A particle is moving in a circle with uniform speed v. In moving from a point to another diametrically opposite point :
the momentum changes by mv
the momentum changes by 2mv
the kinetic energy changes by $\frac{1}{2}{{\mathrm{mv}}}^{{2}}$
the kinetic energy changes by mv^{2}
B.
the momentum changes by 2mv
Initial velocity v_{1} = v
Final velocity v_{2} = -v
Initial momentum p_{1} = mv
Final momentum p= m (-v) = - mv
Change in momentum
Δp = p_{1} - p_{2}
= mv - (-mv)
= 2mv
Assertion: The driver in a vehicle moving with a constant speed on a straight road is in a noninertial frame of reference.
Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.
If both assertion and reason are true and reason is the correct explanation of assertion
If both assertion and reason are true but reason is not the correct explanation of assertion
If assertion is true but reason is false
If both assertion and reason are false
C.
If assertion is true but reason is false
If one takes two frames with origin O_{1} and O_{2} and if a body P is at rest, boith according to O_{1} as well as O_{2}. The body is at rest. If the frame is moving with a constant velocity, the same observation will not be made. Therefore a frame moving with a constant velocity cannot be taken as an inertial frame. Assertion is right.
But the reason given is wrong because the frame on which Newton's laws of motion are applicable is an inertial frame.
Two bodies of mass m and 4m have equal kinetic energy. What is the ratio of their momentum?
1 : 4
1 : 2
1 : 1
2 : 1
B.
1 : 2
Momentum p = $\sqrt{2{\mathrm{mE}}_{\mathrm{k}}}$
Both have equal kinetic energy
$\frac{{\mathrm{p}}_{1}}{{\mathrm{p}}_{2}}{}{=}{}\sqrt{\frac{{\mathrm{m}}_{1}}{{\mathrm{m}}_{2}}}$
$\frac{{\mathrm{p}}_{1}}{{\mathrm{p}}_{2}}{}{=}{}\sqrt{\frac{\mathrm{m}}{4\mathrm{m}}}$
p_{1} : p_{2} = 1 : 2
A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. The speed of the system, will be
3v
$\frac{\mathrm{v}}{2}$
$\frac{\mathrm{v}}{3}$
2v
C.
$\frac{\mathrm{v}}{3}$
According to conservation of momentum
change in moment is eual to final momentum
m_{1}v_{1}+m_{2}v_{2}= (m_{1}+m_{2})v'
mv+2m×0= (m+2m)v'
so, v'=$\frac{\mathrm{v}}{3}$
A force of 49 newtons is just able to move a block of wood weighing 10 kg-on a rough horizontal surface. Its coefficient of friction is :
1
0.7
0.5
zero
C.
0.5
Coefficient of friction
${\mathrm{\mu}}{}{=}{}\frac{\mathrm{F}}{\mathrm{R}}{}{=}{}\frac{\mathrm{F}}{\mathrm{mg}}$
${=}{}\frac{49}{10\times 19}$
= 0.49 ≅ 0.5
A car is moving with speed 30 m/s on a circular path of radius 500 m. If its speed is increasing at a rate of 5m/s^{2}, then resultant acceleration at that instant is
6.83 m/s^{2}
8 m/s^{2}
5.31 m/s^{2}
4 m/s^{2}
C.
5.31 m/s^{2}
The car will experience two types of accelerations :
(i) Radial acceleration due to circular path
a_{r} = $\frac{{\mathrm{\nu}}^{2}}{\mathrm{r}}$
= $\frac{(30{)}^{2}}{500}=1.8\mathrm{m}/{\mathrm{s}}^{2}$
(ii) Tangential acceleration due to tangential speed
a_{T }= $\left(\frac{\u2206\mathrm{\nu}}{\u2206\mathrm{T}}\right)=5\mathrm{m}/{\mathrm{s}}^{2}$
$\therefore $ Resultant acceleration
a = $\sqrt{{\mathrm{a}}_{\mathrm{r}}^{2}+{\mathrm{a}}_{\mathrm{T}}^{2}}$
= $\sqrt{{\left(1.8\right)}^{2}+(5{)}^{2}}$
= $\sqrt{28.24}$
= 5.31 m/s^{2}
300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking =10 m/s^{2}, work done against friction is
200 J
100 J
Zero
Zero
B.
100 J
Net work done in sliding a body up to a height h on inclined plane
= Work done against gravitational force + Work done against frictional force
A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?
2 s
4 s
1 s
6 s
B.
4 s
When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by
$\nu \ge \sqrt{\mathrm{g}\mathrm{r}}\phantom{\rule{0ex}{0ex}}also\nu =r\omega =r\frac{2\mathrm{\pi}}{T}\phantom{\rule{0ex}{0ex}}Hence,\frac{2\mathrm{\pi r}}{T}\ge \sqrt{rg}\phantom{\rule{0ex}{0ex}}T\le \frac{2\mathrm{\pi r}}{\sqrt{rg}}=2\mathrm{\pi}\sqrt{\frac{\mathrm{r}}{\mathrm{g}}}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}\sqrt{\frac{4}{9.8}}\phantom{\rule{0ex}{0ex}}=2\times 3.14\times 0.6389\phantom{\rule{0ex}{0ex}}=4.0\mathrm{sec}\phantom{\rule{0ex}{0ex}}\mathrm{so}{\mathrm{T}}_{\mathrm{max}}=4\mathrm{sec}$
Assertion: Frictional forces are conservative forces.
Reason: Potential energy can be associated with frictional forces.
If both assertion and reason are true and reason is the correct explanation of assertion
If both assertion and reason are true but reason is not the correct explanation of assertion
If assertion is true but reason is false
If both assertion and reason are false
D.
If both assertion and reason are false
Frictional force is resistive force to motion. When two bodies move against each other some of the kinetic energy is heat energy due to friction. This reduces the total kinetic energy in the system.
Frictional work is not completely recoverable. When the force of friction is the total mechanical energy is not conserved. The friction force is therefore called a non conservative or a dissipative force.
A block of mass 1 kg is placed on the wall under a force F. If coefficient of friction ($\mathrm{\mu}$) is 0.5, then minimum value of force applied is :
9.8 N
49 N
19.6 N
1.96 N
C.
19.6 N
Weight of a block = mg
For block to be balanced,
mg = $\mu $F
$\therefore $ F = $\frac{\mathrm{mg}}{\mathrm{\mu}}$
= $\frac{1\times 9.8}{0.5}$ = 19.6 N