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A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g. What is the velocity acquired by the block ? |
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Answer» Here, Mass of the bullet,`m_(1)=10g` `=10/1000`kg =0.01 kg And, Velocity of the bullet, `v_(1)`=400 m/s So, Momentum of bullet=`m_(1)xxv_(1)` `=0.01xx400kg.m//s`…….(1) Now, this bullet of mass 10 g gets embedded into a wooden block of mass 900 g . So ,the mass of wooden blockk alongwith the embedded bullet will become 900+10=910 g. Thus, Mass of wooden block+Bullet, `m_(2)=900+10` =910 g `=910/1000` kg =0.91 kg And, Velocity of wooden block+bullet,`v_(2)=?` (To be calculated) So, Momentum of wooden block+bullet=`m_(2)xxv_(2)` `=0.91xxv_(2)`kg.m/s...(2) Now, according to the law of conservation of momentum, the two momenta as given by equations (1) and (2) should be equal. So, `m_(1)xxv_(1)=m_(2)xxv_(2)` or 0.01`xx400=0.91xxv_(2)` And, `v_(2)=(0.01xx400)/0.91` =4.4 m/s Thus, the velocity acquired by the wooden block (having the bullet embedded in it) is 4.4 metres per second. |
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