Center of Mass and Collision
A uniform thin metal plate of mass $10 \mathrm{~kg}$ with dimensions is shown. The ratio of $\mathrm{x}$ and y coordinates of center of mass of plate in $\frac{n}{9}$. The value of $n$ is ________.
Explanation:
$\begin{array}{ll} x_{\mathrm{cm}}=1.5 & \\ M_{+}=6 \sigma & y_{+}=1 \\ M_{-}=-\sigma & y_{-}=1.5 \end{array}$
$\begin{aligned} & y_{\mathrm{cm}}=\frac{6 \sigma \times 1+(-\sigma) \times 1.5}{6 \sigma-\sigma} \\ & =\frac{6-1.5}{5}=\frac{4.5}{5}=0.9 \\ & \frac{x}{y}=\frac{1.5}{0.9}=\frac{15}{9} \end{aligned}$
In a system two particles of masses $m_1=3 \mathrm{~kg}$ and $m_2=2 \mathrm{~kg}$ are placed at certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance $2 \mathrm{~cm}$. In order to keep the center of mass of the system at the original position, the particle of mass $m_2$ should move towards the center of mass by the distance _________ $\mathrm{cm}$.
Explanation:
To solve this problem, we can make use of the concept of center of mass. The center of mass (CM) of a system remains unchanged if the internal forces act within the system without any external force. When one mass moves toward the CM, to keep the CM at the same position, the other mass must move in a way that the product of each mass with its displacement relative to the CM remains constant.
The formula to ensure that the center of mass remains unchanged can be derived from the principle of conservation of momentum or simply by understanding that the weighted average position (considering masses as weights) does not change.
Let's denote: - $x_1$ as the distance moved by $m_1$ towards the CM, - $x_2$ as the distance $m_2$ needs to move towards the CM, - The total mass of the system as $M = m_1 + m_2$.
Since $m_1$ moves towards the CM by 2 cm, we apply the principle that the weighted sum of displacements (taking mass into account) remains 0 to maintain the center of mass at its original position:
$m_1 \cdot x_1 + m_2 \cdot x_2 = 0$
Given that $m_1 = 3 \, \text{kg}$, $m_2 = 2 \, \text{kg}$, and $x_1 = 2 \, \text{cm}$, we substitute these values into the equation:
$3 \cdot 2 + 2 \cdot x_2 = 0$
Solving for $x_2$ gives:
$6 + 2x_2 = 0$
$2x_2 = -6$
$x_2 = -3 \, \text{cm}$
This means the mass $m_2$ should move $3 \, \mathrm{cm}$ towards the center of mass to keep the center of mass of the system at the original position. The negative sign indicates the direction is towards the center of mass, similar to $m_1$'s movement direction in relation to keeping the CM stationary.
Explanation:
The center of mass (COM) of a system of particles is calculated as:
$X_{COM} = \frac{\sum_{i} m_i x_i}{\sum_{i} m_i}$
$Y_{COM} = \frac{\sum_{i} m_i y_i}{\sum_{i} m_i} $$
where $m_i$ is the mass of the $i$-th particle, and $(x_i, y_i)$ are its coordinates.
2. Coordinate Setup:Let's place the origin at the right angle of the triangle and align the sides along the x and y axes:
- Sphere 1: $(4, 0)$
- Sphere 2: $(0, 4)$
- Sphere 3: $(0, 0)$
Since all spheres have mass $2M$, we can simplify the COM calculations:
$X_{COM} = \frac{2M \cdot 4 + 2M \cdot 0 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$
$Y_{COM} = \frac{2M \cdot 0 + 2M \cdot 4 + 2M \cdot 0} {2M + 2M + 2M} = \frac{4}{3}$
4. Magnitude of Position Vector:The position vector of the COM is $\left(\frac{4}{3}, \frac{4}{3}\right)$. Its magnitude is:
$|\vec{r}_{COM}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$
5. Finding x:We are given that the magnitude of the position vector is of the form $\frac{4\sqrt{2}}{x}$. Comparing this to our result, we find that $x = 3$.
Answer:The value of $x$ is 3.
A solid circular disc of mass $50 \mathrm{~kg}$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 \mathrm{~m} / \mathrm{s}$. The absolute value of work done on the disc to stop it is ________ J.
Explanation:
Using work energy theorem
$\begin{aligned} & \mathrm{W}=\Delta \mathrm{KE}=0-\left(\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\right) \\ & \mathrm{W}=0-\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{K}^2}{\mathrm{R}^2}\right) \\ & =-\frac{1}{2} \times 50 \times 0.4^2\left(1+\frac{1}{2}\right)=-6 \mathrm{~J} \end{aligned}$
Absolute work $=+6 \mathrm{~J}$
$W=-6 J \quad|W|=6 J$
A body starts falling freely from height $H$ hits an inclined plane in its path at height $h$. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $\frac{H}{h}$ for which the body will take the maximum time to reach the ground is __________.
