Physics 5 Marks Questions

$\text{a}=\frac{\text{F}}{2(\text{M}+\text{m})}$

$\text{ma}=\mu_\text{k }\text{R}_1\text{and }\text{R}_1=\text{mg}$

$\Rightarrow\mu=\frac{\text{ma}}{\text{R}_1}=\frac{\text{F}}{2(\text{M}+\text{m})\text{g}}$

Frictional force acting on the smaller block $\text{f}=\mu\text{R}$

$=\frac{\text{F}}{2(\text{M}+\text{m})\text{g}}\times\text{mg}$

$=\frac{\text{m}\times\text{F}}{2(\text{M}+\text{m})}$

Work done w = fs, s = d

$\text{W}=\frac{\text{mF}}{2(\text{M}+\text{m})}\times\text{d}$

$\text{W}=\frac{\text{mFd}}{2(\text{M}+\text{m})}$

$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}_1^2+\text{mgl}$

$\Rightarrow\frac{1}{2}\text{m}(10\text{gl})=\frac{1}{2}\text{mv}_1^2+\text{mgl}$

$\text{v}_1^2=8\text{gl}$

$\text{T}=\frac{\text{mv}^2}{\text{R}}=\text{m}\frac{8 \text{gl}}{1}$

$=8\text{mg}$

$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}^2_2+\text{mg}(2\text{l})$

$\Rightarrow\frac{1}{2}\text{m }10\text{gl}=\frac{1}{2}\text{mv}_2^2+2\text{mgl}$

$\Rightarrow\text{v}_2^2=6\text{gl}$

$$$\text{T}_\text{C}=\frac{\text{mv}_2^2}{1}-\text{mg}=5\text{mg}$

$\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{mv}_3^2+\text{mgl}(1+\cos60^\circ)$

$\Rightarrow\text{v}_3^2=7\text{gl}$

$\text{T}_\text{D}=\frac{\text{mv}_3^2}{1}-\text{mg}\cos60^\circ$

$\text{m}\frac{(7\text{gl})}{1}-0.5\text{mg}$

$=7\text{mg}-0.5\text{mg}$

$=6.5\text{mg}$

1. Radius =R

horizontal speed = v

From the free body diagram, (fig)

N = Normal force $=\text{mg}-\frac{\text{mv}^2}{\text{R}}$

2. When the particle is given maximum velocity so that the centrifugal force balances the weight, the particle does not slip on the sphere.

$\frac{\text{mv}^2}{\text{R}}=\text{mg}$

$\Rightarrow\text{v}=\sqrt{\text{gR}}$

3. If the body is given velocity v₁

$\text{v}_1=\frac{\sqrt{\text{gR}}}{2}$

$\text{v}_1^2-\frac{\text{gR}}{4}$

Let the velocity be v₂ when it leaves contact with the surface, (fig)

So, $\frac{\text{mv}^2}{\text{R}}=\text{mg}\cos\theta$

$\Rightarrow\text{v}_2^2=\text{Rg}\cos\theta\ \dots(1)$

Again, $\frac{1}{2}\text{mv}_2^2-\frac{1}{2}\text{mv}_1^2=\text{mgR}(1-\cos\theta)$

$\Rightarrow\text{v}_2^2=\text{v}_1^2+2\text{gR}(1-\cos\theta)\ \dots(2)$

From equn. (1) and equn (2),

$\text{Rg}\cos\theta=\Big(\frac{\text{Rg}}{4}\Big)+2\text{gR}(1-\cos\theta)$

$\Rightarrow\cos\theta=\Big(\frac{1}{4}\Big)+2-2\cos\theta$

$\Rightarrow3\cos\theta=\frac{9}{4}$

$\Rightarrow\cos\theta=\frac{3}{4}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{3}{4}\Big)$

m₁ = 4kg, m₂ = 1kg, V₂ = 0.3m/sec V₁ = 2 × (0.3) = 0.6 m/sec

(v₁ = 2v₂ in this system)h = 1m = height descent by 1kg block

s = 2 × 1 = 2m distance travelled by 4kg block

u = 0

Applying change in K.E. = work done (for the system)

$\Big[\Big(\frac{1}{2}\Big)\text{m}_1\text{v}_1^2+\Big(\frac{1}{2}\Big)\text{m}_2\text{v}_2^2\Big]-0=(-\mu\text{R})\text{S}+\text{m}_2\text{gh}$ $[\text{R} = 4\text{g} = 40\text{N}]$

