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 \(35 \mathrm{~cm}\) 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

Explain your reasoning. Answer: The change in momentum is the same for both carts. This is because the carts are pushed with the same force for the same distance, and when calculating the change in momentum, both carts have the same final velocity expressions, resulting in the same momentum change value for both carts, \(\Delta p_1 = \Delta p_2 = \sqrt{0.7mF}\).

Step-by-step solution

Write down the given information and the formula for momentum change

We are given two carts with masses \(m\) and \(2m\), and they are pushed with the same force for the same distance of \(35\text{ cm}\). Momentum change is given by the formula: $$\Delta p = p_f - p_i = mv_f - mv_i$$ Initially, both carts are at rest, so their initial velocities, \(v_i\), are 0.

Use Newton's second law of motion to relate force and acceleration

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration: $$F = ma$$ Since both carts are pushed with the same force, we can write their accelerations as: $$a_1 = \frac{F}{m} \quad \text{and} \quad a_2 = \frac{F}{2m}$$

Calculate the final velocities using the formula for displacement

Before finding the change in momentum, we need to find the final velocities of both carts. We can calculate the final velocity using the formula for displacement: $$d = v_it + \frac{1}{2}at^2$$ Since both carts are pushed for the same distance, \(d = 35 \text{ cm} = 0.35 \text{ m}\). Since they start from rest, the initial velocity \(v_i=0\). For the first cart, $$0.35 = 0 + \frac{1}{2}\left(\frac{F}{m}\right)t^2$$ $$t^2 = \frac{0.7m}{F}$$ For the second cart, $$0.35 = 0 + \frac{1}{2}\left(\frac{F}{2m}\right)t^2$$ $$t^2 = \frac{1.4m}{F}$$

Find the change in momentum

Now that we have the time squared expressions for both carts, we can find their final velocities. To do this, we use the second equation of motion: $$v_f = v_i + at$$ For the first cart, $$v_{f1} = 0 + \frac{F}{m}\sqrt{\frac{0.7m}{F}} = \sqrt{\frac{0.7F^2}{m^2}}$$ For the second cart, $$v_{f2} = 0 + \frac{F}{2m}\sqrt{\frac{1.4m}{F}} = \sqrt{\frac{0.7F^2}{4m^2}}$$ Now, we can find the change in momentum for each cart. Change in momentum for the first cart, $$\Delta p_1 = mv_{f1} - 0 = m\sqrt{\frac{0.7F^2}{m^2}} = \sqrt{0.7mF}$$ Change in momentum for the second cart, $$\Delta p_2 = 2m v_{f2} - 0 = 2m\sqrt{\frac{0.7F^2}{4m^2}} = \sqrt{0.7mF}$$

Compare the change in momentum

Now we can compare the change in momentum for both carts: $$\Delta p_1 = \sqrt{0.7mF}$$ $$\Delta p_2 = \sqrt{0.7mF}$$ Since the change in momentum for both carts is the same, the correct answer is: c) The change in momentum is the same for both carts.