Question 1 |
A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly-elastic direct-central impact with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value(s) of m is/are
1 kg only | |
6 kg only | |
1 kg, 6 kg | |
1 kg, 9 kg |
Question 1 Explanation:
Let V_{1} is the speed of 3 kg mass after collision
\mathrm{V}_{2} is the speed of m kg mass after collision
\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}
By linear momentum conservation
\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}
By using (1), (2) \& (3) we get
m = 1 kg (or) 9 kg
\mathrm{V}_{2} is the speed of m kg mass after collision
\begin{array}{l} e=I=\frac{V_{2}-V_{1}}{4} \\ \Rightarrow V_{2}-V_{1}=4 \end{array}
By linear momentum conservation
\begin{array}{l} 3 \times 4=3 V_{1}+m V_{2}\\ \frac{1}{2} \times 3 \times V_{1}^{2}=6 \\ \Rightarrow V_{1}=\pm 2 \end{array}
By using (1), (2) \& (3) we get
m = 1 kg (or) 9 kg
Question 2 |
A point mass is shot vertically up from ground level with a velocity of 4 m/s at time, t_0. It loses 20% of its impact velocity after each collision with the ground. Assuming that the acceleration due to gravity is 10m/s^{2} and that air resistance is negligible, the mass stops bouncing and comes to complete rest on the ground after a total time (in seconds) of
1 | |
2 | |
4 | |
\infty |
Question 2 Explanation:
\begin{aligned} (1) \rightarrow \qquad t&=? \\ v&=u+a t \\ 0&=4-10 t \\ t&=\frac{4}{10}=0.4 s \\ (2) \rightarrow \qquad t^{\prime}&=? \\ u^{\prime}&=0.8 \times u \\ &=0.8 \times 4=3.2 \mathrm{m} / \mathrm{s} \\ \mathrm{v}^{\prime}&=U^{\prime}+\mathrm{al}^{\prime} \\ 0&=3.2-10 t^{\prime} \\ t^{\prime}&=\frac{3.2}{10}=0.32 \mathrm{s} \\ (3) \rightarrow \qquad t^{\prime \prime}&=? \\ U^{\prime \prime} &=0.8 u^{\prime} \\ &=0.8 \times 3.2=2.56 \mathrm{m} / \mathrm{s} \\ V^{\prime \prime} &=U^{\prime \prime}+a t^{\prime} \\ 0 &=2.56-10 t^{\prime \prime} \\ t^{\prime \prime} &=0.256 s \end{aligned}
So, t, t^{\prime}, t^{\prime \prime} are forming a GP series
\begin{array}{l} \text{So, total time }=2\left(t+t^{\prime}+t^{\prime \prime}+\ldots . .0\right)\\ =2[0.4+0.32+0.256+\ldots .0] \\ =2 \times \frac{0.4}{1-0.8}=2 \times 2=4 s \end{array}
Question 3 |
Two disks A and B with identical mass (m) and radius (R) are initially at rest. They roll down from the top of identical inclined planes without slipping. Disk A has all of its mass concentrated at the rim, while Disk B has its mass uniformly distributed. At the bottom of the plane, the ratio of velocity of the center of disk A to the velocity of the center of disk B is.
\sqrt{\frac{3}{4}} | |
\sqrt{\frac{3}{2}} | |
1 | |
\sqrt{2} |
Question 3 Explanation:
m_{A}=m_{B}=m
Outer circumferential radius of both disc A
and B is also same i.e. R
\begin{aligned} I_{A}&=m R^{2} \\ I_{B}&=\frac{m R^{2}}{2} \end{aligned}
\frac{V_{A}}{V_{B}} at bottom of plane =?
m g h=(K \cdot E) of disc A=K \cdot E of disc B
\begin{aligned} \frac{1}{2} m V_{A}^{2}+\frac{1}{2} I_{A} \omega_{A}^{2} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} I_{B} \omega_{B}^{2} \\ \frac{1}{2} m V_{A}^{2}+\frac{1}{2} m R^{2} \frac{V_{A}^{2}}{R^{2}} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} \frac{m R^{2}}{2} \times \frac{V_{B}^{2}}{R^{2}} \\ \frac{V_{A}^{2}}{2}+\frac{V_{A}^{2}}{2} &=\frac{V_{B}^{2}}{2}+\frac{V_{B}^{2}}{4} \\ V_{A}^{2} &=\frac{3}{4} V_{B}^{2} \\ \frac{V_{A}}{V_{B}} &=\sqrt{\frac{3}{4}} \end{aligned}
Outer circumferential radius of both disc A
and B is also same i.e. R
\begin{aligned} I_{A}&=m R^{2} \\ I_{B}&=\frac{m R^{2}}{2} \end{aligned}
\frac{V_{A}}{V_{B}} at bottom of plane =?
