Question

A bullet of mass `10g` travelling horizontally with a velocity of `150 ms^(-1)` strikes a stationary wooden block and come to rest in `0.03 s`. Calculate the distance of penetration of the bullet into the block. Also, Calculate the magnitude of the force exerted by the wooden block on the bullet,

Answer 1

Here, `m = 10g = 10/(1000)kg = 10^(-2) kg`
initial velocity, `u= 150 m//s`, final velocity, v = 0 time taken, `t= 0.03 s`
distance of penetration, s = ?, force applied, F = ?
Acceleration, `a=(v-u)/t=(0-150)/(0.03)= -5xx10^(3) m//s^(2)`
From ` v^(2)-u^(2)= 2` as, `0-(150)^(2)= 2(-5xx10^(3))s` or `s=(150xx150)/(-10xx10^(3))=2.25m`
As, `F= ma = 10^(-2)(-5xx10^(3))= -50N`
Magnitude of force = `F= 50N`