Question

A neutron moving with a speed of 106 ms-1 suffers a head-on collision with a nucleus of mass number 80. What is the fraction of energy retained by the nucleus ?



(Ans. 79/81)

plzz explain step wise step....​

Answer 1

Answer:

Since Neutron collides with the nucleus, momentum before the collision = u(m₁ - m₂) where u is the initial velocity of the protons = 10⁶ m/s.

∴ Momentum = 10⁶(m₁ - m₂)

After collision, let the speed of the Neutron after the collision be v.

∴ Momentum after collision = v(m₁ + m₂)

Now, Momentum will remains conserved during the collision,

∴ u(m₁ - m₂) = v(m₁ + m₂)

⇒ 10⁶(1 - 80) = v(1 + 80)   [Mass of the Proton = 1 u].

∴ v = 10⁶(1 - 80)/(1 + 80)

⇒ v/u = (1 - 80)/(1 + 80)   -------eq(i).

Now, Fraction of the Kinetic energy retained =  0.5mv²/0.5mu²

⇒ 0.5mv²/0.5mu² = (v/u)²

⇒ v²/u² = [(1 - 80)/(1 + 80)]²

⇒ (v/u)² = (-79)²/(81)²

⇒ (v/u)² = 6241/6561

∴ Fraction of the energy retained by the Nucleus is 6241/6561.

Hope it helps.

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Answer 2

Answer:

Here in this case let a neutron of mass m1 and initial speed u1 strikes a nucleus with mass m2 which is at rest so u2 = 0 and undergoes a perfectly elastic collision .Again let us consider that the final velocity of the neutron is v1 .

Now from conservation of kinetic energy and conservation of momentum we can write that ,

v1=(m1−m2)u1+2m2u2(m1+m2)now as u2=0 so , v1=(m1−m2)u1(m1+m2)⇒v1u1=(m1−m2)(m1+m2)⇒v1u1=(1−m2m1)(1+m2m1)⇒v1u1=(1−A)(1+A)  where , A=m2m1