Question

A 3.90 g bullet moving at 300 m/s enters and stops in an initially stationary 2.80 kg wooden block on a horizontal frictionless surface.

What's the speed of the bullet/ block combination?

What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision?

How much work was done in stopping the bullet?

If the bullet penetrated 5.00 cm into the wood, what was the average stopping force?

Answer 1

given that

Mb = 3.90 g = 0.00390 kg

Vb = 300 m/s

Mwb = 2.80 kg

part (a)

using momentum consevation

Mb*Vb = (Mb+Mwb)*V

where V is speed of bullet /block combination.

0.00390 * 300 = ( 0.00390 + 2.80)* V

V = 0.417 m/s

part (b)

initial kinetic energy = 1/2*Mb*(Vb)^2

KEi = 1/2*0.00309*(300^2)

KEi = 175.5 J

final kinetic energy = 0 . ( because when bullet enters in block & stopped, it obviously has no kinetic energy)

so the change in energy is

delta KE= 175.5 joules.

so there is no loss of kinetic energy in this collision .

part(c)

we know that

work done = change in kinetic energy

work done = delta KE

work done = 175.5 J

part(d)

from the third low of motion

v^2-u^2=2*a*s

where v = 0

u = 300 m/s

s = 5cm = 0.05 m/s

0 - 300^2 = 2* a *0.05

a = -90000/0.1 = 900000 m/s^2

F = m*a

F = 0.00390 *900000

F = 3510 N   (this is the average stopping force)