m g h=(K \cdot E) of disc A=K \cdot E of disc B
\begin{aligned} \frac{1}{2} m V_{A}^{2}+\frac{1}{2} I_{A} \omega_{A}^{2} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} I_{B} \omega_{B}^{2} \\ \frac{1}{2} m V_{A}^{2}+\frac{1}{2} m R^{2} \frac{V_{A}^{2}}{R^{2}} &=\frac{1}{2} m V_{B}^{2}+\frac{1}{2} \frac{m R^{2}}{2} \times \frac{V_{B}^{2}}{R^{2}} \\ \frac{V_{A}^{2}}{2}+\frac{V_{A}^{2}}{2} &=\frac{V_{B}^{2}}{2}+\frac{V_{B}^{2}}{4} \\ V_{A}^{2} &=\frac{3}{4} V_{B}^{2} \\ \frac{V_{A}}{V_{B}} &=\sqrt{\frac{3}{4}} \end{aligned}
Question 4 |
A particle of unit mass is moving on a plane. Its trajectory, in polar coordinates, is given by r(t)=t^{2},\theta (t)=t, where t is time. The kinetic energy of the particle at time t=2 is
4 | |
12 | |
16 | |
24 |
Question 4 Explanation:
\begin{aligned} r &=t^{2}: \theta=t \\ \mathrm{K.E} &=\frac{1}{2} m v^{2}=? \text { at } t=2 \mathrm{sec} \\ \Rightarrow m &=1 \mathrm{kg} \\ \vec{V} &=r \omega(\hat{t})+\frac{d r}{d t} \dot{r}=t^{2} \times 1(\hat{t})+2 t \hat{r} \\ \therefore \frac{d \theta}{d t}&=\omega=1
\\ \vec{V} &=t^{2}(\hat{t})+2 t(\hat{r}) \\ \text{at }\qquad t &=2 s \\ \vec{V} &=4(\hat{t})+4(\hat{r}) \\ |\vec{V}| &=\sqrt{16+16}=\sqrt{32} \\ K.E. &=\frac{1}{2}mv^{2}=\frac{1}{2}\times1\times32=16 \end{aligned}
Question 5 |
A circular disc of radius 100 mm and mass 1 kg, initially at rest at position A, rolls without slipping down a curved path as shown in figure. The speed v of the disc when it reaches position B is _________ m/s.
Acceleration due to gravity g = 10 m/s^{2}.
Acceleration due to gravity g = 10 m/s^{2}.
50 | |
30 | |
45 | |
20 |
Question 5 Explanation:
Loss in P.E. = Gain in rotational K . E
+ gain in translational K.E.
\begin{aligned} m g H &=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2} \\ v &=r \omega \\ \omega &=\frac{v}{r} \\ m g H &=\frac{1}{2} \frac{m r^{2}}{2} \times\left(\frac{v}{r}\right)^{2}+\frac{1}{2} m v^{2} \\ 10 \times 30 &=\frac{v^{2}}{4}+\frac{v^{2}}{2} \\ 10 \times 30 \times 4 &=v^{2}+2 v^{2} \\ 3 v^{2} &=30 \times 10 \times 4 \\ v^{2} &=400 \\ \text{or} \qquad v &=20 \mathrm{m} / \mathrm{s} \end{aligned}
+ gain in translational K.E.
\begin{aligned} m g H &=\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2} \\ v &=r \omega \\ \omega &=\frac{v}{r} \\ m g H &=\frac{1}{2} \frac{m r^{2}}{2} \times\left(\frac{v}{r}\right)^{2}+\frac{1}{2} m v^{2} \\ 10 \times 30 &=\frac{v^{2}}{4}+\frac{v^{2}}{2} \\ 10 \times 30 \times 4 &=v^{2}+2 v^{2} \\ 3 v^{2} &=30 \times 10 \times 4 \\ v^{2} &=400 \\ \text{or} \qquad v &=20 \mathrm{m} / \mathrm{s} \end{aligned}
Question 6 |
A system of particles in motion has mass center G as shown in the figure. The particle i has mass mi and its position with respect to a fixed point O is given by the position vector r_{i} . The position of the particle with respect to G is given by the vector \rho _{i} . The time rate of change of the angular momentum of the system of particles about G is
(The quantity \ddot{\rho }_{i} indicates second derivative of {\rho }_{i} with respect to time and likewise for ).
(The quantity \ddot{\rho }_{i} indicates second derivative of {\rho }_{i} with respect to time and likewise for ).
