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A bomb explodes andsplits up into three fragments. Two fragments, each of mass 200g, move away from each other making an angle 120^@ , at a speed of 100 m*s^(-1). Find the direction and velocity of the third fragment whose mass is 500 g. Also find outthe energy released in explosion. |
| Answer» <html><body><p></p>Solution :Fig.1.45 shows the velocity of fragment A and B <a href="https://interviewquestions.tuteehub.com/tag/along-1974109" style="font-weight:bold;" target="_blank" title="Click to know more about ALONG">ALONG</a> OA and OB .A and B have equal mass and speed. From the law of conservation of linear momentum, the third piece must move along OD, <a href="https://interviewquestions.tuteehub.com/tag/inthe-1050050" style="font-weight:bold;" target="_blank" title="Click to know more about INTHE">INTHE</a> direction opposite to the resultant of OA and OB. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XI_P1_U04_C01_SLV_060_S01.png" width="80%"/> <br/> If the velocity of the third <a href="https://interviewquestions.tuteehub.com/tag/pieceis-2928832" style="font-weight:bold;" target="_blank" title="Click to know more about PIECEIS">PIECEIS</a> v, then taking the component along the <a href="https://interviewquestions.tuteehub.com/tag/line-1074199" style="font-weight:bold;" target="_blank" title="Click to know more about LINE">LINE</a> CD in CGS system. <br/> `500v =200xx10^4 cos 60^@ +200xx10^4 cos 60^@` <br/> `or, v=(200xx10^4)/(500)=4xx10^3 cm *s^(-1) =40 m*s^(-1)`. <br/> Hence, the velocity of the third fragment is `40 m*s^(-1)`. it moves so as tomake an angle `<a href="https://interviewquestions.tuteehub.com/tag/120-270396" style="font-weight:bold;" target="_blank" title="Click to know more about 120">120</a>^@` with each of OA and OB. The energy released due to explosion, isthe kinetic energy the three fragment. <br/> `therefore` Energy released <br/> `=1/2 xx200xx(10^4)^2+1/2 xx200xx(10^4)^2+1/2xx500xx(4xx10^3)^2=2400xx10^7 ergs=2400 J`.</body></html> | |