$\Rightarrow\frac{1}{2}\times4\times(0.36)\times\frac{1}{2}\times1\times(0.09)$

$=-\mu\times40\times2+1\times40\times1$

$\Rightarrow0.72+0.045=-80\mu+10$

$\Rightarrow\mu=\frac{9.235}{80}=0.12$

$\mu=0.1,\ \text{v}=0$

1. Work done W = mgh = 0.2 × 10 × 3.2 = 6.4J
2. Work done to slide the block up the incline,

$\text{w}=(\text{mg}\sin\theta)\times\text{s}$

$=(0.2)\times10\times\frac{3.2}{10}\times10=6.4\text{J}$

3. Let, the velocity be v when falls on the ground vertically,

$=\frac{1}{2}\text{mv} -0=6.4\text{J}$

$\Rightarrow\text{v}=8\text{m}/\text{s}$

4. Let V be the velocity when reaches the ground by liding,

$=\frac{1}{2}\text{mV}^2-0=6.4\text{J}$

$\Rightarrow\text{v}=8\text{m}/\text{s}$

$\Rightarrow\text{a}=\frac{\text{T}}{4\text{m}}$

$\Rightarrow\text{A}=\frac{16}{4\text{m}}=\frac{4}{\text{m}}\text{m}/\text{s}^2$

$\text{kx}\cos\theta=\text{mg}$

$\Rightarrow\cos\theta=\frac{\text{mg}}{\text{kx}}$

$\text{V}_\text{m}=60\text{km}/\text{h}=\frac{50}{3}\text{m}/\text{sec},\text{t}_\text{m}=5\text{sec}$

$\text{a}=\frac{\Big(\frac{50}{3}\Big)-0}{5}=\frac{10}{3}$

$\text{F}=\text{ma}=95\times\frac{10}{3}=\frac{950}{3}\text{N}$

$\Rightarrow\text{a}=\frac{\text{g}}{5}\text{m}/\text{sec}^2$

$\text{S}_{4\text{th}}=\frac{\text{a}}{2}(2\text{n}-1)$

$=\frac{\Big(\frac{\text{g}}{5}\Big)}{\text{S}}(2\times4-1)$

$=\frac{7\text{g}}{10}=\frac{7\times9.8}{10}\text{m}$

$\frac{\text{mv}^2}{\text{R}}=\text{mg}\cos\theta$ [Because normal reaction]

$\text{v}^2=\text{Rg}\cos\theta\ \dots(1)$

$\Rightarrow\frac{1}{2}\text{mv}^2-0=\text{mg}(\text{R}-\text{R}\cos\theta)$

$\Rightarrow\text{v}^2=2\text{gR}(1-\cos\theta)\ \dots(2)$

$\Rightarrow\text{Rg}\cos\theta=2\text{gR}(1-\cos\theta)$

$\Rightarrow3\text{gR}\cos\theta=2\text{gR}$

$\Rightarrow\cos\theta=\frac{2}{3}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{2}{3}\Big)$

$\text{v}=\sqrt{3\text{gl}}$

$\frac{1}{2}\text{m}\text{v}^2-\frac{1}{2}\text{m}\text{v}_2^2=-\text{mgh}$

$\text{v}^2=\text{u}^2-2\text{g}(\text{l}+\cos\theta)$

$\text{v}^2=3\text{gl}-2\text{gl}(\text{l}+\cos\theta)\ \dots(1)$

Again,

$\frac{\text{mv}^2}{\text{l}}=\text{mg}\cos\theta$

$\text{v}^2=\text{lg}\cos\theta\ \dots(2)$

From equation (1) and (2), we get

$3\text{gl}-2\text{gl}-2\text{gl}\cos\theta=\text{gl}\cos\theta$

$3\cos\theta=1$

$\cos\theta=\frac{1}{3}$

$\theta=\cos^{-1}\Big(\frac{1}{3}\Big)$

So, angle rotated before the string becomes slack,

$\Rightarrow180^\circ-\cos^{-1}\Big(\frac{1}{3}\Big)=\cos^{-1}\Big(\frac{-1}{3}\Big)$

1. When the bob has an initial height less than the peg and then released from rest (figure), let body travels from A to B.

Since, Total energy at A = Total energy at B

$\therefore$ (K.E)_A = (PE)_A = (KE)_B + (PE)_B

⇒ (PE)_A = (PE)_B [because, (KE)_A = (KE)_B = 0]

So, the maximum height reached by the bob is equal to initial height.

2. When the pendulum is released with $\theta=90^\circ$ and $\text{x}=\frac{\text{L}}{2},$ (figure) the path of the particle is shown in the figure.

