Question

ICSE
I know the solution is in the text book but I didn't understand it. I'd be glad if someone could help.

Answer 1

Ans 1 -

So, if you let the initial kinetic energy be K1 = (1/2) mv² (eq 1), and follow what the question asks i.e., double the mass of the body whose kinetic energy is to be measured, the new kinetic energy that body would possess would be =>(eq 2) K2 = (1/2)× 2m (as the mass would be double) × v².

v² = (K1 × 2)/m (from eq 1)

v² =(K2 ×2)/2m (from eq 2)

Thus,

2K1/m = 2K2/2m

=>2K1/m=K2/m

=>2K1=K2

The new kinetic energy after doubling the mass of the body would be twice the initial kinetic energy of the body.


Ans - 2

The initial kinetic energy K1 = (1/2)×mv² (eq 1).

If you reduce the initial velocity to its third i.e. (1/3)×v², then the new kinetic energy would become K2=(1/2)×m×(1/3)×v² (eq 2).

From eq 1 and eq 2 :-

v² = 2K1/m
v² = 6K2/m

Thus,

2K1/m=6K2/m (dividing by 2/m)

=> K1=3K2
=> K1/3=K2


The new K.E. would be 1/3 of the initial K. E.

Answer 2

Answer:

we see that

k.E=MV^2/2

in this equation we observe that 1/2 is constant and variable is mass (M) or velocity (v)

now we find that mass and velocity is proportional so if mass increase kinetic energy also increase

we see in question 1mass is double from initial mass

suppose intial mass of body be M and velocity be v

final mass =2M

since velocity is cinstant in this question because it show no change so

M initial/M final=KEi/KEf

k.Ef/kE i=2m/m=2

kEf=2 * initial k*E

in second velocity reduce from intial =1/3 of initial

Vi=x

Vc=2x/3

now mass cinstant

so

Vi/Vf=x^2*9/4*x^2=9/4