Question

student 1 is standing on a cart holding a small stone, while student 2 is standing on the ground as shown. the cart is moving at a constant speed (v_1) with respect to the ground in the +x - direction. student 1 then releases the stone from rest with respect to the cart. just before the stone reaches the ground, student 1 measures the stones speed as (2v_1) and its acceleration as (a_1). at the same instant, student 2 measures the stones speed as (v_2) and its acceleration as (a_2). which of the following correctly states the relationships of (v_2) to (v_1) and of (a_2) to (a_1)?
speeds accelerations
a (v_2=sqrt{5}v_1) (a_2 = a_1)
b (v_2=sqrt{5}v_1) (a_2=sqrt{5}a_1)
c (v_2 = 3v_1) (a_2 = a_1)
d (v_2 = 3v_1) (a_2=sqrt{5}a_1)

Explanation:

Step1: Analyze horizontal - velocity component

The cart is moving with a constant speed \(v_1\) in the \(+x\) - direction. The horizontal - velocity component of the stone with respect to the ground is \(v_{x}=v_1\) (because there is no horizontal acceleration for the stone in the absence of non - conservative horizontal forces).

Step2: Analyze vertical - velocity component

Student 1 measures the speed of the stone just before it reaches the ground as \(2v_1\). Since the horizontal component of velocity with respect to the cart is \(0\) (released from rest with respect to the cart), the vertical component of velocity with respect to the cart \(v_{y1}=2v_1\). With respect to the ground, the vertical component of velocity is also \(v_{y1} = 2v_1\) (because the vertical motion is a free - fall motion and is not affected by the horizontal motion of the cart).

Step3: Calculate the speed with respect to the ground

Using the Pythagorean theorem for the velocity components \(v_x\) and \(v_y\), \(v_2=\sqrt{v_{x}^{2}+v_{y}^{2}}\). Substituting \(v_x = v_1\) and \(v_y=2v_1\), we get \(v_2=\sqrt{v_{1}^{2}+(2v_1)^{2}}=\sqrt{v_{1}^{2} + 4v_{1}^{2}}=\sqrt{5}v_1\).

Step4: Analyze acceleration

Both students are observing a free - fall motion of the stone. The acceleration of an object in free - fall near the surface of the Earth is the acceleration due to gravity \(g\). So, \(a_1=a_2 = g\) (acceleration is the same for both observers as it is a property of the free - fall motion and not affected by the relative motion of the observers in non - accelerating horizontal directions).

Answer:

A. \(v_2=\sqrt{5}v_1\), \(a_2 = a_1\)