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. (Figure 1)

Part A

Calculate the x position of the center of mass of the two-stone system the moment after they are set in motion.

Express your answer in meters as an integer.

Part B

Calculate the y position of the center of mass of the two-stone system the moment after they are set in motion.

Part C

Calculate the x position of the center of mass of the two-stone system 2 s after they are set in motion.

Express your answer in meters as an integer.

Part D

Calculate the y position of the center of mass of the two-stone system 2 s after they are set in motion.

Express your answer in meters as an integer.

Part E

Calculate the x position of the center of mass of the two-stone system the moment they hit the ground.

Express your answer in meters as an integer.

Part F

Calculate the y position of the center of mass of the two-stone system the moment they hit the ground.

Express your answer in meters as an integer.

Explanation / Answer

part A:

initial x position of 2 kg stone=0

initial x position of 1 kg stone=0

hence x position of center of mass=(2*0+1*0)/(2+1)=0 m

part B:

initial y position of 2 kg stone=50

initial y position of 1 kg stone=50

hence y position of center of mass=(2*50+1*50)/(2+1)=50 m

part C:

speed of 2 kg mass in x direction=0 m/s

speed of 1 kg mass in x direction=15 m/s

then speed of center of mass in x direction=(2*0+1*15)/(2+1)=5 m/s

as there is no acceleration in x direction, this speed will remain constant

hence x position of center of mass of the system after 2 seconds=2*5=10 m

part D:

initial speed of 2 kg mass in y direction=0

initial speed of 1 kg mass in y direction=0

hence initial speed of center of mass of the system in y direction=0

acceleration due to gravity of the center of mass=-9.8 m/s^2

then y position of the center of mass of the system=initial position+initial speed in y direction*time+0.5*acceleration*time^2

=50+0-0.5*9.8*2^2=30.4 m

part E:

initial speed of center of mass in y direction=0

acceleration in y direction=-9.8 m/s^2

distance covered before they hit ground=-50 m

of time taken is t seconds,

then -50=0*t-0.5*9.8*t^2

==>t=3.1943 seconds

speed of center of mass in x direction=5 m/s

then x position of center of mass at t=3.1943 seconds is =3.1943*5=15.9715 m

part F:

y position of 2kg mass on ground=0

y position of 1 kg mass on ground=0

then y position of center of mass=(2*0+1*0)/(2+1)=0

note:

i have used g=9.8 m/s^2

if g=10 m/s^2 is to be used , then answers would be as follows:

part A :0

part B: 50 m

part C:10 m

part D:30 m

part E:15.81 m

part F: 0 m

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