed of first glider = u₁ = u

initial speed of second glider = u₂ = 0

final speed of both gliders = v₁ = v₂ = v → perfectly inelastic collision

Asked:

the change in kinetic energy = ΔEk =?

Solution:

First, we will apply the Law of Conservation of Momentum as shown below:Next, we'll compute the change in kinetic energy for the first glider:

$\Delta Ek: Ek_1 = ( Ek_1 - Ek ): Ek_1$

$\Delta Ek: Ek_1 = ( \frac{1}{2}mu^2 - \frac{1}{2}(2mv^2)): (\frac{1}{2}mu^2)$

$\Delta Ek: Ek_1 = ( mu^2 - 2mv^2 ): (mu^2)$

$\Delta Ek: Ek_1 = ( mu^2 - 2m(\frac{1}{2}u)^2 ): (mu^2)$

$\Delta Ek: Ek_1 = ( mu^2 - 2m(\frac{1}{4}u^2) ): (mu^2)$

$\Delta Ek: Ek_1 = ( mu^2 - \frac{1}{2}mu^2 ): (mu^2)$

$\Delta Ek: Ek_1 = ( \frac{1}{2}mu^2 ): (mu^2)$

$\Delta Ek: Ek_1 = \frac{1}{2}: 1$

$\boxed {\Delta Ek = \frac{1}{2} Ek_1}$

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Conclusion:

Half of the first glider's initial kinetic energy is converted into thermal energy during this collision.

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Learn more

Impacts of Gravity:
Effect of Earth’s Gravity on Objects:
The Acceleration Due To Gravity:
Newton's Law of Motion:
Example of Newton's Law:

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

Yuliya22 [3.3K] · Answer 2 · 2026-03-24T00:40:50+03:00

Please consult the diagram below.

Initially, the system's kinetic energy (KE) is
KE₁ = (1/2)mu²

After the inelastic collision, both masses merge together.
Momentum conservation states that
m*u = 2m*v
Thus,
v = u/2

The final KE is
KE₂ = (1/2)(2m)v²
= m(u/2)²
= (1/4)mu²
= (1/2) KE₁

The reduction in KE is
KE₁ - KE₂ = (1/2) KE₁.

Energy conservation implies that the reduction in KE must be considered as thermal energy.

Result: 1/2