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A 3.0-kg cart moving to the right with a speed of 1.0 m/s has a head-on collision with a 5.0-kg cart that is initially moving to the left with a speed of 2.0 m/s. After the collision, the 3.0-kg cart is moving to the left with a speed of 1.0 m/s. What is the final velocity of the 5.0-kg cart?

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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