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Chapter 7: Problem 6

Consider two carts, of masses \(m\) and \(2 m,\) at rest on a frictionless air track. If you push the lower-mass cart for \(35 \mathrm{~cm}\) and then the other cart for the same distance 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.

Short Answer

Expert verified
Answer: b) The cart with mass 2m has the larger change.

Step by step solution

01

Recall the concept of momentum

Momentum is a vector quantity which is the product of mass and velocity of an object. Mathematically, momentum (p) is given by: p = m*v Where m is the mass of the object, and v is its velocity.
02

Recall the relation between force, mass, and acceleration:

According to Newton's second law of motion, the force (F) acting on an object is equal to the product of the mass (m) and the acceleration (a) of the object. Mathematically, F = m*a
03

Define the change in momentum

The change in momentum can be defined as the difference between the final momentum (pf) and the initial momentum (pi) of an object. Since both carts are initially at rest, their initial momentum is zero. Therefore, the change in momentum is equal to the final momentum. Mathematically, 螖p = pf - pi
04

Calculate the acceleration of the carts

We know that both carts are pushed with the same force. Let's call this force F. Since the mass of the first cart is m and the mass of the second cart is 2m, we can calculate the acceleration of the first cart (a1) and the second cart (a2) using Newton's second law (F = ma) as follows: a1 = F/m a2 = F/(2m)
05

Calculate the final velocity of the carts

We know that both carts are pushed for the same distance (35 cm or 0.35 m). We can find the final velocity of each cart using the following relation between acceleration, initial velocity, final velocity, and distance (assuming constant acceleration): v^2 = u^2 + 2*a*s where v is the final velocity, u is the initial velocity (0 m/s for both carts), a is the acceleration, and s is the displacement. For cart 1: v1^2 = 0^2 + 2*a1*0.35 For cart 2: v2^2 = 0^2 + 2*a2*0.35
06

Compare the change in momentum

Now that we have the final velocities of the carts, we can calculate their change in momentum and compare them. 螖p1 = m * v1 螖p2 = (2m) * v2 Since a1 = F/m and a2 = F/(2m), the final velocities for cart 1 and cart 2 will not be the same, with v1 being larger than v2. Since mass of cart 1 is m and mass of cart 2 is 2m, the change in momentum of cart 2 will be larger. The answer is: b) The cart with mass 2m has the larger change.

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