A steel ball of mass m falls from a height h
onto a scale calibrated in Newtons. The ball
rebounds repeatedly to nearly the same height
h. The scale is sluggish in its response to the
intermittent hits and displays an average force
Favg, such that
FavgT = F t,
where F t is the brief impulse that the ball
imparts to the scale on every hit, and T is
the time between hits. Calculate this average
force in terms of m, h, and other physical
constants.
1. Favg =mg/T
2. Favg =mg/h
3. Favg = mg
4. Favg = mgT
5. Favg = mgh

The average force F_avg = m*g

Explanation:

Given:

- Mass of the ball = m

- Initially at height = h

- Dropped from rest v_i = 0

Find:

The average force F_avg that the ball imparts to the scale on every hit.

Solution:

- The velocity of the ball just before it hits the scale can be evaluated by the energy conservation as follows:

                                  E.K_1 + E.P_1 = E.K_2 + E.P_2

- Where,                     E.K = 0 initial velocity V_i = 0

                                  E.P = 0 Scale is the datum reference.

Hence,

                                  E.P_1 = E.K_2

                                  m*g*h = 0.5*m*V_o^2

                                  V_o = sqrt ( 2*g*h)

- We will apply the conservation of momentum on the ball just before it hits the scale and just after produces an impulse. The impulse is given by Newton's second law of motion:

                                  Ft = m*(V_f - V_o )

Where, Ft is the impulse. V_f velocity after impact. V_o = V_f (no loss of energy upon impact) - assumption.

- Taking upward direction as +.

                                  Ft = m*(V_o + V_o ) = 2*m*V_o

                                  Ft = 2*m*sqrt ( 2*g*h)

- The total time between two successive hits T can be calculated by using second kinematic equation of motion in vertical direction, as follows:

                                  y = y(0) + V_o*T - 0.5*g*T^2

Where,                       y = y(0) = 0   ... start and end points same.. Datum

Hence,

                                  0 = 0+ V_o*T - 0.5*g*T^2

                                  0 = V_o - 0.5*g*T

                                  T = 2*V_o / g = 2*sqrt ( 2*g*h) / g

- The average force F_avg can now be calculated using the expression given:

                                  F_avg*T = Ft

                                  F_avg = Ft / T

                                  F_avg = [2*m*sqrt ( 2*g*h)] / [2*sqrt ( 2*g*h) / g]

                                  F_avg = m*g

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .