Physics

6.16 The mass of the ball bearing = m

Before the collision, the total K.E. of the system = K.E. of the stationary ball bearing + K.E. of the striking ball bearing = 0 + $\left(\frac{1}{2}\right){mv}^2$

After the collision, the K.E. of the total system is

(a) Case (i) = 0 + $\left(\frac{1}{2}\right)\left(2m\right){\left(\frac{v}{2}\right)}^2$ = $\left(\frac{1}{4}\right){mv}^2$

(b) Case (ii) = 0 + $\left(\frac{1}{2}\right)m{v}^2$

(c) Case (iii) = $\left(\frac{1}{2}\right)\left(3m\right){\left(\frac{v}{3}\right)/}^2$ = $\left(\frac{1}{6}\right)m{v}^2$

Case (ii) is possible since K.E. is conserved in this case.

6.15 Given, the volume of the tank, V = 30 $m^3$

Time required to fill the tank, t = 15 min = 900 s

Height of the tank above the ground, h = 40 m

The efficiency of the pump? = 30%

The density of water,  $?$ = ${10}^3$ kg/ $m^3$

Now, the mass of the water pumped, m = $?V$ = 30 $*{10}^3$ kg = 3 $*{10}^4$ kg

Power consumed = W/t = mgh/t = 3 $*{10}^4*9.8*$ 40 / 900 = 13066 W

P input = Power consumed /? = 43.6 kW

6.14 Momentum is always conserved for an elastic or inelastic collision.

The molecule's initial velocity, u = final velocity v = 200 m/s

Initial kinetic energy = (1/2)m $u^2$

Final kinetic energy = (1/2)m $v^2$ = (1/2)m $u^2$

Therefore, kinetic energy is also conserved

6.12 Electron mass, me = 9.11*10-31 kg

Proton mass, madhya pradesh = 1.67*10–27 kg

Electron's kinetic energy = 10 keV = 10 $*$ ${10}^3$ $*$ 1.60 *10–19 J = 1.60 *10–15 J

Proton's kinetic energy = 100 keV = 100 $*{10}^3$ $*$ 1.60 *10–19 J = 1.60 *10–14 J

The electron kinetic energy is given by Eke = (1/2)m $v_e^2$ where $v_e$ is the velocity of electron

$v_e$ = $?$  { (2 $*$ Eke )/m} = 5.92 $*{10}^7$ m/s

The velocity of proton $v_p$ = $?$  { (2 $*$ Pke )/m} = 4.37 $*{10}^6$ m/s

The speed ratio = $v_e$ / $v_p$ = 13.5

6.11: Force exerted on the body, F =?  $\stackrel{?}{i}+2\stackrel{?}{j}$ +3 $\stackrel{?}{k}$ N

Displacement, s = 4 km

Work done, W = F.s

= (?  $\stackrel{?}{i}+2\stackrel{?}{j}$ +3 $\stackrel{?}{k}).(4\stackrel{?}{k})$

= 0+0-3 $*4$

= 12 J

6.10 Power is given by the relation

P = Fv = mav = mv $\frac{dv}{dt}$ = constant ( say, k)

vdv = $\frac{k}{m}dt$

$\frac{v^2}{2}=\frac{k}{m}t$

v = $\sqrt{\frac{2kt}{m}}$

For displacement x of the body, we have:

v = $\frac{dx}{dt}$ = $\sqrt{\frac{2k}{m}}{t}^{1/2}$

dx = k' $t^{1/2}$ dt where k' = $\sqrt{\frac{2k}{3}}$ = constant

On integrating both sides, we get

x = $\frac{2}{3}k\text{'}{t}^{3/2}$

Therefore x $?t^{3/2}$

6.9 Let us assume

Body mass = m

Acceleration = a

According to Newton's 2nd law F = MA (constant)

We know a = dv/dt = constant. Hence dv = dt $*$ constant

On integrating, v = t + constant

The relation of power is given by P = F $*v$

We have $v=u+at$

Hence,  $P=F*(u+at)$

= $ma(u+at)$

= $mau+m{a}^{2}t$

Therefore P $?t$

6.8 (a) In elastic collision, the initial and final kinetic energy is equal. When the two balls collide, there is no conservation of kinetic energy; it gets converted into potential energy.

(b) The total linear momentum is conserved in an elastic collision.

(c) In case of inelastic condition in case (a), there will be loss of kinetic energy but in case of (b), the total linear momentum will be conserved in inelastic collision also.

(d) It is an elastic collision as the forces involves are conservative forces.

6.7  (a) False. The total momentum and energy is conserved, not the individual.

(b) False. External forces can change the energy of a body.

(c) False. Only work done by conservative force over a closed loop is zero.

(d) True. In an inelastic collision, the final velocity reduces, resulting in loss of the initial kinetic energy.