Question

A stone is dropped at t = 0. A second stone, with 5 times the mass of the first, is dropped from the same point at t = 180 ms. (a) How far below the release point is the center of mass of the two stones at t = 420 ms? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the two-stone system moving at that time?

Answer 1

Answer:

a. 0.379m

b.2.646 m/s

Step-by-step explanation:

Let mass of first stone=$m_1$

Mass of second stone=$5m_1$

Both stones are in  free fall.

The first one free fall during $t_1=420 ms=420\times 10^{-3}=0.42 s$

$1 s=1000 ms$

The second one free fall during $t_2=t_1-180 ms=0.42-0.18=0.24 s$

Both stone start motion  at the same point with initial velocity=0

a.We have to find the  distance between the release point  and  below the release point where is the center of mass of the two stones.

$y=y_0+v_0t+\frac{1}{2}gt^2$

$y_0=0,v_0=0$

Substitute the values

$y_1=\frac{1}{2}(9.8)(0.42)^2=0.864 m$

$y_2=\frac{1}{2}\times (9.8)(0.24)^2=0.282 m$

$y_{com}=\frac{m_1y_1+m_2y_2}{m_1+m_2}=\frac{m_1y_1+5m_1y_2}{m_1+5m_1}=\frac{0.864+5(0.282)}{6}=0.379 m$

Hence, center of mass  from the release point at distance=0.379 m

b.We have to find $v_com$

$v_{com}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_{com}=\frac{m_1v_1+5m_1v_2}{m_1+5m_1}=\frac{v_1+5v_2}{6}$

$v_1=v_01+gt=0+(9.8)(0.42)=4.116 m/s$

$v_2=v_02+gt=(9.8)(0.24)=2.352 m/s$

Now, $v_{com}=\frac{4.116+5(2.352)}{6}=2.646 m/s$