Explanation:
Total time of flight $=\mathrm{T}$
$T=\sqrt{\frac{2 h}{g}}+\sqrt{\frac{2(H-h)}{g}}$
For max. time $=\frac{\mathrm{dT}}{\mathrm{dh}}=0$
$\begin{aligned} & \sqrt{\frac{2}{\mathrm{~g}}}\left(\frac{-1}{2 \sqrt{\mathrm{H}-\mathrm{h}}}+\frac{1}{2 \sqrt{\mathrm{h}}}\right)=0 \\ & \sqrt{\mathrm{H}-\mathrm{h}}=\sqrt{\mathrm{h}} \\ & \mathrm{h}=\frac{\mathrm{H}}{2} \Rightarrow \frac{\mathrm{H}}{\mathrm{h}}=2 \end{aligned}$
The momentum of a body is increased by $50 \%$. The percentage increase in the kinetic energy of the body is ___________ $\%$.
Explanation:
The momentum (p) and kinetic energy (K) of a body are related by the equations:
$p = mv$,
$K = \frac{1}{2}mv^2$,
where m is the mass and v is the velocity of the body.
We can express v in terms of p and m:
$v = \frac{p}{m}$,
and substitute this into the equation for K to get:
$K = \frac{p^2}{2m}$.
So, the kinetic energy is proportional to the square of the momentum.
If the momentum is increased by 50%, the new momentum is 1.5p, and the new kinetic energy is:
$K' = \frac{(1.5p)^2}{2m} = \frac{2.25p^2}{2m} = 2.25K$.
The percentage increase in the kinetic energy is then:
$\frac{K'-K}{K} \times 100 = \frac{2.25K - K}{K} \times 100 = 1.25 \times 100 = 125\%$.
So, the percentage increase in the kinetic energy of the body is 125%.
Explanation:
$h^{\prime}=(0.5)^{2} \times 20 m=5 m$
A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2 m/s. The initial speed of the smaller body before collision is ___________ ms$^{-1}$.
Explanation:
Before collision
After collision
Momentum conservation
u + 0 = 3 v – 2
3v - u = 2 …(1)
$ \begin{aligned} & \frac{v+2}{u}=1 \Rightarrow v+2=u \\\\ & u-v=2 \quad ....(2) \end{aligned} $
Adding (1) and (2)
$ \begin{aligned} & 2 v=4 \\\\ & v=2 \mathrm{~m} / \mathrm{s} \\\\ & \therefore u=4 \mathrm{~m} / \mathrm{s} \end{aligned} $
The distance of centre of mass from end A of a one dimensional rod (AB) having mass density $\rho=\rho_{0}\left(1-\frac{x^{2}}{L^{2}}\right) \mathrm{kg} / \mathrm{m}$ and length L (in meter) is $\frac{3 L}{\alpha} \mathrm{m}$. The value of $\alpha$ is ___________. (where x is the distance from end A)
Explanation:
$\rho = {\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)$ kg/m
${x_{cm}} = {{A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)x\,dx} } \over {A\int\limits_0^L {{\rho _0}\left( {1 - {{{x^2}} \over {{L^2}}}} \right)\,dx} }}$
${x_{cm}} = {{{{{L^2}} \over 2} - {{{L^2}} \over 4}} \over {L - {L \over 3}}} = {{{{{L^2}} \over 4}} \over {{{2L} \over 3}}} = {{3L} \over 8}$
$ \Rightarrow \alpha = 8$
Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt x$ m. The value of x is ____________.
Explanation:
${d_{cm}} = 3\sin 45^\circ = {3 \over {\sqrt 2 }}$
${d_{cm}} = {2 \over 3} \times {3 \over {\sqrt 2 }} = \sqrt 2 = \sqrt x $
$x = 2$
A man of 60 kg is running on the road and suddenly jumps into a stationary trolly car of mass 120 kg. Then, the trolly car starts moving with velocity 2 ms$-$1. The velocity of the running man was ___________ ms$-$1, when he jumps into the car.
Explanation:
Initially man was moving with velocity v1 and trolley was at rest, finally both were moving with velocity 2 ms$-$1 after man jumps on the trolley.
So,
$ \Rightarrow \quad m_{1} v_{1}+0=\left(m_{1}+m_{2}\right) v_{2} $
$ \text { Here, } m_{1}=\text { mass of man }=60 \mathrm{~kg} $
$ m_{2}=\text { mass of trolley }=120 \mathrm{~kg} $
$ v_{1}=\text { speed of } \text { man } $
$ v_{2}=\text { speed of man and trolley }=2 \mathrm{~m} / \mathrm{s} $
$ \Rightarrow 60 \times v_{1}=(60+120) \times 2 $
$ \Rightarrow v_{1}=\frac{(60+120) \times 2}{60}=6 \mathrm{~m} / \mathrm{s} $
A batsman hits back a ball of mass 0.4 kg straight in the direction of the bowler without changing its initial speed of 15 ms$-$1. The impulse imparted to the ball is ___________ Ns.
