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Object X travels across a horizontal surface and collides with object Y. The velocity as a function of time for object X and object Y before, during, and after the
collision is shown in the graph. Both objects have a mass of 2 kg. Which of the following correctly describes the momentum p of the system and the kinetic
energy K of the system?

Object X travels across a horizontal surface and collides with object Y The velocity as a function of time for object X and object Y before during and after the class=

Respuesta :

Answer:

The correct option is;

(B) [tex]\underset{P}{\rightarrow}[/tex] is conserved, K Not conserved

Explanation:

From the diagram related to the question and from the question, we have;

m₁ = m₂ = 2 kg

v₁ = 4 m/s, v₂ = -2 m/s, v₃ = 1 m/s, therefore, we have;

The total initial momentum of the system = m₁ × v₁ + m₂ × v₂

m₁ (v₁ + v₂) = 2×(4 - 2) = 4 kg·m/s

The total final momentum of the system = m₁ × v₃ + m₂ × v₃ = 2 ×1 +2 ×1  = 4 kg·m/s

Given that the total initial momentum = The total final momentum, we have that the momentum, p, is conserved

The kinetic energy of the system before collision  = 1/2×m₁×v₁² + 1/2×m₂×v₂²

∴ The kinetic energy of the system before collision  = 1/2*2*(4)² + 1/2*2*(-2)² = 20 J

∴ The kinetic energy of the system after collision  = 1/2×m₁×v₃² + 1/2×m₂×v₃²

1/2*2*(1)² + 1/2*2*(1)² = 2 J

The kinetic energy of the system before collision ≠ The kinetic energy of the system after collision

Therefore, the kinetic energy of the system is not conserved.

onyebuchinnaji

If the velocities of object X and object Y after collision is the same (inelastic collision), then only momentum is conserved but if the velocities of both objects after collision are different (elastic collision), then both momentum and kinetic energy are conserved.

According to principle of conservation of linear momentum, the total momentum before and after collision is conserved.

For inelastic collision, only momentum is conserved while kinetic energy is not conserved.

[tex]m_1u_1 \ + \ m_2u_2 = v(m_1 + m_2)[/tex]

[tex]\frac{1}{2} m_1u_1^2 \ + \frac{1}{2} m_2u_2^2 = \frac{1}{2}v^2(m_1+m_2)[/tex]

For elastic collision, both momentum and kinetic energy are conserved;

[tex]m_1u_1 \ + \ m_2u_2 =m_1v_1 \ + \ m_2v_2[/tex]

[tex]\frac{1}{2} m_1u_1^2 \ + \frac{1}{2} m_2u_2^2 = \frac{1}{2}m_1v_1^2 \ +\ \frac{1}{2}m_2v_2^2[/tex]

Thus, we can conclude that, if the velocities of object X and object Y after collision is the same (inelastic collision), then only momentum is conserved but if the velocities of both objects after collision are different (elastic collision), then both momentum and kinetic energy are conserved.

Learn more here:https://brainly.com/question/24424291

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