Everything Science Grade 11

12.5 CHAPTER 12. FORCE,MOMENTUM AND IMPULSE

has come out the otherside.

SOLUTION

Step 1 : Draw rough sketch of the situation

2 kg

BEFORE

vi 1 = 500 m·s−^1

vf 1 = 200 m·s−^1

vf 2 =? m·s−^1

vi 2 = 0 m·s−^1 (stationary)

AFTER

50 g = 0,05 kg

Step 2 : Choose a frame of reference

We will choose to the right as positive.

Step 3 : Apply the Law of Conservation of momentum

pi = pf

m 1 vi 1 + m 2 vi 2 = m 1 vf 1 + m 2 vf 2

(0,05kg)(+500m· s−^1 ) + (2kg)(0m· s−^1 ) = (0,05kg)(+200m· s−^1 ) + (2kg)(vf 2 )

25 + 0− 10 = 2 vf 2

vf 2 = 7,5m· s−^1 in the same direction asthe bullet

Physics in Action: Impulse ESBFL

A very important application of impulse is improving safety and reducing injuries. In many cases, an
object needs to be brought to rest from a certaininitial velocity. This means there is a certain specified
change in momentum.If the time during whichthe momentum changescan be increased then the
force that must be applied will be less and so itwill cause less damage.This is the principle behind
arrestor beds for trucks,airbags, and bending your knees when you jump off a chair and land on the
ground.