Energy Of A Bullet Fired Through Planks

Classical Mechanics Level 3

A bullet is fired into a row of planks in such a way that it loses the same amount of kinetic energy for every plank it passes through. If the bullet loses $5\%$ of its velocity after passing through the first plank, then what is the minimum number of planks required to stop the bullet?

The answer is 11.

1 solution

Hari Krishna Sahoo
Feb 2, 2016

Energy lost by bullet when it strikes 1 plank=1/2(mv^2)-1/2[m{(19v/20)}^2=1/2mv^2(1-361/400)=1/2mv^2(39/400) no of planks required=total energy of bullet/energy lost by striking 1 plank =1/2(mv^2)/1/2mv^2(39/400) =400/39 =10.256 *But,no of planks can't be fraction so we round it to 11.

I have a doubt, how did you get 400/39 from 1/(361/400)

Siddhant Kishore - 5 years, 4 months ago

There is still a typo in the answer. The numerical term in first equation is 1 - 361/400 = 39/400 This term then goes into the denominator . 1/(39/400) = 400/39 = 10.256 ~ 11

Ajit Deshpande - 5 years, 4 months ago

Thanks a lot. I now understand clearly

Siddhant Kishore - 5 years, 4 months ago

sry,for my mistake I have edited my solution .hope you will understand.

Hari Krishna Sahoo - 5 years, 4 months ago

I think there is a mistake in you solution. It says that "1/2(mv^2)-1/2[m{v-(19v/20)}^2]=1/2mv^2(361/400)."

But: 1/2(mv^2)-1/2[m{v-(19v/20)}^2]

= 1/2(mv^2) - 1/2[m(20v/20 - 19v/20)^2]

= 1/2(mv^2) - 1/2[m(v/20)^2]

= 1/2(mv^2) - (1/400)(1/2mv^2)

= (400/400)1/2(mv^2) - (1/400)(1/2mv^2)

= (399/400)(1/2mv^2)

One four hundredth of the KE is lost, so it will take 400 planks to stop the bullet.

I think the mistake that you make is that you say that (1 - 19/20)^2 = (1 - 361/400) = 9.75%. And 100% / 9.75% = 10.256 planks. But you can't distribute exponents over subtraction. Correct me if I'm mistaken.

Nathan Smith - 5 years, 4 months ago

Energy Of A Bullet Fired Through Planks

The answer is 11.

1 solution

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