Q.When two bodies make a head-on elastic collision, show that the coefficient of restitution equals 1.
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Start your 14-day free trial to unlock the full solution →Combining momentum and KE conservation for an elastic collision gives (v2-v1) = (u1-u2), so e = 1.
Consider two bodies of masses m1 and m2, moving along the same straight line with initial velocities u1 and u2 (u1 > u2, so body 1 catches up with body 2), undergoing a head-on (one-dimensional) ELASTIC collision, after which their velocities become v1 and v2.
By definition, an elastic collision conserves BOTH total momentum and total kinetic energy.
Step 1: Conservation of momentum.
m1 u1 + m2 u2 = m1 v1 + m2 v2
=> m1 (u1 - v1) = m2 (v2 - u2) ... (i)
Step 2: Conservation of kinetic energy.
(1/2) m1 u1^2 + (1/2) m2 u2^2 = (1/2) m1 v1^2 + (1/2) m2 v2^2
=> m1 (u1^2 - v1^2) = m2 (v2^2 - u2^2)
=> m1 (u1 - v1)(u1 + v1) = m2 (v2 - u2)(v2 + u2) ... (ii)
Step 3: Divide equation (ii) by equation (i).
The common factors m1(u1 - v1) and m2(v2 - u2) cancel from both sides, leaving:
(u1 + v1) = (v2 + u2)
=> u1 - u2 = v2 - v1
Step 4: Interpret this result.
The left side, (u1 - u2), is the relative velocity of approach of the two bodies before collision (how fast body 1 is closing in on body 2).
The right side, (v2 - v1), is the relative velocity of separation after collision (how fast body 2 is pulling away from body 1).
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