Question

A ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is three fourths of the original. Find the coefficient of restitution.

Answer 1

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ANSWER::

Mass of first ball = m

Speed of first ball = v

Mass of second ball = m

Let final velocities of first and second ball are v₁ and v₂ respectively.

Using law of conservation of momentum,

m(v₁ + v₂) = mv

v₁ + v₂ = v Equation 1

And , v₁ - v₂ = ev Equation 2

Given:-

Final Kinetic Energy = 3/4 Initial Kinetic Energy

(1/2)mv₁² + (1/2)mv₂² = (3/4) x 1/2 mv²

v₁² + v₂² = (3/4) v²

[(v₁ + v₂)² + (v₁ - v₂)² ] / 2 = 3v² / 4

(1 + e²)v² / 2 = 3v² / 4

1 + e² = 3/2

e² = 1/2

e = 1 / √2

Hope it helps!

Answer 2

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➠ Mass of 1st ball => m
➠ speed => v
➠ Mass of 2nd ball => m

➧ Let final velocities of 1st and 2ndball are v1 and v2 respectively.

➧ Using law of conservation of momentum,

m (v1 + v2) = mv
➠ v1 + v2 = V ..........①

➧ Also,

v1 - v2 = ev ..............②

➧ Given that,

➧ Final K.E = ¾ Initial K.E

➠ ½ mv1 ^2 + ½ mv2 ^2 = ¾ × ½

➠ v1 ^2 + v2 ^2 = ¾ v2

➠ (v1 + v2)^2 + (v1 - v2)^2 / 2 = 3/4 v2

➠ (1 + e^2) v^2 / 2 = 3/4

➠ 1 + e^2 = 3 / 2

➠ e^2 = 1 / 2

➠ e = 1 / √2...✔