System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 22

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Question 1
1 / -0

A hollow sphere rolls down a $$30^o$$ incline of length 6m without slipping. The speed of center of mass a the bottom of plane is

A
$$6 m/s^{-1}$$

B
$$ 3 ms^{-1}$$

C
$$6\sqrt{2} ms^{-1}$$

D
$$3\sqrt{2} ms^{-1}$$

Solution

apply consecration of energy

$${E_i} = mgh$$

$${E_i} = m10(6\sin 30^\circ )$$

$${E_i} = 30\,m$$

$${E_f} = {1 \over 2}m{v^2} + {1 \over 2}l{\omega ^2}$$

$${E_f} = {1 \over 2}m{R^2}{\omega ^2} + {1 \over 2}\left( {{2 \over 3}m{R^2}} \right){\omega ^2}$$

$${E_f} = {5 \over 6}m{R^2}{\omega ^2}$$

$${E_i} = {E_f}$$

$${R^2}{\omega ^2} = 36$$

$${v_{cm}} = 6$$
Question 2
1 / -0

A body has its centre of mass at the origin. The x-coordinates of the particles

A
may be all positive

B
must be positive for some particles and negative in other particles

C
Must be all non - negative

D
may be positive for some particles and negative in other particles

Solution

The centre of mass of the body is at origin this means that some coordinates of particles in body are positive and some are negative. They cannot be purely positive or negative separately.

Answer: May be positive for some cases and negative in other cases.
Question 3
1 / -0

Moment of inertia about axis 1 is :

A
$$\dfrac { { M2 }^{ 2 } }{ 12 } $$

B
$$\dfrac { { Mb }^{ 2 } }{ 12 } $$

C
$$\dfrac { M1^{ 2 } }{ 3 } $$

D
$$\dfrac { Mb^{ 2 } }{ 3 } $$
Question 4
1 / -0

The centre of mass of a system of particles is at the origin. It follows that:

A
the number of particles to the right of the origin is equal to the left of origin.

B
the total mass of the particles to the right of the origin is same as total mass to the left of the origin.

C
the number of particles on the X-axis should be equal to the number of particles on the Y-axis .

D
if there is a particle on the +ve X-axis, there should be atleast one particle on the -ve X-axis.

E
None of these.

Solution

Center of mass of a system of particles is at the origin. It follows that:
(a) Number of particles to the right of origin=Number of particles to the left of origin
(b) Total mass of particles to the right of origin=Total mass of particles to the left of origin
Question 5
1 / -0

In classical system:

A
the varying mass system is not considered

B
the varying mass system must be considered

C
the varying mass system may be considered

D
only varying of mass due to velocity is considered

Solution

In classical systems varying mass may be considered. One common example is the burning of fuel in a rocket due to which its mass keeps on decreasing with time. In such a situation newton's second law is modified a little.
$$F= \dfrac{dP}{dt}$$
$$\Rightarrow F=\dfrac{d(mv)}{dt}$$
$$\Rightarrow F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$$
$$\Rightarrow F=ma+v\dfrac{dm}{dt}$$
Question 6
1 / -0

The momentum of a body in two perpendicular directions at any instant of time 't' is given by
$$P_x=2t^2+6$$ and $$p_y=\cfrac {3t^2}2 +3$$. The force acting on the body at t=2 sec is:

A
$$5 units$$

B
$$2 units$$

C
$$10 units$$

D
$$7 units$$

Solution

$${ F }_{ x }=\cfrac { d{ P }_{ x } }{ dt } =4t\Longrightarrow { F }_{ x }=4(2)=8\\ { F }_{ y }=\cfrac { d{ P }_{ y } }{ dt } =3t\Longrightarrow { F }_{ y }=3(2)=6\\ F=\sqrt { { { F }_{ x } }^{ 2 }+{ { F }_{ y } }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ 6 }^{ 2 } } =\sqrt { 64+36 } =\sqrt { 100 } =10$$
Question 7
1 / -0

Internal forces can change.

A
the linear momentum but not the kinetic energy

B
the kinetic energy but not the linear momentum

C
linear momentum as well as kinetic energy

D
neither the linear momentum nor the kinetic energy.