Explanation:
$l = m\Delta v$
$ = 0.4 \times 2 \times 15 = 12$ Ns
Explanation:
V' = 5 m/s
m1 V = m2 $\times$ 5 $-$ m1 $\times$ 100
${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$
V = 400 m/s
Explanation:
2 $\times$ 4 = 2 $\times$ 1 + m2 $\times$ v2
m2v2 = 6 ..... (i)
by coefficient of restitution
$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$ m/s
by (i)
m2 $\times$ 5 = 6
m2 = 1.2 kg
${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$
${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$
x = 25
Explanation:
$\Rightarrow$ | x | = 2
Explanation:
Before collision,
As the collision is perfectly elastic, therefore momentum is conserved, i.e.
pinitially = pfinally
$\Rightarrow$ mu = Mv ..... (i)
Angular momentum will also be conserved about point O.
$ \Rightarrow mv\,.\,{L \over 2} = {{M{L^2}} \over {12}}\omega $
$ \Rightarrow \omega = {{6mv} \over {ML}}$ .... (ii)
$\because$ Coefficient of restitution,
$e = {{{\mathop{\rm Relative}\nolimits} \,velocity\,after\,collision} \over {{\mathop{\rm Relative}\nolimits} \,velocity\,before\,collision}}$
$ \Rightarrow 1 = {{v + {{\omega L} \over 2}} \over u}$
$ \Rightarrow v + {{\omega L} \over 2} = u$ .... (iii)
From Eqs. (ii) and (iii), we get
$v + {{3mu} \over M} = u$
$ \Rightarrow {{mu} \over M} + {{3mu} \over M} = u$ [using Eq. (i)]
$ \Rightarrow {{4mu} \over M} = u \Rightarrow {m \over M} = {1 \over 4}$ .... (iv)
According to question,
ratio of masses $\left( {{m \over M}} \right) = {1 \over x}$.
Comparing it with Eq. (iv), we get x = 4.
The initial velocity of the particle is $5\sqrt 2 $ ms-1 and the air resistance is assumed to be negligible. The magnitude of the change in momentum between the points A and B is x $\times$ 10-2 kgms-1. The value of x, to the nearest integer, is __________.
Explanation:
$|\overrightarrow u | = |\overrightarrow v |$ .... (1)
$\overrightarrow u = u\cos 45\widehat i + u\sin 45\widehat j$ ..... (2)
$\overrightarrow v = v\cos 45\widehat i - v\sin 45\widehat j$ ..... (3)
$|\Delta \overrightarrow P | = |m(\overrightarrow v - \overrightarrow u )|$ .... (4)
$\Delta P = 2mu\sin 45^\circ $
$ = 2 \times 5 \times {10^{ - 3}} \times 5\sqrt 2 \times {1 \over {\sqrt 2 }}$
= 50 $\times$ 10$-$3
= 5 $\times$ 10$-$2
Explanation:
initial total momentum of system = $10 \times 10\sqrt {3\widehat i} $
Final total momentum of system
$ = 10 \times 10\widehat j + 10 \times x(\cos 30^\circ \widehat i - \sin 30^\circ \widehat j)$
Now by conservation of momentum
$10 \times 10\sqrt {3\widehat i} = 10 \times 10\widehat j + 10 \times x\left( {{{\sqrt 3 } \over 2}\widehat i - {1 \over 2}\widehat j} \right)$
$ \Rightarrow $ x = 20
[a is an area as shown in the figure]
Explanation:
$dm = \sigma {1 \over 2}R \times Rd\theta $
$dm = {{\sigma {R^2}d\theta } \over 2}$
${x_{cm}} = {{\int {x\,dm} } \over {\int {dm} }} = {{\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d\theta \left( {{{2R} \over 3}\cos \theta } \right)} } \over {\int\limits_0^{\pi /2} {{{\sigma {R^2}} \over 2}d\theta } }}$
$ = {{2R} \over 3}{{\int\limits_0^{\pi /2} {\cos \theta d\theta } } \over {\int\limits_0^{\pi /2} {d\theta } }}$
$ = {{2R} \over 3}\left( {{2 \over \pi }} \right)$
$ = {{4R} \over {3\pi }}$
So the value of x = 4
The configuration of pieces after collision is shown in the figure.
The value of $\theta$ to the nearest integer is ____________.
