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