Question

1.
A body, of mass 'm' and energy E, collides elastically with mass 2m at rest. The change in K E.of mass m. 1].-E/9. 2]-8E/9
(4) -5E/9
(3) E/4​

Answer 1

Answer:

Given:

Body of mass "m" and energy E collides with a body of mass "2m" at rest.

To find:

Changes in Kinetic energy of mass "m"

Calculation:

Since this is an MCQ, I will tell you the formula for calculation of final velocity in case of Elastic collision.

Let final velocity be "v".

For Elastic collision of 2 balls, the final velocity of 1st mass is given by the following formula:

Refer to the attached photo to understand better.

v ={(m1-m2) × u1/(m1+m2)} +

{(2m2u2)/(m1+m2)}.

So putting the values as following:

m1 = m,

m2 = 2m,

u1 = u ,

u2 = 0,

we get :

v = {(m-2m) × u/(m+2m)} +

{(2m × 0)/(m+2m)}

=> v = (-m) × u/(3m) + 0

=> v = -u/3.

Now, initial KE = 1/2 (mu²) = E

Final KE = 1/2 (mv²) = 1/2 {m(u/3)²}

= E/9.

So change in KE = final KE - initial KE

∆KE = E/9 - E = -8E/9.

Answer 2

$\huge\bf{Answer:-}$

Refer the attachment (1) & (2)