Question

evaluate
$( \frac{1}{2} ) {}^{2} \times ( \frac{ - 3}{2} ) {}^{3} \times ( \frac{ - 2}{3} ) {}^{2} \times ( \frac{ - 3}{5} ) {}^{2}$

Answer 1

(1/4)×(-27/8)×(4/9)×(9/25)= {-27/(25×8)} = -27/200

Answer 2

Consider two frames of references S and S'. Further, S' is moving with constant velocity v along X-direction. To explain the variation of mass with velocity, consider the collision of two exactly similar balls A and B, each of mass m, moving in opposite direction along X-axis with equal speed u' in frame S'. After collision they coalesce into one body.

Applying the, law of conservation of momentum on the collision of the balls in frame s', we have

imagesimages

After collision, the coalesced mass must be at rest in frame S'. Hence, it moves with velocity v in frame S. Let u1,u2 be the velocities and m1, m2 be the masses of balls A and B, respectively, in frame S. Using the law of addition of velocities, the above velocities can be written as

......(1)

.....(2)

Applying the law of conservation of momentum on the collision of the balls in frame s, we have

m1u1 + m2u2 = (m1 + m2)v....(3)

Substituting u1 and u2 values from Equations (1) and (2), we have

.........(4)

The above equation makes a relationship between the masses of balls in frame S and their velocities in frame S'. Now, to obtain relation between masses of balls and their velocities in frame S, we proceed as follows. Squaring Equation (1)

images

and using the above equation, the value images is

Therefore,

...(5)

Similarly, using equation (2) we get

...(6)

Dividing Equation (6) by Equation (5) and taking square root throughout, we have

...(7)

Comparing Equations (4) and (7), we have

....(8)

Suppose, m2 is at rest in frame s, then u2 = 0 and m2 = m0 (say) where m0 is the rest mass of the ball B, thenEquation (19.48) becomes

....(9)

As both the balls are similar, hence the rest masses of both balls are the same, so we can write the rest mass of m2 is equal to rest mass of m1, that is equal to mo. Then, Equation (9) becomes

....(10)

Here, m1 is the mass of ball A when it is moving with velocity u1 in frame s. After collision, the coalescent mass containing mass of ball A moves with velocity v in frame s.

In general, if we take the mass of ball A as m, when it is moving with velocity v in frame s, then

.......(11)

where m0 is the rest mass of the body and m is the effective mass.

Equation (11) is the relativistic formula for the variation of mass with velocity. Here, we see some special cases:

Case (i): When the velocity of the body, v is very small compared to velocity of height, c, then v2/c2 is negligible compared to one. Therefore,

m = m0*

Case (ii): If the velocity of the body v is comparable to the velocity of light c, then images is less than one, so, m> mo.

The mass of a moving body appears greater than its rest mass.

Case (iii): Suppose the velocity of a body is equal to velocity of light, c, then, it possess infinite mass.

The effective mass of particles has been experimentally verified by using particle accelerators in case of electrons and protons by increasing their velocities very close to velocity of light.