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Physics / Chapter 42: Nuclear Physics / Section 42-1

Discovering the Nucleus — Question 8

Section 42-1 · Physics pages 7006–7007

Question 8 Target recoil limits the fixed-nucleus approximation A nonrelativistic projectile of mass m collides head-on elastically with a stationary target of mass M = 20m. Its initial kinetic energy is 5.00 MeV. m, v M V Schematic; labels, not drawing lengths, specify the data. (a) Use momentum and kinetic-energy conservation to find the final projectile velocity relative to its initial velocity. (b) Find the recoil energy fraction and the target recoil energy.
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Question 8 - Solution Target recoil limits the fixed-nucleus approximation m, v M V (a) Method and calculation mv = mv′ + MV, mv2 = m(v′)2 + MV 2. v′/v = (m −M)/(m + M) = −19/21 . (b) Interpretation and further calculation ER/E = 1 −(v′/v)2 = 4mM/(m + M)2 = 80/441 = .1814 . ER = .9070 MeV, E′ = 4.093 MeV . Physical check An infinitely heavy nucleus would take momentum with negligible recoil energy. Even a target twenty times heavier produces an appreciable head-on energy loss.

Question 8

Target recoil limits the fixed-nucleus approximation

A nonrelativistic projectile of mass m collides head-on elastically with a stationary target of mass M=20mM=20m. Its initial kinetic energy is 5.00 MeV.

Diagram shown in the original SVG worksheet.

  1. Use momentum and kinetic-energy conservation to find the final projectile velocity relative to its initial velocity.

  2. Find the recoil energy fraction and the target recoil energy.

Question 8 - Solution

Target recoil limits the fixed-nucleus approximation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

mv=mv′+MV,mv2=m(v′)2+MV2.mv=mv\prime+MV,\quad mv^2=m(v\prime)^2+MV^2.v′/v=(m−M)/(m+M)=−19/21.\boxed{v\prime/v=(m-M)/(m+M)=-19/21}.

(b) Interpretation and further calculation

ER/E=1−(v′/v)2=4mM/(m+M)2=80/441=.1814.\boxed{E_R/E=1-(v\prime/v)^2=4mM/(m+M)^2=80/441=.1814}.ER=.9070MeV,E′=4.093MeV.\boxed{E_R=.9070\,\mathrm{MeV},\quad E\prime=4.093\,\mathrm{MeV}}.

Physical check

An infinitely heavy nucleus would take momentum with negligible recoil energy. Even a target twenty times heavier produces an appreciable head-on energy loss.

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