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Answer Questions · Q23

Q.Discuss the following as special cases of elastic collisions and obtain their exact or approximate final velocities in terms of their initial velocities.

(i) Colliding bodies are identical.
(ii) A very heavy object collides on a lighter object, initially at rest.
(iii) A very light object collides on a comparatively much massive object, initially at rest.
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Starting from the general head-on elastic-collision formulas (4.8.4), v1=m1−m2m1+m2u1+2m2m1+m2u2v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2 and v2=2m1m1+m2u1+m2−m1m1+m2u2v_2=\frac{2m_1}{m_1+m_2}u_1+\frac{m_2-m_1}{m_1+m_2}u_2:

  1. IDENTICAL colliding bodies, m1=m2m_1=m_2: the first fraction in each formula vanishes (m1−m2m1+m2=0\frac{m_1-m_2}{m_1+m_2}=0) and the second becomes 1, giving v1=u2v_1=u_2 and v2=u1v_2=u_1 -- the two bodies simply EXCHANGE their velocities exactly.
  2. A VERY HEAVY object (m1m_1) striking a much LIGHTER, initially-at-rest object (m2≪m1m_2\ll m_1, u2=0u_2=0): with m1≫m2m_1\gg m_2, m1−m2m1+m2≈1\frac{m_1-m_2}{m_1+m_2}\approx 1 and 2m1m1+m2≈2\frac{2m_1}{m_1+m_2}\approx 2, giving v1≈u1v_1\approx u_1 (the heavy striker is essentially UNAFFECTED, continuing at almost its original speed) and v2≈2u1v_2\approx 2u_1 (the light struck object flies off at roughly DOUBLE the striker's speed). …

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