Question

two spheres, a and b, of identical size and surface material, but different masses, are dropped from rest near the surface of earth. while falling, each sphere experiences a resistive force which is proportional to the sphere’s velocity. what are the relationships of the magnitude of the initial acceleration (a_0) of each sphere and of the terminal speed (v_t) of each sphere if (m_a < m_b)?
initial acceleration\tterminal speed
a\t(a_{0,a}=a_{0,b})\t(v_{t,a}=v_{t,b})
b\t(a_{0,a}=a_{0,b})\t(v_{t,a}<v_{t,b})
c\t(a_{0,a}<a_{0,b})\t(v_{t,a}=v_{t,b})
d\t(a_{0,a}<a_{0,b})\t(v_{t,a}<v_{t,b})

Explanation:

Step1: Analyze initial - acceleration

At the moment of release ($v = 0$), the resistive force $F_r=kv = 0$ (where $k$ is the proportionality constant and $v$ is the velocity). According to Newton's second - law $F = ma$, and the only force acting on the spheres is the gravitational force $F = mg$. So, $mg=ma_0$, and $a_0 = g$. Since $a_0$ is independent of mass, $a_{0,A}=a_{0,B}=g$.

Step2: Analyze terminal - speed

At terminal speed, the net force on the sphere is zero, i.e., $mg=F_r$. Given $F_r = kv_T$, we have $mg=kv_T$, and $v_T=\frac{mg}{k}$. Since $m_A\lt m_B$ and $k$ is the same for both spheres (identical size and surface material), $v_{T,A}\lt v_{T,B}$.

Answer:

B. $a_{0,A}=a_{0,B}$, $v_{T,A}\lt v_{T,B}$