To find the time it takes for both carts to have the same speed, we can use the concept of acceleration and the constant contact force between the carts during the collision.
First, let's calculate the acceleration of the carts using Newton's second law, F = ma, where F is the net force exerted on the cart and m is the mass of the cart.
For Cart 1:
Mass (m1) = 0.10 kg
Acceleration (a1) = 13 m/s^2
Therefore, the net force acting on Cart 1 (F1) can be calculated as:
F1 = m1 * a1
F1 = 0.10 kg * 13 m/s^2
F1 = 1.3 N
For Cart 2:
Mass (m2) = 0.24 kg
Acceleration (a2) = 5.4 m/s^2
Similarly, the net force acting on Cart 2 (F2) can be calculated as:
F2 = m2 * a2
F2 = 0.24 kg * 5.4 m/s^2
F2 = 1.3 N
Since the average contact force during the collision is the same for both carts (1.3 N), we can equate the two forces:
F1 = F2
m1 * a1 = m2 * a2
Now, to find the time it takes for both carts to have the same speed, we can use the relationship between acceleration, time, and change in velocity:
a = (v_f - v_i) / t
Since the initial velocity for Cart 1 is 1.2 m/s and Cart 2 is at rest, the equation can be written as:
a1 * t = -a2 * t
Simplifying the equation, we have:
(a1 + a2) * t = 0
Since a1 = 13 m/s^2 and a2 = 5.4 m/s^2:
(13 + 5.4) * t = 0
18.4 * t = 0
Since the acceleration sum is zero, the time it takes for both carts to have the same speed is 0 seconds. This implies that they have the same speed instantaneously once the contact force vanishes.
To find the final speed of the carts after the collision, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision since there is no external force acting on the system:
m1 * v1i + m2 * v2i = (m1 + m2) * vf
Substituting the given values:
(0.10 kg * 1.2 m/s) + (0.24 kg * 0 m/s) = (0.10 kg + 0.24 kg) * vf
0.12 kg m/s = 0.34 kg * vf
vf = 0.12 kg m/s / 0.34 kg
vf ≈ 0.35 m/s
Therefore, the final speed of the carts after the collision is approximately 0.35 m/s