# 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%

### 5 Answers

- knrLv 68 years agoBest Answer
the percentage of increase in momentum = 10[ ( square root of (300+100 ) -10 ] = 100 %

so the answer is (1)

- Steve4PhysicsLv 78 years ago
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).

- 8 years ago
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

- IvanLv 68 years ago
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

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- cavittLv 43 years ago
ok.E preliminary = 0.5 * m * v1* v1 very final ok.E = 0.5 * m * v2* v2 ok.E preliminary/ok.E very final = a million/2 so, v1^2/v2^2 = 0.5 v2/v1 = a million/0.seventy one = a million.4 Momentum preliminary = m*v1 Momentum very final = m*v2 = m * a million.4 * v1 So momentum is better by way of a million.4 cases