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A rubber cord has a cross -sectional area {1}{m}{m}^{{{2}}}and

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A rubber cord has a cross -sectional area $1 m m^{2}$ and total unstretched length $10.0 c m$ . It is streched to $12.0 c m$ and then released to project a missile of mass 5.0 g.Taking young's modulus Y for rubber as $5.0 \times 10^{8} N / m^{2}$ .Calculate the velocity of projection .

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A rubber cord has a cross -sectional area

1 m m^{2}

and total unstretched length

10.0 c m

. It is streched to

12.0 c m

and then released to project a missile of mass 5.0 g.Taking young's modulus Y for rubber as

5.0 \times 10^{8} N / m^{2}

.Calculate the velocity of projection .

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Text solutionVerified

Answer: A::B::CSolution: Equivalent force constant of rubber cord

k = Y \frac{A}{l} = \frac{( 5.0 \times 18 ^{8} ) ( 1.0 \times 10 ^{- 6} )}{( 0.1 )} = 5.0 \times 10^{3} N / m

Now,from conservation of mechanical energy,
elastic potential energy of cord=Kinetic energy of missile
:.

\frac{1}{2} k (Δ l)^{2} = \frac{1}{2} m v^{2} \Rightarrow ∴ v = (\frac{k}{m}) Δ l

= ⎝ ⎛ \frac{5.0 \times 10 ^{3}}{5 \cdot 0 \times 10 ^{- 3}} ⎠ ⎞ (12.0 - 10.0) \times 10^{- 2} = 20 m / s

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Question Text	A rubber cord has a cross -sectional area $1 m m^{2}$ and total unstretched length $10.0 c m$ . It is streched to $12.0 c m$ and then released to project a missile of mass 5.0 g.Taking young's modulus Y for rubber as $5.0 \times 10^{8} N / m^{2}$ .Calculate the velocity of projection .
Answer Type	Text solution:1
Upvotes	152