contestada

A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. When a compressed spring attached to one of the carts is released, the carts are pushed apart. After the spring is released, the speed of cart B is 6.0 meters per second, as represented in the diagram belowWhat is the speed of cart A after the spring is released?

Respuesta :

meerkat18
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.
Lanuel

The speed of cart A after the spring is released is 2.57 m/s.

Given the following data:

  • Mass of cart A = 7.0 kg
  • Initial velocity of cart A and B = 0 m/s (since they were held at rest).
  • Mass of cart B = 3.0 kg
  • Final velocity of cart B = 6.0 m/s

To find the speed of cart A after the spring is released, we would apply the law of conservation of momentum:

[tex]M_AV_A - M_BV_B = M_AV_{fA} + M_BV_{fB}[/tex]

Where:

  • [tex]M_A[/tex] is the mass of the first cart.
  • [tex]M_B[/tex] is the mass of the second cart.
  • [tex]V_A \;and\; V_B[/tex] are the initial velocities.
  • [tex]V_{fA}[/tex] is the initial velocity of the first cart.
  • [tex]V_{fB}[/tex] is the final velocity of the second cart.

Substituting the given parameters into the formula, we have;

[tex]7 \times 0 - 3 \times 0 = 7 \times -V_{fA} + 3 \times 6\\\\0 = -7V_{fA} + 18\\\\V_{fA} = \frac{18}{7}\\\\V_{fA} = 2.57 m/s[/tex]

Note: The negative sign indicate that cart A is moving to the left direction.

Ver imagen Lanuel
Q&A Education