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Solve the following problem. A marble of mass 2m travelling at 6 cm/s is directly followed by another marble of mass m with double speed.

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Question

Solve the following problem.

A marble of mass 2m travelling at 6 cm/s is directly followed by another marble of mass m with double speed. After a collision, the heavier one travels with the average initial speed of the two. Calculate the coefficient of restitution.

Sum
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Solution

Given: m1 = 2m, m2 = m, u1 = 6 cm/s,

u2 = 2u1 = 12 cm/s,

v1 = `("u"_1 + "u"_2)/2 = 9` cm/s

To find: Coefficient of restitution (e)

Formulae:
i. m1u1 + m2u2 = m1v1 + m2v

ii. e = `("v"_2 - "v"_1)/("u"_1 - "u"_2)`

Calculation:

From formula (i),

[(2m) × 6] + (m × 12) = (2m × 9) + mv2

∴ 12 + 12 = 18 + v2

∴ v2 = 6 cm/s

From formula (ii),

e = `(6 - 9)/(6 - 12) = (- 3)/(-6) = 0.5`

The coefficient of restitution is 0.5.

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Chapter 4: Laws of Motion - Exercises [Page 77]

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Balbharati Physics [English] Standard 11 Maharashtra State Board
Chapter 4 Laws of Motion
Exercises | Q 3. (xiii) | Page 77

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