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

hence, lost in kinetic energy will be 140.625 J

[ 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

= 300 - 159.375

= 140.625 J

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Answer 2

$\huge\bold\purple{Answer:-}$

140.625 J