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Grade: 11
        
a bullet loses 1/20 of its velocity in passing through a plank .the least number of planks required to stop the bullet is
4 months ago

Answers : (3)

Arun
22779 Points
							
Let the thickness of one plank = d
and the acceleration provided by the plank = a
v^2 = vo^2 + 2ad
If n planks are required to stop the bullet, then
0^2 = vo^2 + 2a*nd
2and = -vo^2
n = vo^2/(-2ad) -----------------(1)
v = vo - vo/20 = 19 vo/20 in passing through one plank
(19 vo/20)^2 = vo^2 + 2ad
361/400 * vo^2 = vo^2 + 2ad
-2ad = vo^2(1 - 361/400) 
-2ad = vo^2 * 39/400
Substituting this value of -2ad into equation (1):
n = vo^2/(vo^2 * 39/400) = 400/39 
The minimum number of planks needed = smallest integer greater than 400/39 = 11.
 
4 months ago
Khimraj
3008 Points
							
No. of planks
➾ n² / 2n - 1

➧ Where n is loss in velocity.

➧ There your answer will not consume time as time is marks in such exam,

➾ 20² / 2 × 20 - 1
➾ 10.25
➾ 11 planks ...✔
 
3 months ago
aswanth nayak
66 Points
							
Dear Student,
Let v be the speed of the bullet incident on the first plank.
Its speed after it passes the plank = 19/20 v
If x is the thickness of the plank, the deceleration a due to resistance of plank is given by,
            (19/20)² v² - v² = 2ax
 => 2ax = -39/400 v²
Let the bullet is stopped after passing through n such planks. Then the distance covered by bullet is x
       => 0 - (19/20)² v² = 2anx
      => -(19/20)v² = n × -39/400 v²
      => n = 361/39 = 9.25 ≈ 9
 
==> 9 Planks
 
Hope it helps you
 
regards
2 months ago
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