Solution

Answer should be B because
1)Internal forces can never ever change the total mechanical energy(KE+PE) of an object, but they can just convert the energy from one form to another. This means they can convert a part of KE to PE, which implies they can change the KE but the total (KE+PE) remains conserved.
2)We have Fext=(dP/dt), which means ONLY external forces can change the linear momentum of an object and NOT the internal forces.
Question 8
1 / -0

Unit of which physical quantity is same as that of unit of impulse of force?

A
Force

B
Acceleration

C
Momentum

D
Velocity

Solution

Momentum,
Both momentum and impulse of force have the same unit, i.e. kg m/s.
Question 9
1 / -0

A constant torque acting on a uniform circular wheel changes its angular momentum from $$L$$ to $$4L$$ in $$4$$ seconds. The torque acting on it is:

A
$$4L$$

B
$$12L$$

C
$$\dfrac{3}{4} L$$

D
$$\dfrac{4}{3} L$$

Solution

As we know that, $$\tau=\dfrac{\Delta{L}}{{\Delta}t}$$

So, $$\tau=\dfrac{4L-L}{4}=\dfrac{3L}{4}$$
Question 10
1 / -0

A particle of mass $$m$$ moves uniformly with a speed $$v$$ along a circle of radius $$r$$. A and B are two points on the circle, such that the arc $$AB$$ subtends an angle $$\theta$$ at the center of the circle. The magnitude of change in momentum as the particle moves from $$A$$ to $$B$$ is given by:

A
$$2 mv sin (\dfrac{\theta}{2})$$

B
$$2 mv cos (\dfrac{\theta}{2})$$

C
$$zero$$

D
$$2 mv tan (\dfrac{\theta}{2})$$

Solution

Assume particle is at lowest point of the circle

$$P_i= mv\hat{i}$$

After it rotates $$\theta$$ angle

$$P_f= mv\cos \theta \hat{i}+mv\sin \theta \hat{j}$$

Change in momentum $$=P_f-P_1 =$$ $$mv\cos \theta \hat{i}+mv\sin \theta \hat{j} -mv\hat{i}$$

$$ = mv(-1+\cos \theta) \hat{i}+mv\sin \theta \hat{j} $$

$$ |\Delta P|= mv\sqrt{(-1+\cos \theta)^2+\sin^2 \theta \ }$$

$$ |\Delta P|= mv\sqrt{1+\cos^2 \theta-2\cos \theta+\sin^2 \theta }$$

$$ |\Delta P|= 2mv\sin (\dfrac{ \theta }{2}) $$

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System of Particles and Rotational Motion Test - 22

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System of Particles and Rotational Motion Test - 22

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Question 1

$$6 m/s^{-1}$$

$$ 3 ms^{-1}$$

$$6\sqrt{2} ms^{-1}$$

$$3\sqrt{2} ms^{-1}$$

Solution

Question 2

may be all positive

must be positive for some particles and negative in other particles

Must be all non - negative

may be positive for some particles and negative in other particles

Solution

Question 3

$$\dfrac { { M2 }^{ 2 } }{ 12 } $$

$$\dfrac { { Mb }^{ 2 } }{ 12 } $$

$$\dfrac { M1^{ 2 } }{ 3 } $$

$$\dfrac { Mb^{ 2 } }{ 3 } $$

Question 4

the number of particles to the right of the origin is equal to the left of origin.

the total mass of the particles to the right of the origin is same as total mass to the left of the origin.

the number of particles on the X-axis should be equal to the number of particles on the Y-axis .

if there is a particle on the +ve X-axis, there should be atleast one particle on the -ve X-axis.

None of these.

Solution

Question 5

the varying mass system is not considered

the varying mass system must be considered

the varying mass system may be considered

only varying of mass due to velocity is considered

Solution

Question 6

$$5 units$$

$$2 units$$

$$10 units$$

$$7 units$$

Solution

Question 7

the linear momentum but not the kinetic energy

the kinetic energy but not the linear momentum

linear momentum as well as kinetic energy

neither the linear momentum nor the kinetic energy.

Solution

Question 8

Force

Acceleration

Momentum

Velocity

Solution

Question 9

$$4L$$

$$12L$$

$$\dfrac{3}{4} L$$

$$\dfrac{4}{3} L$$

Solution

Question 10

$$2 mv sin (\dfrac{\theta}{2})$$

$$2 mv cos (\dfrac{\theta}{2})$$

$$zero$$

$$2 mv tan (\dfrac{\theta}{2})$$

Solution

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