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A bullet when fired at a target with a velocity of 100 m/sec penetrates one metre into it. If thebullet is fired at a similar target with a thickness0.5 metre, then it will emerge from it with avelocity of: |
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Answer» tion:The following equation holds for constant acceleration:v1^2=V2^2+2as where:v1 = initial velocityv2 final velocity ( after acceleration or deceleration period)\s = distance traveledSo in our case v1 = 100 m/s ( in the firs case, the bullet comes to a STOP).s= 1mSo v1^2 = 2as Therefore, a = v1^2/2s = 100^2/(2*1) = 5000 m/s^2 v1 = 100 m/s , and a = 5000 m/s^2 ( assuming the same material as in the firs case so the DECCELERATION is the same)then :v1= 100 , v2 =?s= 0.5mand 100^2= v2^2+2*a*s=v2^2+2*5000*0.5 = v2^2+5000Hence: v2^2 = 10000-5000 = 5000and v2= SQRT(5000) = 71 m/s.The emerging velocity for 0.5 m is 71 m/s. |
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