Answer:b
Explanation:
Given
mass of first cart [tex]m_1=6 kg[/tex]
mass of second cart [tex]m_2=3 kg[/tex]
velocity of first cart [tex]v_1=3 m/s[/tex]
conserving momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]6\times 3+3\times 0=(9)\cdot v[/tex]
[tex]v=2 m/s[/tex]
Initial kinetic Energy [tex]K.E._1=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/tex]
[tex]K.E._1=\frac{1}{2}\cdot 6\cdot 3^2+0[/tex]
[tex]K.E._1=27 J[/tex]
Final Kinetic Energy
[tex]K.E._2=\frac{1}{2}(m_1+m_2)v^2[/tex]
[tex]K.E._2=\frac{1}{2}(6+3)\cdot 2^2=18 J[/tex]
Ratio of initial Kinetic Energy to the Final Kinetic Energy
[tex]=\frac{27}{18}=1.5[/tex]
The ratio of the initial kinetic energy would be:
1.5
According to the question,
First cart's mass, m₁ = 6 kg
Second cart's mass, m₂ = 3 kg
First cart's velocity, v₁ = 3 m/s
By using Conserving momentum, we get
→ m₁v₁ + m₂v₂ = (m₁ + m₂)v
By substituting the values,
6 × 3 + 3 × 0 = 9 × v
18 + 0 = 9v
v = [tex]\frac{18}{9}[/tex]
= 2 m/s
The Initial Kinetic energy will be:
→ K.E₁ = [tex]\frac{1}{2}[/tex]m₁v₁² + m₂v₂²
= [tex]\frac{1}{2}[/tex] × 6 × (3)² + 0
= 27 J
The Final Kinetic energy will be:
→ K.E₂ = [tex]\frac{1}{2}[/tex] (m₁ + m₂) v²
= [tex]\frac{1}{2}[/tex] (6 + 3) (2)²
= 18 J
hence,
The ratio will be:
= [tex]\frac{27}{18}[/tex]
= 1.5
Thus the above answer is correct.
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