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Question Number 23518 by Tinkutara last updated on 01/Nov/17

A ball of mass 1 kg moving with velocity 3 m/s collides with spring of natural length 2 m and force constant 144 N/m. What will be length of compressed spring?

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{velocity} \\ $$$$\mathrm{3}\:\mathrm{m}/\mathrm{s}\:\mathrm{collides}\:\mathrm{with}\:\mathrm{spring}\:\mathrm{of}\:\mathrm{natural} \\ $$$$\mathrm{length}\:\mathrm{2}\:\mathrm{m}\:\mathrm{and}\:\mathrm{force}\:\mathrm{constant}\:\mathrm{144}\:\mathrm{N}/\mathrm{m}. \\ $$$$\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{length}\:\mathrm{of}\:\mathrm{compressed} \\ $$$$\mathrm{spring}? \\ $$

Commented by Tinkutara last updated on 01/Nov/17

Book′s answer: 1.5 m Mine: 1.75 m Which is right?

$$\mathrm{Book}'\mathrm{s}\:\mathrm{answer}:\:\mathrm{1}.\mathrm{5}\:\mathrm{m} \\ $$$$\mathrm{Mine}:\:\mathrm{1}.\mathrm{75}\:\mathrm{m} \\ $$$$\mathrm{Which}\:\mathrm{is}\:\mathrm{right}? \\ $$

Commented by mrW1 last updated on 01/Nov/17

case 1: (1/2)mv^2 =(1/2)kΔ^2 Δ=v(√(m/k))=3(√(1/(144)))=(3/(12))=(1/4)=0.25 m L_2 =L−Δ=2−0.25=1.75 m case 2: (1/2)mv^2 +mgΔ=(1/2)kΔ^2 kΔ^2 −2mgΔ−mv^2 =0 ⇒Δ=((2mg+(√(4m^2 g^2 +4kmv^2 )))/(2k)) ⇒Δ=((mg+(√(m^2 g^2 +kmv^2 )))/k)=((10+(√(100+144×9)))/(144))=0.32 m L_2 =L−Δ=1.68 m

$$\mathrm{case}\:\mathrm{1}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{k}\Delta^{\mathrm{2}} \\ $$$$\Delta=\mathrm{v}\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=\mathrm{3}\sqrt{\frac{\mathrm{1}}{\mathrm{144}}}=\frac{\mathrm{3}}{\mathrm{12}}=\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0}.\mathrm{25}\:\mathrm{m} \\ $$$$\mathrm{L}_{\mathrm{2}} =\mathrm{L}−\Delta=\mathrm{2}−\mathrm{0}.\mathrm{25}=\mathrm{1}.\mathrm{75}\:\mathrm{m} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{mv}^{\mathrm{2}} +\mathrm{mg}\Delta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{k}\Delta^{\mathrm{2}} \\ $$$$\mathrm{k}\Delta^{\mathrm{2}} −\mathrm{2mg}\Delta−\mathrm{mv}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\Delta=\frac{\mathrm{2mg}+\sqrt{\mathrm{4m}^{\mathrm{2}} \mathrm{g}^{\mathrm{2}} +\mathrm{4kmv}^{\mathrm{2}} }}{\mathrm{2k}} \\ $$$$\Rightarrow\Delta=\frac{\mathrm{mg}+\sqrt{\mathrm{m}^{\mathrm{2}} \mathrm{g}^{\mathrm{2}} +\mathrm{kmv}^{\mathrm{2}} }}{\mathrm{k}}=\frac{\mathrm{10}+\sqrt{\mathrm{100}+\mathrm{144}×\mathrm{9}}}{\mathrm{144}}=\mathrm{0}.\mathrm{32}\:\mathrm{m} \\ $$$$\mathrm{L}_{\mathrm{2}} =\mathrm{L}−\Delta=\mathrm{1}.\mathrm{68}\:\mathrm{m} \\ $$

Commented by Tinkutara last updated on 01/Nov/17

Thank you very much Sir! What is the reason for case 2?

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reason}\:\mathrm{for}\:\mathrm{case}\:\mathrm{2}? \\ $$

Commented by mrW1 last updated on 01/Nov/17

case 2 is the case in that the spring is compressed mostly. that is when the mass hits the spring vertically with a speed v. i just wanted to see if the spring could be compressed 0.5m even in the worst case.

$$\mathrm{case}\:\mathrm{2}\:\mathrm{is}\:\mathrm{the}\:\mathrm{case}\:\mathrm{in}\:\mathrm{that}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{is} \\ $$$$\mathrm{compressed}\:\mathrm{mostly}.\:\mathrm{that}\:\mathrm{is}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{mass}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{spring}\:\mathrm{vertically}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{speed}\:\mathrm{v}.\:\mathrm{i}\:\mathrm{just}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{see}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{spring}\:\mathrm{could}\:\mathrm{be}\:\mathrm{compressed}\:\mathrm{0}.\mathrm{5m}\:\mathrm{even} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{worst}\:\mathrm{case}. \\ $$

Commented by mrW1 last updated on 01/Nov/17

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