A 2 kg stone is dropped from a 50-m-tall building. Simultaneously, a second 1 kg

Question

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.

Explanation / Answer

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

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