Problem 7 - University Physics with Modern Physics

Question

Consider two carts, of masses \(m\) and \(2 m\), at rest on a frictionless air track. If you push the lower-mass cart for \(3 \mathrm{~s}\) and then the other cart for the same length of time and with the same force, which cart undergoes the larger change in momentum? a) The cart with mass \(m\) has the larger change b) The cart with mass \(2 m\) has the larger change. c) The change in momentum is the same for both carts. d) It is impossible to tell from the information given.

Step-by-Step Solution

Answer

(a) The lighter cart (b) The heavier cart (c) The change in momentum is the same for both carts Answer: (c) The change in momentum is the same for both carts.

Step 1: Define the variables

Let's define the following variables: - \(F\): the force exerted on the carts - \(t\): duration of time the force is exerted, \(3\,\text{s}\) - \(m\): mass of the first cart - \(2m\): mass of the second cart - \(v_1\): final velocity of the first cart after being pushed - \(v_2\): final velocity of the second cart after being pushed Note that since both carts start at rest, their initial velocities are zero.

Step 2: Apply Newton's Second Law

Newton's second law states that the net force acting on an object is equal to its mass times its acceleration: \(F_{net} = ma\). In this exercise, the applied force, \(F\), is the only force involved, so \(F_{net} = F\). Therefore, we can write the equation for the acceleration of each cart as follows: - Cart 1: \(a_1 = \frac{F}{m}\) - Cart 2: \(a_2 = \frac{F}{2m}\)

Step 3: Determine the final velocities

Using the equations for acceleration, we can determine the final velocities of each cart after being pushed for time \(t\), using the equation \(v = at\). Thus, we have: - Cart 1: \(v_1 = a_1 t = \frac{F}{m} (3\,\text{s}) = \frac{3F}{m}\) - Cart 2: \(v_2 = a_2 t = \frac{F}{2m} (3\,\text{s}) = \frac{3F}{2m}\)

Step 4: Calculate the change in momentum

Since both carts start at rest, their initial momentum is zero. Thus, the change in momentum for each cart equals its final momentum. The momentum of an object is given by \(p=mv\), so we have: - Cart 1: \(\Delta p_1 = m v_1 = m(\frac{3F}{m}) = 3F\) - Cart 2: \(\Delta p_2 = 2m v_2 = 2m(\frac{3F}{2m}) = 3F\)

Step 5: Compare the changes in momentum

We found that: - Cart 1: \(\Delta p_1 = 3F\) - Cart 2: \(\Delta p_2 = 3F\) Since \(\Delta p_1 = \Delta p_2\), we conclude that the change in momentum is the same for both carts. Therefore, the correct answer is (c) The change in momentum is the same for both carts.

Key Concepts

Newton's Second Law

When exploring physics problems, understanding Newton's Second Law is crucial. This law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration, represented by the equation: \[ F_{net} = ma \]In the context of the given problem, Newton's Second Law tells us how to relate the force applied to the carts on the air track to their acceleration. Since both carts experience the same force but have different masses, the acceleration they undergo will differ accordingly.The heftier cart, with mass \(2m\), experiences half the acceleration of the lighter cart because the same force is spread over twice as much mass. This principle underpins the counterintuitive outcome that despite the difference in mass, both carts experience the same change in momentum when acted upon by the same force for the same duration. The equation encapsulates the fundamental relationship between an object's mass, its acceleration, and the force exerted upon it, which is the cornerstone of many physics problems, including this momentum question.

Momentum Conservation

The conservation of momentum is a fundamental concept in physics that states that the total momentum of a closed system is constant if no external forces are acting on it. In the case of our two carts, they are considered a closed system during the time the forces are exerted.

However, we're examining the effect of an external force on the individual carts, so in each case, the system is not closed. The key insight here is that when the same force is applied for the same time, it imparts the same change in momentum to both carts. This is reflected in the solution, which shows that despite their different masses, both carts end up with an equal change in momentum due to the force applied. The conservation principle can be illustrated by the equation:\[ \text{Initial momentum} + \text{Impulse} = \text{Final momentum} \]Since the initial momentum of both carts is zero (they start at rest), and the impulse (force × time) is the same for both, the final momenta are indeed equal. This beautifully showcases the conservation of momentum in action where each action (push) results in a proportionate and predictable reaction (change in momentum).

Frictionless Air Track

The idea of a frictionless air track in physics problems is crucial for simplifying real-world complexity. It allows us to focus on the fundamental principles at play without getting bogged down by the unpredictable nature of friction. In a perfect world without friction, objects move in response to forces without losing energy to heat and other forms of energy dissipation. Therefore, the air track concept assumes that the only forces at work are those we choose to investigate, such as the force being applied to the carts.In the given problem, this assumption means that the only forces we have to consider are the force pushing on the carts and the force of gravity. Since the force of gravity is being countered by the normal force from the track, its effect in the horizontal direction of the force being applied is negligible. Hence, we only concern ourselves with the applied force, which makes analyzing the problem using Newton's laws and momentum conservation much clearer and more straightforward.

Mass and Acceleration

Diving into the relationship between mass and acceleration exposes the core of dynamics in physics. According to Newton's Second Law, acceleration is inversely proportional to mass when force is constant; in simpler terms, a heavier object will accelerate less under the same force than a lighter one.In our problem, the smaller mass cart \(m\) accelerates twice as much as the larger mass cart \(2m\) when the same force is applied. The reason lies within the formula for acceleration derived from Newton's Second Law:\[ a = \frac{F}{m} \]This formula indicates that for a given force, acceleration diminishes with increasing mass. A practical consequence of this distinct connection is that a force applied to multiple objects with varying masses will result in different accelerations but, quite intriguingly, the same change in momentum if the duration of the applied force remains constant. Understanding this relationship is vital in explaining the outcome of the problem and bridging the gap between theoretical foundations and their real-world applications.