Answer:
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
