After falling from rest from a height of 30 m, a 0.48-kg ball rebounds upward, reaching a height of 20 m. If the contact between ball and ground lasted 1.8 ms, what average force was exerted on the ball?
2 answers
I worked it out and got 4.8N
I used the equations of motion for constant acceleration (The object is assumed to fall with the acceleration due to gravity which I took to be 10 m/s^2)
Equations of motion:
V^2=U^2 + 2*a*s
V=U+at
Note: when objects rebounds, acceleration = - 10m/s^2 since the latter is a vector quantity; thus when object falls down and rebounds, sign changes.
Moreover, once the speed with which object rebounds has been found, I used the concept of workdone= force X Distance moved in the direction of the force. Kinetic energy = Workdone
therefore, F X distance = 0.5 X m X V^2
Solve for F and you'll get 4.8 N
Equations of motion:
V^2=U^2 + 2*a*s
V=U+at
Note: when objects rebounds, acceleration = - 10m/s^2 since the latter is a vector quantity; thus when object falls down and rebounds, sign changes.
Moreover, once the speed with which object rebounds has been found, I used the concept of workdone= force X Distance moved in the direction of the force. Kinetic energy = Workdone
therefore, F X distance = 0.5 X m X V^2
Solve for F and you'll get 4.8 N
1 answer