Answer
(a) It takes 0.743 ms for the bullet to stop.
(b) The tree exerts a force of 848 N on the bullet.
Work Step by Step
We can find the rate of deceleration of the bullet.
$a = \frac{v^2-v_0^2}{2x} = \frac{0 - (350~m/s)^2}{(2)(0.130~m)}$
$a = -4.71\times 10^5~m/s^2$
The magnitude of acceleration is $4.71\times 10^5~m/s^2$
(a) $t = \frac{v-v_0}{a} = \frac{0-350~m/s}{-4.71\times 10^5~m/s^2}$
$t = 7.43\times 10^{-4}~s = 0.743~ms$
It takes 0.743 ms for the bullet to stop.
(b) $F = ma$
$F = (1.80\times 10^{-3}~kg)(4.71\times 10^5~m/s^2)$
$F = 848~N$
The tree exerts a force of 848 N on the bullet.