At point C, the string will become slack and so the particle will start making projectile motion.

$\Rightarrow\Big(\frac{1}{2}\Big)\text{mv}_\text{c}^2-0=\text{mg}\Big(\frac{\text{L}}{2}\Big)(1-\cos\alpha)$

Because, distance between A nd C in the vertical direction is $\frac{\text{L}}{2}(1-\cos\alpha)$

$\Rightarrow\text{v}_\text{c}^2=\text{gl}(1-\cos\alpha)\ \dots(1)$

Again, form the freebody diagram,

$\frac{\text{mv}_\text{c}^2}{\frac{\text{L}}{2}}=\text{mg}\cos\alpha$ [because T_c = 0]

So, $\Rightarrow\text{v}_\text{c}^2=\frac{\text{gl}}{2}\cos\alpha\ \dots(2)$

From Eqn.(1) and equn (2),

$\text{gl}(1-\cos\alpha)=\frac{9\text{L}}{2}\cos\alpha$

$\Rightarrow(1-\cos\alpha)=\frac{1}{2}\cos\alpha$

$\Rightarrow\frac{3}{2}\cos\alpha=1$

$\Rightarrow\cos\alpha=\frac{2}{3}\ \dots(3)$

To find highest position C, before the string becomes slack.

$\text{BF}=\frac{\text{L}}{2}+\frac{\text{L}}{2}\cos\theta$

$\text{BF}=\frac{\text{L}}{2}+\frac{\text{L}}{2}\times\frac{2}{3}=\text{L}\Big(\frac{1}{2}+\frac{1}{3}\Big)$

So, $\text{BF}=\frac{5\text{L}}{6}$

3. If the particle has to complete a vertical circle, at the point C.

$\frac{\text{mv}_\text{c}^2}{(\text{L}-\text{x})}=\text{mg}$

$\Rightarrow\text{v}_\text{c}^2=\text{g}(\text{L}-\text{x})\ \dots(4)$

Again, applying energy principle between A and C,

$\frac{1}{2}\text{mv}_\text{c}^2-0=\text{mg}(\text{OC})$

$\Rightarrow\frac{1}{2}\text{v}_\text{c}^2-0=\text{mg}[\text{L}-2(\text{L}-\text{x})]$

$\Rightarrow\text{mg}(2\text{x}-\text{L})$

$\Rightarrow\text{v}_\text{c}^2=2\text{g}(2\text{x}-\text{L})\ \dots(5)$

From equn. (4) and equn (5),

$\text{g}(\text{L}-\text{x})=2\text{g}(2\text{x}-\text{L})$

$\Rightarrow\text{L}-\text{x}=4\text{x}-2\text{L}$

$\Rightarrow5\text{x}=3\text{L}$

$\therefore\ \frac{​​​​\text{x}}{\text{L}}=\frac{3}{5}=0.6$

So, the rates $\Big(\frac{\text{x}}{\text{L}}\Big)$ should be 0.6

Let us consider a small element which makes angle $'\text{d}\theta'$ at the centre.

$\therefore\ \text{dm}=\Big(\frac{\text{m}}{\ell}\Big)\text{Rd}\theta$

1. Gravitational potential energy of ‘dm’ with respect to centre of the sphere,

$=(\text{dm})\text{gR}\cos\theta$

$=\Big(\frac{\text{mg}}{\ell}\Big)\text{R}\cos\theta\text{d}\theta$

So, Total G.P.E. $=\int\limits_0^\frac{\ell}{\text{r}}\frac{\text{mgR}^2}{\ell}\cos\theta\text{d}\theta$ $\Big[\alpha=\Big(\frac{\ell}{\text{R}}\Big)\Big]$(angle subtended by the chain at the centre)

$=\frac{\text{mR}^2\text{g}}{\ell}[\sin\theta]\Big(\frac{\ell}{\text{R}}\Big)$

$=\frac{\text{mR}\text{g}}{\ell}\sin\theta\Big(\frac{\ell}{\text{R}}\Big)$

2. When the chain is released from rest and slides down through an angle $\theta,$ the K.E. of the chain is given,

K.E. = Change in potential energy.

