A solid rubber ball of radius 1.5cm is dropped from a height of 1 m onto a hard

Question

A solid rubber ball of radius 1.5cm is dropped from a height of 1 m onto a hard surface. it bounces. the objective of this problem is to figure out the magnitude of the contact force on the ball during the bounce.

a) the contact force pushing the ball upwards will be far larger in magnitude than the gravitational force pushing the ball downwards. explain why in words.

b) estimate the mass m of the ball and also the speed v at which the ball wil be traveling just before it hits the floor. use the internet or the library of the density of rubber.

c) if the ball is in contact with the floor for about one millisecond, and if it bounces back upwards with about the equal and opposite velocity to that it had before the contact, what is the mean acceleration a of the ball during the bounce?

Explanation / Answer

All these problems can be solved using the "Kinematic Equatiions" and Newtons 2nd Law : Force=mass*accelleration

a) The force of contact will be greater than the gravitational force because the ball is "accellerating" from 1m above the ground till it hits the ground. gravity is defined as 9.81m/s^s and Force = mass* accelleration. Since the ball is accellerating by 9.81m/s^2 for 1 m and it weights 16.67*10^3kg...the force of contact at the ground will be greater than the gravitational constanstant of accelleration. I wouldn't say the force of contact will be "FAR" greater than the force of gravity because it is only dropped from one meter

^^^ this question is kinda weird^^^

questions B) and C) are generic ones...

b) To find the mass of the ball. You need to find a given value of density () of rubber (ie 1.2* 10^3 kg/m3) and also the volume of the ball since it is a sphere, the volume is given by V= (4/3)* r ^3 (but first convert cm to m (ie 1.5 cm = .015 m ) so volume of the ball is 14.137 * 10^6 m^3..... density = mass/volume soo... the mass is 16.97 *10^3 kg or .01697 grams.

The speed of the ball as it hits the ground is found by the kinematic equation (or many other ways) Vf2 = Vi2 + 2* a * d Vi =0 since the ball is "dropped" (meaning initial velocity is zero) so the velocity as the ball hits the ground is 4.43 m/s.


c) If the ball returns to its initial position after the bounce, the mean accelleration will be 0. It accellerates going down, and decellerates going back up....this is assuming the ball returns to the intial position it was dropped at.

Hope this helps

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