Question

Solve the following problem.

A ball of mass 100 g dropped on the ground from 5 m bounces repeatedly. During every bounce, 64% of the potential energy is converted into kinetic energy. Calculate the following:

1. Coefficient of restitution.
2. The speed with which the ball comes up from the ground after the third bounce.
3. The impulse was given by the ball to the ground during this bounce.
4. Average force exerted by the ground if this impact lasts for 250 ms.
5. The average pressure exerted by the ball on the ground during this impact if the contact area of the ball is 0.5 cm².

Answer 1

Given that, for every bounce, 64% of the initial energy is converted to final energy.

a) Coefficient of restitution in case of inelastic collision is given by,

e = `- "v"_"s"/"u"_"a" = - "v"/"u"`    ....(1)

∴ `"e"^2 = "v"^2/"u"^2`

∴ v² = e² × u²

∴ `1/2 "mv"^2 = "e"^2 xx 1/2 "mu"^2`

∴ (K.E.)_f = e² × `1/2` mu²

∴ `("K.E.")_"f"/("K.E.")_"i" = "e"^2`

∴ `64/100 = "e"^2`

∴ e = 0.8

b) From equation (1),

v = – eu

∴ After first bounce,

v₁ = - eu

after second bounce,

v₂ = - ev₁ = - e(- eu) = e²u

and after third bounce,

v₃ = - ev₂ = - e(- e²u) = - e³u

But u = `sqrt(2"gh")`

∴ `"v"_3 = - "e"^3 xx sqrt(2"gh") = -(0.8)^3 xx sqrt(2 xx 10 xx 5)`    .....(∵ h = 5 m given)

= - (0.8)³ × 10 = - 5.12 m/s

c) Impulse given by the ball during third bounce, is,

J = Δp = mv₃ - mv₂ 

= m × (- e³u - e²u)

= - m × e²u × (e + 1)

= - 100 × 10^-3 × (0.8)² × 10 × (0.8 + 1)

= - 1.152 Ns

d) Average force exerted in 250 ms,

F = `"J"/"t" = (- 1.152)/(250 xx 10^-3)`

`= (- 0.1152)/25 xx 10^3`

= – antilog{log (0.1152) – log (25)} × 10³

= – antilog{`bar1` .0615 – 1.3979} × 10³

= – antilog{`bar3` .6636} × 10³

= - 4.609 × 10^-3 × 10³

= - 4.609 N

e) Average pressure for area 

0.5 cm² = 0.5 × 10^-4 m²

P = `"F"/"A" = (4.608)/(0.5 xx 10^-4)`

= 9.216 × 10⁴ N/m²