In: Physics
A 2 kg stone is dropped from a 50-m-tall building. Simultaneously, a second 1 kg stone is thrown horizontally from the building at a speed of 15 m/s, as shown in the figure.
Calculate the x position of the center of mass of the two-stone system the moment after they are set in motion.
Calculate the y position of the center of mass of the two-stone system the moment after they are set in motion.
Calculate the x position of the center of mass of the two-stone system 2 s after they are set in motion.
Calculate the y position of the center of mass of the two-stone system 2 s after they are set in motion.
Calculate the x position of the center of mass of the two-stone system the moment they hit the ground.
Calculate the y position of the center of mass of the two-stone system the moment they hit the ground.
let the top of tower is choosen as the origin (0,0)
(a) x =( m1x1 +m2x2)/(m1+m2)
x= (2x0 +1x0)/(2+1) =0
(b) y=( m1y1 +m2y2)/(m1+m2)
y= (2x0 +1x0)/(2+1) =0
(c) and (d) After 2 seconds,
for stone with mass 2kg which is dropped
y= Voy t - 1/2gt2 ; Voy=0
y= -1/2 (9.8)(2)2 = -19.6 m
x=0
for stone of mass 1kg which is thrown horizontally at 15 m/s
x= 15x2= 30m
y= Voy t - 1/2gt2 ; Voy=0
y= -1/2 (9.8)(2)2 = -19.6 m
Xcom=( m1x1 +m2x2)/(m1+m2)
Xcom=( 2x0 + 1x30)/(2+1)= 10 m
Ycom= ( m1y1 +m2y2)/(m1+m2)
Ycom = (2x-19.6 +1x-19.6)/(2+1) =-19.6m
(e) and(f)
let t is the time taken by both bodies to reach ground
-50= 1/2 (-9.8) t2
t=3.2 seconds
for stone with mass 2kg which is dropped
x=0 and y= -50 m on reaching ground
for stone of mass 1kg which is thrown horizontally at 15 m/s
x= 15x3.2= 48 m and y= -50m
Xcom=( m1x1 +m2x2)/(m1+m2)
Xcom=( 2x0 + 1x48)/(2+1)= 16m
Ycom= ( m1y1 +m2y2)/(m1+m2)
Ycom = (2x-50 +1x-50)/(2+1) =-50m