A body of mass 1.0 kg strikes elastically with another body at rest and continues to move in the same direction with one-fourth of its initial velocity. The mass of the other body is

A body of mass 1.0 kg strikes elastically with another body at rest an

1.	A body of mass 1.0 kg strikes elastically with another body at rest and continues to move in the same direction with one-fourth of its initial velocity. The mass of the other body is
Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.6 kg<br/>2.4 kg <br/>3.0 kg<br/>4.0 kg</p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C04_E01_150_S01.png" width="80%"/> <br/> According to the law of convervation of <a href="https://interviewquestions.tuteehub.com/tag/linear-1074458" style="font-weight:bold;" target="_blank" title="Click to know more about LINEAR">LINEAR</a> <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> , we get `m_1 u_1 + m_2 xx 0 = m_1 v_1 + m_2v_2` <br/> or `m_1 u_1 = m_1 xx (u_1)/(4) + m_2v_2 " or " m_2v_2 = 3/4 m_1u_1 ""….(i)` <br/> According to the law of conservation of kinetic energy, we get <br/> `1/2 m_1 u_1^2 + 1/2 m_2 xx 0^2 = 1/2 m_1v_1^2 + 1/2 m_2v_2^2` <br/> or `m_2v_2^2 = 15/16 m_1u_1^2 "".....(ii)` <br/> Divide (ii) by (i), we get <br/> `(m_2 v_2^2)/(m_2v_2) = (15/16 m_1 u_1^2)/(3/4 m_1 u_1) " or " v_2 = 5/4 u_1` <br/> On substituting this value of `v_2` in (i), we get <br/> `m_2 5/4 u_1 = 3/4 m_1 u_1 " or " m_2 = 3/5 m_1 = 3/5 xx 1.0 kg = 0.6 kg`.</body></html>