Explanation:
Conserving momentum along x-axis
$\overrightarrow {{p_i}} $ = $\overrightarrow {{p_f}} $
10 $\times$ 10$\sqrt 3 $ $=$ 10 $\times$ 20 cos$\theta$
$ \Rightarrow $ cos$\theta$ = ${{\sqrt 3 } \over 2}$
$ \therefore $ $\theta$ = 30$^\circ$
Explanation:
$P = \sqrt {2mKE} $ ($\because$ KE = same)
$ \Rightarrow $ ${{{p_1}} \over {{p_2}}} = \sqrt {{{{m_1}} \over {{m_2}}}} $
$ \Rightarrow $ ${n \over 2} = \sqrt {{4 \over {16}}} $
$ \Rightarrow $ $n = 1$
Explanation:
Also, p1 = p2 = p
$ \Rightarrow $ M1V1 = M2V2 = p
Also, we know that
$K = {{{p^2}} \over {2M}} \Rightarrow {K_1} = {{{p^2}} \over {2{M_1}}}$ & ${K_2} = {{{p^2}} \over {2{M_2}}}$
$ \Rightarrow {{{K_1}} \over {{K_2}}} = {{{p^2}} \over {2{M_1}}} \times {{2{M_2}} \over {{p^2}}} \Rightarrow {{{K_1}} \over {{K_2}}} = {{{M_2}} \over {{M_1}}} = {2 \over 1}$
$ \Rightarrow {A \over 1} = {2 \over 1}$
$ \Rightarrow $ $ \therefore $ A = 2
Explanation:
Using conservation of linear momentum in y-direction,
pi = pf
As, pi = 0
and pf = mv1 sin30$^\circ$ $-$ mv2 sin30$^\circ$
$\Rightarrow$ 0 = m $\times$ ${1 \over 2}$v1 $-$ m $\times$ ${1 \over 2}$v2
$\Rightarrow$ v1 = v2 or v1 : v2 = 1 : 1
Since, v1 : v2 = x : y (given)
$\therefore$ x = 1
Explanation:
$ \therefore $ x = ${{3R} \over 8}$ = ${{3 \times 8} \over 8}$ = 3 cm
Explanation:
Momentum conservation along x,
mv0 × cos $\theta $ × 2 = 2m $ \times $ $\left( {{{{v_0}} \over 2}} \right)$
$ \Rightarrow $ cos $\theta $ = ${1 \over 2}$
$ \Rightarrow $ $\theta $ = 60o
$ \therefore $ 2$\theta $ = 120o
Explanation:
= ${{ - {{{a^2}} \over 4}.d} \over {\pi {a^2} - {{{a^2}} \over 4}}}$
= ${{ - d} \over {4\pi - 1}}$ = ${{ - a} \over {2\left( {4\pi - 1} \right)}}$
$ \therefore $ $ - {a \over X}$ = ${{ - a} \over {2\left( {4\pi - 1} \right)}}$
$ \Rightarrow $ X = ${2\left( {4\pi - 1} \right)}$ = ${\left( {8\pi - 2} \right)}$ = 23.12
So, nearest integer value of X = 23
Explanation:
${1 \over 2}mv_1^2 = {1 \over 2}\left( {{1 \over 2}m{u^2}} \right)$
$ \Rightarrow $ $v_1^2 = {{{u^2}} \over 2}$
$ \Rightarrow $ ${v_1} = {u \over {\sqrt 2 }}$ ......(1)
Also collision is elastic : ki = kf
${{1 \over 2}m{u^2} = {1 \over 2}mv_1^2 + {1 \over 2}\left( {10m} \right)v_2^2}$
$ \Rightarrow $ ${{1 \over 2} \times {1 \over 2}m{u^2} = {1 \over 2}\left( {10m} \right)v_2^2}$
$ \Rightarrow $ ${v_2^2 = {{{u^2}} \over {20}}}$
$ \Rightarrow $ ${v_2} = {u \over {\sqrt {20} }}$
By momentum conservation along y :
m1v1sin $\theta $1 = m2v2sin $\theta $2
$ \Rightarrow $ mv1sin $\theta $1 = 10mv2sin $\theta $2
$ \Rightarrow $ ${u \over {\sqrt 2 }}$sin $\theta $1 = 10 $ \times $ ${u \over {\sqrt {20} }}$sin $\theta $2
$ \Rightarrow $ sin $\theta $1 = $\sqrt {10} $ sin $\theta $2
$ \therefore $ n = 10
Explanation:
(0.1)(3$\widehat i$) + (0.1)(5$\widehat j$) = (0.1)(4)($\widehat i$ + $\widehat j$) + (0.1)$\overrightarrow v $
$ \Rightarrow $ $\overrightarrow v = - \widehat i + \widehat j$
$ \therefore $ $\left| {\overrightarrow v } \right|$ = $\sqrt 2 $
KEB = ${1 \over 2} \times 0.1 \times {\left( {\sqrt 2 } \right)^2}$ = ${1 \over {10}}$ J
$ \therefore $ x = 1