Question

A 7.80 g bullet has a speed of 620 m/s when it hits a target, causing the target to move 6.30 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force (in N) that stops the bullet. (Enter the magnitude.) N (b) Assuming the force is constant, determine how much time elapses (in s) between the moment the bullet strikes the target and the moment it stops moving. s

Answer 1

Answer:

$23796.19\ \text{N}$

$0.0002032\ \text{s}$

Explanation:

F = Force

s = Displacement = 6.3 cm

m = Mass of bullet = 7.8 g

v = Velocity of bullet = 620 m/s

t = Time taken

Work done is given by

$W=Fs$

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2$

Using work energy considerations we get

$Fs=\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{1}{2s}mv^2\\\Rightarrow F=\dfrac{1}{2\times 0.063}\times 7.8\times 10^{-3}\times 620^2\\\Rightarrow F=23796.19\ \text{N}$

The average force that stops the bullet is $23796.19\ \text{N}$.

Force is given by

$F=m\dfrac{v-u}{t}\\\Rightarrow t=m\dfrac{v-u}{F}\\\Rightarrow t=7.8\times 10^{-3}\times \dfrac{620}{23796.19}\\\Rightarrow t=0.0002032\ \text{s}$

The time taken to stop the bullet is $0.0002032\ \text{s}$.