InterviewSolution
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InterviewSolution
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A wooden block of mass `10 gm` is dropped from the top of a tower `100m` high. Simultaneously, a bullet of mass `10 gm` is fired from the foot of the tower vertically upwards with a velocity of `100m//sec`, figure. If the bullet is embedded in it, how high will it rise above the tower before it starts falling ? (Consider `g=10m//sec^(2))` A. `80m`B. `85 m`C. `75 m`D. `10 m` |
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Answer» Correct Answer - C Ignoring `g` compared to initial velocity, the bullet and block will meet after `t=(h)/(u)=(100)/(100)=1sec` Durinig this time, distance travelled by the block `s_(1)=(1)/(2)g t^(2)=(1)/(2)xx10xx1^(2)=5m` Distance travelled by the bullet, `s_(2)=100-s_(1)=100-5=95m` velocity of bullet just before collision, `u_(1)=100-10xx1=90 m//s,` upwards velocity of block just before collision, `u_(2)=10xx1=10 m//s, ` downwards According to law of conservation of linear momentum , `m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))V` `0.01(-10+90)=0.02V` `V=40 m//s` Maximum height risen by the block `=(V^(2))/(2g)=(40xx40)/(2xx10)=80m` Height reached above the top of the tower `=80-s_(1)=80-5` `=75m` |
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