A rubber ball of mass 0.1 Kg is dropped on the ground from a height of

1.	A rubber ball of mass 0.1 Kg is dropped on the ground from a height of 2.5 m and it rises to a height of 0.4m. Assuming the time of contact with the ground to be 0.01 s, calculate the force exerted by the ground on the wall, g = 10ms-2.
Answer» Mass of the ball m = 0.1 kg Time in contact with the ground, t = 0.01s 1. When the ball is dropped on ground, u = 0, s = 2.5m, g = 10ms^-2 From the equation, v² = u² + 2gs v²= 0 + 2 × 10 × 2.5 v = 7.07ms^-1 , downwards. 2. When the ball rises up from the ground, v = 0, s = 0.4m, g = 10ms^-2 From the equation v² = u² + 2gs 0 = u²+ 2(- 10)0.4 u²= 8 u =2.83ms^-1 , upwards. Assuming the upward velocity +ve & downward velocity -ve, change of velocity = 2.83 – (- 7.07) i.e v – u = 9.9ms^-1 ∴ Force exerted by the ground on the ball is, F = m \(\frac {v-u}{t}\) = 0.1 \(\frac {9.9}{0.01}\) = 99 N.

A rubber ball of mass 0.1 Kg is dropped on the ground from a height of 2.5 m and it rises to a height of 0.4m. Assuming the time of contact with the ground to be 0.01 s, calculate the force exerted by the ground on the wall, g = 10ms-2.

Answer»

Mass of the ball m = 0.1 kg

Time in contact with the ground, t = 0.01s

1. When the ball is dropped on ground,

u = 0, s = 2.5m, g = 10ms^-2

From the equation, v² = u² + 2gs

v²= 0 + 2 × 10 × 2.5

v = 7.07ms^-1 , downwards.

2. When the ball rises up from the ground,

v = 0, s = 0.4m, g = 10ms^-2

From the equation v² = u² + 2gs

0 = u²+ 2(- 10)0.4

u²= 8

u =2.83ms^-1 , upwards.

Assuming the upward velocity +ve & downward velocity -ve, change of velocity

= 2.83 – (- 7.07)

i.e v – u = 9.9ms^-1

∴ Force exerted by the ground on the ball is,

F = m \(\frac {v-u}{t}\) = 0.1 \(\frac {9.9}{0.01}\) = 99 N.