Anonymous
Anonymous asked in Science & MathematicsPhysics · 8 years ago

# Kinetic energy and momentum relation?

If kinetic energy of a body is increased by 300% then percentage change in momentum will be

1) 100%

2) 150%

3) 265%

4) 73.2%

Relevance

the percentage of increase in momentum = 10[ ( square root of (300+100 ) -10 ] = 100 %

• Pretend the mass is 2kg and the object is travelling at 1m/s.

KE = ½ x 2 x 1² = 1J

Momentum = 2 x 1 = 2kgm/s

If the kinetic energy is increased by 300%, that means it is increased by 3J, and the new kinetic energy is 1+3 = 4J.

The new speed is given by ½ x 2 x v² = 4 giving v= 2m/s.

New momentum = 2x2 = 4kgm/s

The momentum has doubled (from 2 to 4), which is an increase of 100%. So the correct answer is 1).

• let the initial KE be x.

as the mass of the body is fixed, let it be m.

now new KE is 3x. the relation between KE and momentum is p=root(2mKE)

chang in momentum is {root(6mKE) -root(2mKE)}. so the percentage will be root(3)-1 X 100 = 73 % proximate . . hit maximum points cuz i did lot of hard work for it! have fun

• KE E=p^2/2m p is momentum m=mass

taking log and differentiating

dE/E=2Dp/p

or DE/E %=2Dp/p%

300=2x(%Dp)

or %Dp=300/2=150