Laws of Motion Test

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Laws of Motion Test - 50

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Question 1
1 / -0

A box of mass m is made to move from A to B as shown in figure. Where will the normal force be larger:

A
A

B
B

C
Same at both A and B

D
None of the above

Solution

At A the normal force will be larger than B, since it has to support the weight of the box + it also has to turn the box upwards

Hence the correct option is (a)
Question 2
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Complete the statement of the first law of motion. "A body at rest stays at ____ and a body in motion stays in ____ unless an ____ is applied"

A
Motion; Rest; External Force

B
Rest; Motion; External Force

C
Rest; Motion; Internal Force

D
None of these

Solution

The first law of motion states "A body at rest stays at rest and a body in motion stays in motion unless an external force is applied"
Question 3
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Banking of roads is done due to

A
provide enough friction for circular motion of the vehicle

B
provide necessary centripetal force required for circular motion of the vehicle

C
provide enough radius of curvature for circular motion of the vehicle

D
provide enough area for navigating in the circular motion of the vehicle

Solution

This case is similar to bending of cyclist but bending cars at an angle is a dangerous stunt and hence banking of roads is done.
Here a component of N will provide necessary centripetal force.
Question 4
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A block of mass $$m$$ is on an inclined plane of angle of angle $$\theta$$. The coefficient of friction between the block and the plane is $$\mu$$ and $$\tan { \theta } >\mu$$. The block is held stationary by applying a force $$P$$ parallel to the plane. The direction of force pointing up the plane is taken to be positive. As $$P$$ is varied from $${ P }_{ 1 }=mg\left( \sin { \theta } -\mu \cos { \theta } \right)$$ to $${ P }_{ 2 }=mg\left( \sin { \theta } -\mu \cos { \theta } \right)$$, the frictional force $$f$$ versus $$P$$ graph will look like

A

B

C

D

Solution
Question 5
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A 0.25kg ball attached to a 1.5 m rope moves with a constant speed of 15 m/s around a vertical circle. Calculate the tension force on the rope at the middle of the circle:

A
37.5 N

B
137.5 N

C
2.5 N

D
25 N

Solution

The centripetal force acts along the direction of the tension and hence $$T=mv^2/R = 0.25 \times 15^2/1.5=37.5 N$$
Question 6
1 / -0

Directions For Questions

Two smooth balls A and B each of mass $$m$$ and radius $$R$$, have their centres at $$(0,0,R)$$ and $$(5R,-R,R)$$ respectively, in a coordinate system as shown. Ball A, moving along positive x axis, collides with ball B. Just before collision, speed of ball A is $$4m/s$$ and ball B is stationary. The collision between the balls is elastic.

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Impulse of the force exerted by A and B during the collision, is equal to

A
$$\left( \sqrt { 3 } mi+3mj \right) kg-m/s$$

B
$$\left( \cfrac { \sqrt { 3 } }{ 2 } mi-\sqrt { 3 } mj \right) kg-m/s$$

C
$$\left( 3mi-\sqrt { 3 } mj \right) kg-m/s$$

D
$$\left( 2\sqrt { 3 } mi+3mj \right) kg-m/s\quad $$

Solution

$${ \overrightarrow { J } }_{ Aon B }=m{ \overrightarrow { v } }_{ { B }_{ f } }-{ \overrightarrow { v } }_{ { B }_{ i } }=m\left[ 4\cos { { 30 }^{ o } } \left( \cos { 30\hat { i } } -\sin { { 30 }^{ o }j } \right) -0 \right] =\left( 3mi-\sqrt { 3 } mj \right) kg-m/s$$
Question 7
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The figure shows the position -time (x- t) graph of one-dimensional motion of a body of mass $$0.4$$ kg. The magnitude of each impulse is :

A
$$0.4$$ Ns

B
$$0.8$$ Ns

C
$$1.6$$ Ns

D
$$0.2$$ Ns

Solution

B. $$0.8Ns$$

Impulse= change in momentum

= final momentum- initial momentum

Initial velocity, $$u=\dfrac{2-0}{2-0}=1m/s^{-1}$$

Final velocity, $$v=\dfrac{0-2}{4-2}=-1m/s^{-1}$$

Initial momentum, $$P_i= mu=0.4\times1=0.4Ns$$

Final momentum,$$P_f=mv= 0.4\times (-1)=-0.4Ns$$

Impulse$$=P_f-P_i$$

$$=-0.4-0.4=-0.8Ns$$

The magnitude of each impulse is $$0.8Ns$$
Question 8
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A ball is moving in the direction as shown. What will be the direction of momentum?

A
$$W$$

B
$$N$$

C
$$S$$

D
$$E$$

Solution

Since momentum $$p=mv$$

Direction of momentum is same as the direction of velocity. Hence, the direction of momentum of ball in the given case is East.
Question 9
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A cyclist leans with the horizontal at an angle of $$30^{o}$$, while negotiating round a circular road of radius $$20\sqrt{3}$$ meters. Speed of the cycle should be :

A
$$7\sqrt{3}\ m/s$$

B
$$14\ m/s$$

C
$$7\sqrt{6}\ m/s$$

D
$$14\sqrt{3}\ m/s$$

Solution

Balancing the forces:
$$N\sin { 30° } =mg\longrightarrow (1)$$
and
$$ N\cos { 30 } °=\cfrac { m{ v }^{ 2 } }{ R } \longrightarrow (2)$$
dividing $$(2)$$ by $$(1)$$
$$\tan { 30° } =\cfrac { gR }{ { v }^{ 2 } } \\ \Rightarrow { v }^{ 2 }=\cfrac { gR }{ \tan { 30° } } \\ \Rightarrow { v }^{ 2 }=10\times 20\sqrt { 3 } \times \sqrt { 3 } \\ \quad \quad =200\times 3=600\\ v=\sqrt { 600 } =24.5\quad m/s\\ \quad \quad \quad \quad =14\sqrt { 3 } m/s$$
Question 10
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A car is moving in a circular horizontal track of radius $$10m$$ with a constant speed of $$10m/s$$. A plumb bob is suspended from the roof of the car by a light rigid rod of length $$1m$$. The angle made by the rod with track is

A
zero

B
$${30}^{o}$$

C
$${45}^{o}$$

D
$${60}^{o}$$

Solution

T cos$$\theta$$ = mg..........(1)
T Sin$$\theta$$ = $$\dfrac {mv^2}r$$.....(2)
$$Equ^n\ (2)/(1)$$
$$tan \theta = \dfrac{v^2}{rg} = \dfrac{100}{100} =1$$
I used $$g=10$$ and we get $$\theta =45^0$$