A 20 g bullet moving at 300 m/s stops after penetrating 2 cm of bone. Calculate the average force exerted by the bullet .

Explanation

Given
Mass of the bullet , m=` 20 xx 10^(-3) kg `
Initial velocity , u= 300 m/s
Final velocity ,v=0
Distance ,`s=2 cm = 2 xx 10^(-2)m`
To find : Force exerted by the bullet f=?
Solution
Using equations of motion ,
`v^(2)= u^(2)+ 2as `
`0 = (300)^(2)+ 2(a) xx 2 xx 10^(-2)`
` a=-((300)^(2))/(4 xx 10^(-2)) =-(9 xx 10^(4))/(4 xx 10^(-2))`
`a= - 2.25 xx 10^6 m//s^(2)`
` a= 2.25 xx 10^(6) m//s^(2)`
Average force ,
F= ma
`= 20 xx 10^(-3) xx 2.25 xx 10^(6)`
`F=45 xx 10^(3)= 4.5 xx 10^4 N`