$=\frac{\text{mR}^2\text{g}}{\ell}\sin\Big(\frac{\ell}{\text{R}}\Big)-\text{m}\int\frac{\text{gR}^2}{\ell}\cos\theta\text{d}\theta​​$

$=\frac{\text{mR}^2\text{g}}{\ell}\Big[\sin\Big(\frac{\ell}{\text{R}}\Big)+\sin\theta-\sin\Big\{\theta+\Big(\frac{\ell}{\text{R}}\Big)\Big\}\Big]$

3. Since, K.E. $=\frac{1}{2}\text{mv}^2=\frac{\text{mR}^2\text{g}}{\ell}\Big[\sin\Big(\frac{\ell}{\text{R}}\Big)+\sin\theta-\sin\theta\Big\{\theta+\Big(\frac{\ell}{\text{R}}\Big)\Big\}\Big]$

Taking derivative of both sides with respect to ‘t’

$\Big(\frac{1}{2}\Big)\times2\text{v}\times\frac{\text{dv}}{\text{dt}}$

$=\frac{\text{R}^2\text{g}}{\ell}\Big[\cos\theta-\cos\Big(\theta+\Big(\frac{\ell}{\text{R}}\Big)\Big)\Big]$

$\therefore\ \Big(\text{R}\frac{\text{d}\theta}{\text{dt}}\Big)\frac{\text{dv}}{\text{dt}}$

$=\frac{\text{R}^2\text{g}}{\ell}\times\frac{\text{d}\theta}{\text{dt}}\Big[\cos\theta-\cos\Big(\theta+\Big(\frac{\ell}{\text{R}}\Big)\Big)\Big]$

When the chain starts sliding down, $\theta=0.$

So, $\frac{\text{dv}}{\text{dt}}=\frac{\text{Rg}}{\ell}\Big[1-\cos\Big(\frac{\ell}{\text{R}}\Big)\Big]$

1. $\text{m}=2\text{kg},\ \theta=37^\circ,\ \text{F}=20\text{N}$

From the free body diagram

$\text{F}=(2\text{g}\sin\theta)+\text{ma}$

$\Rightarrow\text{a}=(20-20\sin\theta)/\text{s}=4\text{m}/\text{sec}^2$

$\text{S}=\text{ut}+\frac{1}{2}\text{at}^2$ $(\text{u}=0,\text{t}=1\text{s},\text{a}=1.66)$

$=2\text{m}$

So, work, done W = Fs = 20 × 2 = 40J

2. If W = 40J

$\text{S}=\frac{\text{W}}{\text{F}}=\frac{40}{20}$

h = 2 sin 37° = 1.2m

So, work done W = -mgh = -20 × 1.2 = -24J

3. v = u + at = 4 × 10 = 40 m/sec

So, K.E. $=\frac{1}{2}\text{mv}^2=\frac{1}{2}\times2\times16=16\text{J}$

1. Net force on the particle between A & B, F $=\text{mg}\sin\theta$

work done to reach B, W = FS $=\text{mg}\sin\theta\ell$

Again, work done to reach B to C = mgh $=\text{mgR}(1-\cos\theta)$

So, Total workdone $=\text{mg}[\ell\sin\theta+\text{R}(1-\cos\theta)]$

Now, change in K.E. = work done

$\Rightarrow\frac{1}{2}\text{mv}_0^2=\text{mg}[\ell\sin\theta+\text{R}(1-\cos\theta)]$

$\Rightarrow\text{v}_0=\sqrt{2\text{g}[\ell\sin\theta+\text{R}(1-\cos\theta)]}$

2. When the block is projected at a speed 2v₀.

Let the velocity at C will be V_c.

Applying energy principle,

$\Rightarrow\frac{1}{2}\text{mv}_\text{c}^2-\frac{1}{2}\text{m}(2\text{v}_0)^2=-\text{mg}[\ell\sin\theta+\text{R}(1-\cos\theta)]$

$\Rightarrow\text{v}_\text{c}^2=4\text{v}_0-2\text{g}[\ell\sin\theta+\text{R}(1-\cos\theta)]$

$\Rightarrow4.2\text{g}[\ell\sin\theta+\text{R}(1-\cos\theta)]-2\text{g}[\ell\sin\theta+\text{R}(1-\cos\theta)]$

$\Rightarrow6\text{g}[\ell\sin\theta+\text{R}(1-\cos\theta)]$

So, force acting on the body,

$\Rightarrow\text{N}=\frac{\text{mv}_\text{c}^2}{\text{R}}=6\text{mg}\Big[\Big(\frac{\ell}{\text{R}}\Big)\sin\theta+1-\cos\theta\Big]$

3. Let the block loose contact after making an angle $\theta$

$\frac{\text{mv}^2}{\text{R}}=\text{mg}\cos\theta$

$\Rightarrow\text{v}^2={\text{R}}\text{g}\cos\theta\ \dots(1)$

Again, $\frac{1}{2}\text{mv}^2=\text{mg}(\text{R}-\text{R}\cos\theta)$

$\Rightarrow\text{v}^2=2\text{gR}(1-\cos\theta)\ \dots(2)$

From (1) and (2),

$\cos\theta=\frac{2}{3}$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{2}{3}\Big)$

When the particle is released from rest (fig), the centrifugal force is zero. N force is zero $=\text{mg}\cos\theta$

$=\text{mg}\cos30^\circ=\frac{\sqrt3\text{mg}}{2}$

When the particle leaves contact with the surface (fig), N = 0.

