Answer :
Answer:
[tex]v = 0.8 m/s[/tex] towards left
Explanation:
As we know that there is no external force on the system of two cart so total momentum of the system is conserved
so we will say
[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
now plug in all data into the above equation
[tex]3(1) + 5(-2) = 3(-1) + 5 v[/tex]
here we assumed that left direction of motion is negative while right direction is positive
so we can solve it for speed v now
[tex]3 - 10 = - 3 + 5 v[/tex]
[tex]5 v = -4[/tex]
[tex]v = -0.8 m/s[/tex]
The final velocity of the second cart (5kg) is 0.8 m/s towards the left.
The given parameters;
- mass of the first cart, m₁ = 3.0 kg
- initial speed of the first cart, u₁ = 1 m/s
- mass of the second object, m₂ = 5 kg
- initial speed of the second object, u₂ = 2 m/s
- final velocity of the first object, v₁ = 1 m/s
The final velocity of the second object is determined by applying the principle of conservation of linear momentum as shown below;
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2 v_2\\\\3(1) \ + 5(-2) = 3(-1) + 5v_2\\\\-7 = -3 + 5v_2\\\\-4 = 5v_2\\\\v_2 = \frac{-4}{5} = -0.8 \ m/s[/tex]
Thus, the final velocity of the second cart (5kg) is 0.8 m/s towards the left.
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