\Sigma_{i}r_{i} \times m_{i}\ddot{\rho }_{i} | |
\Sigma_{i}\rho _{i} \times m_{i}\ddot{r}_{i} | |
\Sigma_{i} r_{i} \times m_{i}\ddot{r}_{i} | |
\Sigma_{i} \rho _{i} \times m_{i}\ddot{\rho }_{i} |
Question 6 Explanation:
Time rate of change of angular momentum
\begin{aligned} &=\frac{d \vec{L}}{d t} \\ \bar{L}_{G} &=\vec{P} \times \vec{r} \end{aligned}\\ \;[\text{For particle }m_{i} \text{ about G }]\\ \begin{aligned} \Rightarrow \quad G & =m_{i} \times \vec{v}_{i} \times \vec{P}_{r} \\ & \vec{P}=m_{i} \times \vec{V}_{i} \\ \text { But } & \vec{V}_{i}=\frac{d}{d t} \vec{r}_{i}=\dot{r}_{i} \end{aligned}
V_{i}= rate of change o positions Position vector
is given with respect to "0"
\begin{aligned} \underset{{L}_{G}}{\rightarrow} &=m_{i} i_{i} \vec{\rho}_{i} \\ \text { Now } \frac{d}{d t} L_{G} &=\frac{d}{d t}\left[p_{i} m_{i} f_{i}\right] \\ &=\rho_{i} m_{i} \ddot{r}_{i} \end{aligned}
For whole body
\frac{d L}{d t}=\sum \frac{d L_{G}}{d t}=\sum \rho_{i} m_{i} \ddot{r}_{i}
Question 7 |
A mass of 2000 kg is currently being lowered at a velocity of 2 m/s from the drum as shown in the figure. The mass moment of inertia of the drum is 150 Kg-m^{2} On applying the brake, the mass is
brought to rest in a distance of 0.5 m. The energy absorbed by the brake (in kJ) is __________
12.65 | |
14.11 | |
45.5 | |
58.5 |
Question 7 Explanation:
E_{\text {intial }}-E_{\text {brake }}=E_{\text {final }}
\frac{1}{2}l\omega_{1}^{2}+\frac{1}{2} m v_{1}^{2}+m g H_{1}-E_{\mathrm{brake}}
=\frac{1}{2} I \omega_{2}^{2}+\frac{1}{2} m v_{2}^{2}+m g H_{2}
E_{\mathrm{brake}}=\frac{1}{2} I \omega_{1}^{2}+\frac{1}{2} m v_{1}^{2}+m g\left(H_{1}-H_{2}\right)
H_{1}-H_{2}=0.5 \mathrm{m}
E_{\mathrm{brake}}=\frac{1}{2} I \omega_{1}^{2}+m g h+\frac{1}{2} m v_{1}^{2}
=\frac{\frac{1}{2} \times 150 \times 2^{2}+2000 \times 9.81 \times 0.5+\frac{1}{2} \times \frac{2000 \times 2^{2}}{1000}}{1000}
=14.11 \mathrm{kJ}
Question 8 |
A point mass having mass M is moving with a velocity V at an angle \theta to the wall as shown in the figure. The mass undergoes a perfectly elastic collision with the smooth wall and rebounds. The total change (final minus initial) in the momentum of the mass is
-2MV \cos \theta \hat{J} | |
2MV \sin \theta \hat{J} | |
2MV \cos \theta \hat{J} | |
-2MV \sin \theta \hat{J} |
Question 8 Explanation:
\begin{aligned} P_{1}&=M V \cos \theta \hat{i}+M V \sin \theta \hat{j} \\ P_{f}&=M V \cos \theta \hat{i}-M V \sin \theta \hat{j} \\ P_{f}-P_{i}&=-2 M V \sin \theta \hat{j} \end{aligned}
Question 9 |
A point mass M is released from rest and slides down a spherical bowl (of radius R) from a height H as shown in the figure below. The surface of the bowl is smooth (no friction). The velocity of the mass at the bottom of the bowl is
\sqrt{gH} | |
\sqrt{2gR} | |
\sqrt{2gH} | |
0 |
Question 9 Explanation:
Loss of P.E. = Gain in K.E.
\begin{aligned} M g h &=\frac{1}{2} M v^{2} \\ v &=\sqrt{2 g H} \end{aligned}
Question 10 |
A small ball of mass 1 kg moving with a velocity of 12 m/s undergoes a direct central impact with a stationary ball of mass 2 kg. The impact is perfectly elastic. The speed (in m/s) of 2 kg mass ball after the impact will be ____________
8m/s | |
9m/s | |
9m/s | |
4m/s |
Question 10 Explanation:
1. Conserving linear momentum
\begin{aligned} 1 \times 12 &=1 \times V_{1}+2 \times V_{2} \\ 12 &=V_{1}+2 V_{2}\qquad \ldots(i) \end{aligned}
2. Velocity of approach = Velocity of seperation
12-0=v_{2}-v_{1}
V_{2}-V_{1}=12 \qquad \ldots(ii)
From (i) and (ii), we get
v_{2}=8 \mathrm{m} / \mathrm{s}
There are 10 questions to complete.