Two carts, one twice as heavy as the other, are at rest on a horizontal frictionless track. A person pushes each cart with the same force for 4.00 s.
If the magnitude of the momentum of the lighter cart after the push is P, the magnitude of the momentum of the heavier cart is
A. 1/4 P
B. 1/2 P
C. P
D. 2P
E. 4P
To solve this problem, we need to understand the concept of momentum. Momentum is defined as the product of an object's mass and its velocity. In equation form, momentum (P) can be expressed as:
P = m * v
where P is the momentum, m is the mass, and v is the velocity.
Given that the two carts are at rest initially and are pushed with the same force for the same amount of time, we can assume that the force is constant for both carts.
We know that the momentum of an object can be calculated using the equation P = m * v. However, in this case, we don't know the individual velocities of the carts after the push.
However, we can use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after any interaction.
Since the system is closed and there are no external forces acting on the system (no friction or external pushes), the total momentum before the push is zero (as the carts are at rest).
After the push,
P_total = P_lighter + P_heavier
Since the carts have the same force applied for the same duration, the change in momentum should be equal for both carts. Therefore,
P_lighter = -P_heavier
The negative sign indicates that the direction of the momentum for the lighter cart is opposite to that of the heavier cart. This makes sense because the lighter cart will experience a greater acceleration than the heavier cart due to its smaller mass for the same force applied.
Since the question asks for the magnitude of the momentum of the heavier cart, we can solve for that value by substituting -P_heavier for P_lighter in the equation.
P_total = P_lighter + P_heavier
0 = -P_heavier + P_heavier
0 = 0
This shows that the total momentum of the system before and after the push is zero. As a result, the magnitude of the momentum of the heavier cart is 0. Therefore, the answer is 0.
The correct option is not listed among the choices provided.
To find the answer, we need to understand the relationship between force, time, and momentum. The formula for momentum is given by:
Momentum (p) = mass (m) x velocity (v)
When a person pushes an object with a force (F) for a certain amount of time (t), the change in momentum (Δp) can be calculated using Newton's second law:
Δp = F x t
In this case, we have two carts: one is twice as heavy as the other. Let's denote the mass of the lighter cart as m1 and the mass of the heavier cart as m2. Since the force and time values are the same for both carts, we can compare their changes in momentum.
Now, let's calculate the change in momentum for the lighter cart:
Δp1 = F x t
Since the force and time are the same for both carts, we can simply express the mass of the lighter cart in relation to the mass of the heavier cart:
m1 = (1/2) x m2
So, substituting this relation into the equation above:
Δp1 = F x t = (1/2) x m2 x v1
Next, let's calculate the change in momentum for the heavier cart:
Δp2 = F x t = m2 x v2
Now, comparing the changes in momentum for the two carts, we have:
Δp2/Δp1 = (m2 x v2) / ((1/2) x m2 x v1)
Simplifying this equation:
Δp2/Δp1 = 2 x (v2/v1)
Δp2/Δp1 = 2 x (v2/v1) = 2
This means that the magnitude of the momentum of the heavier cart is twice (2 times) that of the lighter cart after the push. Therefore, the correct answer is option D: 2P.