Question

A block of mass 2.0 kg moving at 2.0 m/s collides head-on with another block of equal mass kept at rest.

(a) Find the maximum possible loss in kinetic energy due to the collision.
(b) If the actual loss in kinetic energy is half of the maximum, find the coefficient of restitution.

Answer 1

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ANSWER::

Mass of first block , m₁ = 2 kg

Speed of block , v₁ = 2 m/s

Mass of second block , m₂ = 2 kg

Let final velocity of second block = v₂

Using law of conservation of momentum ,

2 x 2 = (2+2) v

v = 1 m/s

Therefore ,

(a) Loss in Kinetic Energy in elastic collision = (1/2) x 2 x 2² v - (1/2)(2+2) x 1²

= 4 - 2

= 2 J

(b) Actual loss = Maximum Loss / 2 = 2/2 = 1 J

(1/2) x 2 x 2² - (1/2) x 2 x v₁² + (1/2) x 2 x v₂² = 1

4 - (v₁² + v₂²) = 1

4 - [(1 + e²) x 4] / 2 = 1

2(1 + e²) = 3

1 + e² = 3/2

e² = 1/2

e = 1 / √2

Hope it helps!

Answer 2

(a) maximum kinetic energy is lost in inelastic collision when the two masses move together with same velocity `v2` after collision.

By conservation of momentum mv1=(m+m)v2 v2=1 m/s. Initail KE=1/2 2 2^2=4 J.

Final KE=1/2 4 1^2= 2 J.

Loss in KE= 4-2=2 J (b) Loss in KE=1 J.

Final KE=4-1=3 J. 3=1/2 m(v1)^2 + 1/2 m(v2)^2, 3= (v1)^2 + (v2)^2 also , 4=mv1 +mv2, v1+v2 =2, v1=2 -v2(put this value in the KE equation and solve) v1=1 +(1/2)^1/2, v2= 1 - (1/2)^1/2, velocity of separation= v1-v2 =2^1/2, velocity of approach =2 e=(2^1/2) / 2 =1/(2^1/2).