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A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 22 m/s and the other ball, of mass 1.3 kg, is moving downward at 11 m/s. How high do the combined two balls of putty rise above the collision point

Sagot :

Answer:

The height balls rise above the collision point, is approximately 7.37 meters

Explanation:

The given parameters just before the collision are;

The mass, m₁ and velocity, v₁ of the ball moving upward are;

m₁ = 3.0 kg, v₁ = 22 m/s

The mass, m₂ and velocity, v₂ of the ball moving downward are;

m₂ = 1.3 kg, v₂ = -11 m/s (downward motion)

The type of collision = Inelastic collision

We note that the momentum is conserved for inelastic collision

Let, [tex]v_f[/tex], represent the final velocity of the balls after collision, we have;

∴ Total initial momentum = Total final momentum

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·[tex]v_f[/tex]

Therefore, we get;

m₁·v₁ + m₂·v₂ = 3.0 kg × 22 m/s + 1.3 kg × (-11) m/s = 51.7 kg·m/s

(m₁ + m₂)·[tex]v_f[/tex] = (3.0 kg + 1.3 kg) ×

∴ 51.7 kg·m/s = 4.3 kg × [tex]v_f[/tex]

[tex]v_f[/tex] = (51.7 kg·m/s)/4.3 kg ≈ 12.023 m/s

The final velocity, [tex]v_f[/tex] ≈ 12.023 m/s

The maximum height, h, the combined balls will rise from the point of collision, moving upward at a velocity of [tex]v_f[/tex] ≈ 12.023 m/s, is given from the kinetic equation of motion, v² = u² - 2·g·h, as found follows

At maximum height, we have;

[tex]h_{max} = \dfrac{v_f^2}{2 \cdot g }[/tex]

Therefore;

[tex]h_{max} \approx \dfrac{12.023^2}{2 \times 9.81 } \approx 7.37[/tex]

The height the combined two balls of putty rise above the collision point, [tex]h_{max}[/tex] ≈ 7.37 m.

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