So, $\frac{\text{mv}^2}{\text{R}}\text{mg}\cos\theta$

$\Rightarrow\text{v}^2=\text{Rg}\cos\theta\ \dots(1)$

Again, $\frac{1}{2}\text{mv}^2=\text{mgR}(\cos30^\circ-\cos\theta)$

$\Rightarrow\text{v}^2=2\text{Rg}\Big(\frac{\sqrt3}{2}-\cos\theta\Big)\ \dots(2)$

From equn. (1) and equn. (2),

$\text{Rg}\cos\theta=\sqrt3\text{Rg}-2\text{Rg}\cos\theta$

$\Rightarrow3\cos\theta=\sqrt3$

$\Rightarrow\cos\theta=\frac{1}{\sqrt3}$

$\Rightarrow\theta=\cos^{-1}\frac{1}{\sqrt3}$

So, the distance travelled by the particle before leaving contact,

$\ell=\text{R}\Big(\theta-\frac{\pi}{6}\Big)$ $\Big[\text{because}30^\circ=\frac{\pi}{6}\Big]$

putting the value of $\theta,$ we get $\ell=0.43\text{R}$

Let the sphere move towards left with an acceleration ‘a'

Let m = mass of the particle

The particle ‘m’ will also experience the inertia due to acceleration ‘a’ as it is on the sphere. It will also experience the tangential inertia force $\Big(\text{m}\Big(\frac{\text{dv}}{\text{dt}}\Big)\Big)$ and centrifugal force $\Big(\frac{\text{mv}^2}{\text{R}}\Big).$

$\text{m}\frac{\text{dv}}{\text{dt}}=\text{ma}\cos\theta+\text{mg}\sin\theta$

$\Rightarrow\text{mv}\frac{\text{dv}}{\text{dt}}=\text{ma}\cos\theta\Big(\text{R}\frac{\text{d}\theta}{\text{dt}}\Big)+\text{mg}\sin\theta\Big(\text{R}\frac{\text{d}\theta}{\text{dt}}\Big)$

Because, $\text{v}=\text{R}\frac{\text{d}\theta}{\text{dt}}$

$\Rightarrow\text{vdv}=\text{aR}\cos\theta\text{ d}\theta+\text{gR}\sin\theta\text{ d}\theta$

Integrating both sides we get,

$\frac{\text{v}^2}{2}=\text{aR}\sin\theta-\text{gR}\sin\theta+\text{C}$

Given that, at $\theta=0,\text{v}=0$

So, $\text{C}=\text{gR}$

So, $\frac{\text{v}^2}{2}=\text{aR}\sin\theta-\text{gR}\sin\theta+\text{gR}$

$\therefore\ \text{v}^2=2\text{R}(\text{a}\sin\theta+\text{g}-\text{g}\cos\theta)$

$\Rightarrow\text{v}=[2\text{R}(\text{a}\sin\theta+\text{g}-\text{g}\cos\theta)]^\frac{1}{2}$

Let, x length of chain is on the table at a particular instant.

So, work done by frictional force on a small element ‘dx’.

$\text{dW}_\text{f}=\mu\text{Rx}=\mu\Big(\frac{\text{M}}{\text{L}}\text{dx}\Big)\text{gx}$ $[$ where $\text{dx}=\frac{\text{M}}{\text{L}}\text{dx}]$

Total work don by friction,

$\text{W}_\text{f}=\int\limits_\frac{2\text{L}}{3}^0\mu\frac{\text{M}}{\text{L}}\text{gx dx}$

$\therefore\ \text{W}_\text{f}=\mu\frac{\text{M}}{\text{L}}\text{g}\Big[\frac{\text{x}^2}{2}\Big]_\frac{2\text{L}}{3}^0$

$=\mu\frac{\text{M}}{\text{L}}\Big[\frac{4\text{L}^2}{18}\Big]$

$=2\mu\text{Mg}\frac{\text{L}}{9}$