Respuesta :
Answer:
Explanation:
Given
cart 1 has twice the speed of cart 2
cart 2 has twice the mass of cart 1
Suppose m is the mass of cart 1 so mass of cart 2 is 2 m
suppose u is the initial velocity of car 2 so initial velocity of car 1 is u
For Elastic collision velocity after collision is given by
velocity of cart 2 is [tex]v_2=\frac{2m_1}{m_1+m_2}\cdot u_1-\frac{m_1-m_2}{m_1+m_2}\cdot u_2[/tex]
velocity of cart 1 is [tex]v_1=\frac{m_1-m_2}{m_1+m_2}\cdot u_1+\frac{2m_2}{m_1+m_2}\cdot u_2[/tex]
here [tex]m_1=m,m_2=2 m[/tex]
[tex]u_1=2u,u_2=u[/tex]
substituting values we get
[tex]v_1=\frac{m-2m}{3m}\cdot 2u+2\cdot \frac{2m}{3m}\cdot u[/tex]
[tex]v_1=\frac{2u}{3}[/tex]
[tex]v_2=\frac{2m}{3m}\cdot (2u)-\frac{m-2m}{3m}\cdot u[/tex]
[tex]v_2=\frac{5u}{3}[/tex]
The Speed of Cart 1 right after the collision if the collision is elastic is;
v₁ = ¹/₃u₁ or expressed as one-third of its' initial speed before the collision
Let the speeds be as follows;
Initial speed of Cart 1 = u₁
final speed of Cart 1 = v₁
Initial speed of Cart 2 = u₂
Final speed of Cart 2 = v₂
Since the collision is elastic, then we can say that;
u₁ + v₁ = u₂ + v₂ ---(eq 1)
Also, we can say that;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ ---(eq 2)
Let us make v₂ the subject in eq 1 and put in eq 2;
v₂ = u₁ - u₂ + v₁ ----(eq 3)
Thus;
m₁u₁ + m₂u₂ = m₁v₁ + m₂(u₁ - u₂ + v₁)
making v₁ the subject gives;
v₁ = [(m₁ - m₂)/(m₁ + m₂)]u₁ + [2m₂/(m₁ + m₂)]u₂
We are told that Cart 1 has a speed twice that of cart 2. Thus;
u₂ = ¹/₂v₁ = ¹/₂u₁
Also, we see that cart 2 has twice the inertia of cart 1.
Thus, m₂ = 2m₁
Plugging in the values gives;
v₁ = [(m₁ - 2m₁)/(m₁ + 2m₁)]u₁ + [2(2m₁)/(m₁ + 2m₁)]( ¹/₂u₁)
Simplifying this gives;
v₁ = -¹/₃u₁ + ²/₃u₁
v₁ = ¹/₃u₁
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