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A bullet of mass 50 g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g=10ms_(-2)

Answer» <html><body><p></p>Solution :`m_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)50g=0.05kg,m_(2)=450g=0.45g` <br/>The speed of the bullet is up. The second <a href="https://interviewquestions.tuteehub.com/tag/body-900196" style="font-weight:bold;" target="_blank" title="Click to know more about BODY">BODY</a> is at rest`(u_(2)=0)`. Let the common velocity of the bullet and the object after the bullet is embedded into the object is <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>. <br/> `v=(m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2))` <br/> `v(00.5u_(1)+(0.45xx0))/((00.5+0.45))=(0.05)/(0.50)u_(1)` <br/> The combined velocity is the <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> velocity for the vertical upward motion of the combine bullet and the object. From second equation of motion, <br/> `v=<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(2gh)` <br/> `v=sqrt(2xx10xx1.8)=sqrt(36)` <br/> `v=6ms^(-1)` <br/> Substituting this in the above equation, the value of `U_(1)` is <br/> `6=(0.05)/(0.50)u_(1) "or" u_(1)(0.50)/(0.05)xx6=10xx6` <br/> `v=60ms_(-1)`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_V01_C04_SLV_021_S01.png" width="80%"/></body></html>


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