Question

. A ball of mass 6 kg, moving with speed 10 m/s,
collides head-on with a ball of identical mass at
rest. If the coefficient of restitution is e=1/4,then
the loss of kinetic energy in the collision will be
(1) 1125 J
(2) 1125
(3) 2500 J
(4) 2689 J
​

Answer 1

Answer:

the answer will be 4)- 2689J

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Answer 2

Answer:

       =140.626

Explanation:

[ here your options are wrong. because lost in kinetic energy can't be greater than initial kinetic energy ]

it is given that, coefficient of restitution , e = 1/4

we know, coefficient of restitution is the ratio of velocity of seperation to velocity of approach.

i.e., e = (v₂ - v₁)/(u₁- u₂)

initial velocity of first ball, u₁ = 10m/s

initial velocity of 2nd ball, u₂ = 0 m/s

so,

e = 1/4 = (v₂ - v₁)/(10 - 0)

⇒v₂ - v₁ = 2.5 .......(1)

form law of conservation of linear momentum,

m₁u₁+ m₂u₂ = m₁v₁ + m₂v₂

⇒(6kg)(10m/s) + 0 = (6kg)(v₁) + (6kg)(v₂)

⇒v₁+ v₂ = 10 ......(2)

from equations (1) and (2) we get,

v₂ = 6.25 m/s , v₁ = 3.75 m/s

now, loss of kinetic energy, K.E = initial kinetic energy - final kinetic energy

= 1/2 × 6 × 10² - 1/2 × 6kg × (6.25)² - 1/2 × 6kg × (3.75)²

= 300- 117.1875 - 42.1875

=140.626