Question

. A .22 rifle bullet, traveling at 350 m/s, strikes a block of soft wood, which it penetrates to a depth of 0.130 m. The block of wood is clamped in place and doesn’t move. The mass of the bullet is 1.80 g. Assume a constant retarding force. How much time is required for the bullet to stop? What force, in Newtons, does the wood exert on the bullet? 1kg = 1000g

Answer 1

@Michele_Laino
will help u with this

m very sleepy atm

Answer 2

@rvc alright thanks tho

Answer 3

here the work \(W\) by such retarding force \(F\) has to be equal to the kinetic energy \(T\) of the bullet, so we can write:
\[\huge T = \frac{1}{2}m{v^2} = Fd\]
where \(m\) is the mass of the bullet, and \(v\) is its speed

Answer 4

thanks @Michele_Laino

Answer 5

of course, \(d\) is the depth

Answer 6

so it would be like
t=1/2(1.80)*(350)^2= F*0.130

right?

Answer 7

more precisely, we have to measure the bullet mass using Kg, so we can write:
\[\Large \frac{1}{2} \cdot 0.0018 \cdot {350^2} = F \cdot 0.130\]

Answer 8

I got this:
\[\large T = 110.25\;Joules\]

Answer 9

and:
\[\Large F = \frac{{0.0018 \cdot {{350}^2}}}{{2 \cdot 0.13}} = ...?\]

Answer 10

14.3?

Answer 11

I got this:
\[\Large F = \frac{{0.0018 \cdot {{350}^2}}}{{2 \cdot 0.13}} = 848\;newtons\]

Answer 12

ohh ok
got it thanks

Answer 13

finally the requested time, in order to stop the bullet, is fiven by the subsequent computation:
\[\Large \tau = \frac{{mv}}{F} = \frac{{0.0018 \cdot 350}}{{848}} = ...?\]

Answer 14

oops.. is given*

Answer 15

0.000742

Answer 16

that's right!

Answer 17

it is \(\tau=0.000743 \) seconds

Answer 18

okay, thank you so much for your help