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

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To find the ratio of the magnitude of the momentum of the heavier cart to the magnitude of the momentum of the lighter cart, we need to understand the concept of momentum.

Momentum (p) is defined as the product of an object's mass (m) and its velocity (v):

p = m * v

In this case, both carts are being pushed with the same force for the same amount of time. According to Newton's second law of motion, force (F) is equal to the rate of change of momentum:

F = Δp / Δt

Since the force is the same for both carts, we can rewrite the equation as:

F * Δt = Δp = m * Δv

The force, Δt (change in time), and acceleration are the same for both carts. However, the masses of the carts are different. Let the mass of the lighter cart be m and the mass of the heavier cart be 2m. Therefore, the change in velocity, Δv, will be different for each cart.

Now, let's consider the impulse-momentum principle, which states that the change in momentum of an object is equal to the impulse applied to it:

Δp = J

Where J is the impulse and is equal to the product of the force and the time:

J = F * Δt

Since the force and change in time are the same for both carts, their impulses are equal:

J1 = J2

For the lighter cart:

J1 = F * Δt = Δp1

For the heavier cart:

J2 = F * Δt = Δp2

Since the impulses are equal, we can set the two expressions for impulse equal to each other:

Δp1 = Δp2

Now, substituting the equation for momentum (p = m * v) into the previous equation:

m1 * v1 = m2 * v2

v1 / v2 = m2 / m1

Since the mass of the heavier cart is twice the mass of the lighter cart (m2 = 2m1), we can substitute this into the equation:

v1 / v2 = 2m1 / m1

v1 / v2 = 2

This means that the velocity of the lighter cart is twice the velocity of the heavier cart.

Now, let's consider the relationship between momentum and velocity. Since momentum is directly proportional to velocity, if the velocity of the lighter cart is twice that of the heavier cart, then the momentum of the lighter cart will be twice the momentum of the heavier cart.

Therefore, the magnitude of the momentum of the heavier cart is 1/2 P.

The answer is B.