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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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