If there be perfectly elastic collision between two bodies of the same mass moving along a straight line, the

Question 27 of 31

Q. If there be perfectly elastic collision between two bodies of the same mass moving along a straight line, the two bodies will

(a) stay static

(b) stick together

(d) move with interchanged speeds

West Bengal WbchseWest Bengal HS First Year (WBCHSE Class XI) Annual Examination 2018MCQ· 1mImportance★★★★★

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Equal-mass, perfectly elastic, head-on collision ⇒ the two bodies simply SWAP velocities.

For a 1-D elastic collision between masses $m_1=m_2=m$ with initial velocities $u_1,u_2$ , both momentum and kinetic energy are conserved:

$mu_1+mu_2=mv_1+mv_2$

$\tfrac12 mu_1^2+\tfrac12 mu_2^2=\tfrac12 mv_1^2+\tfrac12 mv_2^2$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Q. Two equal masses m1 and m2 are moving along the same straight line with velocities 5 ms^-1 and -9 ms^-1 respectively. If the collision is elastic, then calculate the velocities after the collision of m1 and m2 respectively.$
(a) -9 ms^-1 and 5 ms^-1
(b) -4 ms^-1 and 10 ms^-1
(c) 5 ms^-1 and 1 ms^-1
(d) 10 ms^-1 and 0 ms^-1
›Reveal solutionSolution

When two equal masses undergo a 1D elastic collision, they exchange velocities exactly: m1 comes out with m2's original velocity and vice versa. Here that gives m1 = -9 ms^-1 and m2 = 5 ms^-1.

For a general 1D elastic collision between masses m1 (initial velocity u1) and m2 (initial velocity u2), the standard results derived from simultaneously conserving momentum and kinetic energy are:

v1 = [(m1-m2)/(m1+m2)] u1 + [2m2/(m1+m2)] u2

v2 = [2m1/(m1+m2)] u1 + [(m2-m1)/(m1+m2)] u2

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Two equal masses m1 and m2 are moving along the same straight line with velocities 5 ms^-1 and -9 ms^-1 respectively. If the collision is elastic, then calculate the velocities after the collision of m1 and m2 respectively.$
(a) -9 ms^-1 and 5 ms^-1
(b) -4 ms^-1 and 10 ms^-1
(c) 5 ms^-1 and 1 ms^-1
(d) 10 ms^-1 and 0 ms^-1
›Reveal solutionSolution

A special, well-known result of 1-D elastic collision theory: when the two colliding masses are equal, the bodies exchange their velocities exactly — the faster one takes on the slower one's speed and vice versa.

For a 1-D elastic collision, both momentum and kinetic energy are conserved. Solving the two conservation equations for two masses m1 and m2 with initial velocities u1 and u2 gives the general final velocities:

v1 = [(m1−m2)u1 + 2m2·u2] / (m1+m2)

v2 = [(m2−m1)u2 + 2m1·u1] / (m1+m2)

Here m1 = m2 = m (equal masses), so (m1−m2) = 0 and (m2−m1) = 0. The formulas simplify to:

v1 = 2m·u2 / 2m = u2

v2 = 2m·u1 / 2m = u1

…

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Q.If there be perfectly elastic collision between two bodies of the same mass moving along a straight line, the two bodies will

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Q.If there be perfectly elastic collision between two bodies of the same mass moving along a straight line, the two bodies will

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