Question Number 53325 by mr W last updated on 20/Jan/19
Commented by mr W last updated on 20/Jan/19
$${Two}\:{balls}\:{have}\:{equal}\:{size}.\:{Ball}\:{A}\:{has} \\ $$$${mass}\:{m}\:{and}\:{rests}.\:{Ball}\:{B}\:{has}\:{mass} \\ $$$${nm}\:\left({n}\:{is}\:{integer}\:{greater}\:{than}\:\mathrm{1}\right).\:{Ball} \\ $$$${B}\:{moves}\:{with}\:{velocity}\:{v}\:{towards}\:{ball} \\ $$$${A}\:{and}\:{behind}\:{ball}\:{A}\:{there}\:{is}\:{a}\:{wall}. \\ $$$${Assume}\:{there}\:{is}\:{no}\:{friction}\:{and}\:{all} \\ $$$${collisions}\:{are}\:{elastic}\:{and}\:{the}\:{wall}\:{is} \\ $$$${far}\:{away}\:{enough}. \\ $$$${How}\:{many}\:{times}\:{will}\:{both}\:{balls}\: \\ $$$${collide}\:{with}\:{each}\:{other}? \\ $$
Commented by ajfour last updated on 20/Jan/19
$${too}\:{good}\:{a}\:{question}\:{Sir},\:{let}\:{me}\:{two} \\ $$$${days}\:{at}\:{least}. \\ $$
Commented by mr W last updated on 20/Jan/19
$${thanks}\:{sir}!\:\:{waiting}\:{for}\:{your}\:{result}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
$${i}\:{have}\:{seen}\:{similar}\:{question}\:{in}\:{you}\:{tube}…{sharing} \\ $$$${the}\:{link}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
Commented by mr W last updated on 20/Jan/19
$${thanks}\:{for}\:{this}\:{result}\:{but}\:{which}\:{i}\:{can}\:{not} \\ $$$${follow}\:{yet}.\:{i}'{ll}\:{go}\:{to}\:{watch}\:{the}\:{video}, \\ $$$${but}\:{at}\:{first}\:{i}'{ll}\:{try}\:{to}\:{solve}\:{in}\:{my} \\ $$$${own}\:{way}.\:{my}\:{understanding}\:{is}: \\ $$$${during}\:{the}\:{first}\:{x}\:{collisions}\:{the} \\ $$$${energy}\:{will}\:{be}\:{transfered}\:{from}\:{ball}\:{B} \\ $$$${to}\:{ball}\:{A}.\:{during}\:\:{the}\:{next}\:{y}\:{collisions} \\ $$$${the}\:{energy}\:{will}\:{be}\:{tranfered}\:{back}\:{to} \\ $$$${ball}\:{B}\:{and}\:{ball}\:{B}\:{will}\:{be}\:{finally}\:{faster} \\ $$$${than}\:{ball}\:{A}\:{and}\:{no}\:{further}\:{collision} \\ $$$${is}\:{possible}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
![tanθ=(√(m_L /m_H )) N_c θ=π N_c =(π/θ)=(π/(tan^(−1) (√(m_L /m_H )))) [N_c =number of collissiin] 1)example..(m_L /m_H )=(1/(100)) tan^(−1) ((√(1/(100))) ) tan^(−1) ((1/(10)))≈(1/(10)) N_c =(π/(1/(10)))=10π≈31 number of collision 2)example two (m_L /m_H )=(1/(10000)) tan^(−1) ((√(m_L /m_H )) ) tan^(−1) ((1/(100))) N_c =(π/(1/(100)))=100π=314 number of collission 3)(m_L /m_H )=(1/n) tan^(−1) ((√(1/n)))≈(1/( (√n))) N_c =(π/(1/( (√n))))=π(√n) number of collission.. sir pls check the video in youtube...also check my answer.. number of collission N_c =π(√n)](https://www.tinkutara.com/question/Q53344.png)
$${tan}\theta=\sqrt{\frac{{m}_{{L}} }{{m}_{{H}} }}\: \\ $$$${N}_{{c}} \theta=\pi \\ $$$${N}_{{c}} =\frac{\pi}{\theta}=\frac{\pi}{{tan}^{−\mathrm{1}} \sqrt{\frac{{m}_{{L}} }{{m}_{{H}} }}}\:\:\left[{N}_{{c}} ={number}\:{of}\:{collissiin}\right] \\ $$$$\left.\mathrm{1}\right){example}..\frac{{m}_{{L}} }{{m}_{{H}} }=\frac{\mathrm{1}}{\mathrm{100}} \\ $$$${tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}}{\mathrm{100}}}\:\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{10}}\right)\approx\frac{\mathrm{1}}{\mathrm{10}} \\ $$$${N}_{{c}} =\frac{\pi}{\frac{\mathrm{1}}{\mathrm{10}}}=\mathrm{10}\pi\approx\mathrm{31}\:{number}\:{of}\:{collision} \\ $$$$\left.\mathrm{2}\right){example}\:{two}\:\:\frac{{m}_{{L}} }{{m}_{{H}} }=\frac{\mathrm{1}}{\mathrm{10000}} \\ $$$${tan}^{−\mathrm{1}} \left(\sqrt{\frac{{m}_{{L}} }{{m}_{{H}} }}\:\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$${N}_{{c}} =\frac{\pi}{\frac{\mathrm{1}}{\mathrm{100}}}=\mathrm{100}\pi=\mathrm{314}\:{number}\:{of}\:{collission} \\ $$$$\left.\mathrm{3}\right)\frac{{m}_{{L}} }{{m}_{{H}} }=\frac{\mathrm{1}}{{n}} \\ $$$${tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}}{{n}}}\right)\approx\frac{\mathrm{1}}{\:\sqrt{{n}}} \\ $$$${N}_{{c}} =\frac{\pi}{\frac{\mathrm{1}}{\:\sqrt{{n}}}}=\pi\sqrt{{n}}\:{number}\:{of}\:{collission}.. \\ $$$${sir}\:{pls}\:{check}\:{the}\:{video}\:{in}\:{youtube}…{also} \\ $$$${check}\:{my}\:{answer}.. \\ $$$$\boldsymbol{{number}}\:\boldsymbol{{of}}\:\boldsymbol{{collission}}\:\boldsymbol{{N}}_{{c}} =\pi\sqrt{{n}}\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
$${thank}\:{you}\:{sir}…{i}\:{have}\:{just}\:{tried}\:{to}\:{understand} \\ $$$${the}\:{video}…{you}\:{pls}\:{go}\:{through}\:{it}…{am}\:{i}\:{really} \\ $$$${understood}\:{it}\:{yet}…{you}\:{pls}\:{cross}\:{check}… \\ $$$${hiwever}\:{your}\:{logic}\:{regarding}\:{collissin}\:{and} \\ $$$${energy}\:{transfer}\:{really}\:{excellent}… \\ $$$${thank}\:{you}\:{sir}… \\ $$
Commented by mr W last updated on 22/Jan/19
$${very}\:{nice}\:{method}\:{in}\:{video}! \\ $$