Laws of Motion Test

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Laws of Motion Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Total impulse imposed by ball of mass 2 kg to the ball of mass 4 kg has the magnitude equal to

A
10 N-s

B
5.3 N-s

C
40 N-s

D
16 Ns

Solution
Question 2
1 / -0

A gun of mass M fires a bullet of mass m with a velocity v relative to the gun. The average force required to bring the gun to rest in 0.5 sec. is

A
$$

\dfrac{2 \mathrm{Mmv}}{\mathrm{M}+\mathrm{m}}

$$

B
$$

\dfrac{M m v}{2(M+m)}

$$

C
$$

\dfrac{3 \mathrm{Mmv}}{2(\mathrm{M}+\mathrm{m})}

$$

D
$$

\dfrac{\mathrm{Mmv}}{(\mathrm{M}+\mathrm{m})}

$$

Solution

Velocity of gun $$=v_1$$

Velocity of bullet $$=v-v_1$$

Then,

$$m(v-v_1)=Mv_1$$

$$v_1=\dfrac{mv}{m+M}$$

Distance travelled by gun in $$0.5\,s$$

$$0=v_1-a\times 0.5$$

$$\implies a=2v_1$$

Also,

$$S=v_1\times 0.5-\dfrac 12 a0.5^2$$

$$\implies S=\dfrac{v_1}{4}$$

From work energy theorem,

$$F\times S=\dfrac 12 Mv_1^2-0$$

$$S=2Mv_1$$

$$F=\dfrac{2Mmv}{m+M}$$
Question 3
1 / -0

A ball of mass 250 g moving with 20 m/s strikes a vertical wall and rebounds along the same line with a velocity of 15 m/s. If the time of contact is 0.1 s, the force exerted by the wall on the ball is

A
$$87.5 N$$

B
$$12.5 N$$

C
$$-12.5 N$$

D
$$-87.5 N$$

Solution
Question 4
1 / -0

Total impulse imposed by a ball of mass $$2 kg$$ moving with the speed of $$10 m/s$$ in rightward direction to the ball of mass $$4 kg$$ moving with the speed of $$8 m/s$$ in the same direction is (The collision is perfectly elastic)

A
$$10 Ns$$

B
$$5.3 Ns$$

C
$$40 Ns$$

D
$$16 Ns$$
Question 5
1 / -0

Calculate the impulse necessary to stop the car of mass 500 kg travelling at 72 kmph.

A
10000 Ns

B
20000 Ns

C
30000 Ns

D
40000 Ns

Solution
Question 6
1 / -0

An impulse $$\vec { 1 }$$ changes the velocity of a particle from $$\vec { v } _ { 1 }$$ to $$\vec { v } _ { 2 }$$ . Kinetic energy gained by the particle is

A
$$\frac { 1 } { 2 } \vec { I } \left( \vec { v } _ { 1 } + \vec { v } _ { 2 } \right)$$

B
$$\frac { 1 } { 2 } \vec { I } \left( \vec { v } _ { 1 } - \vec { v } _ { 2 } \right)$$

C
$$\vec { 1 } \cdot \left( \vec { v } _ { 1 } - \vec { v } _ { 2 } \right)$$

D
$$\vec { I } \left( \vec { v } _ { 1 } + \vec { v } _ { 2 } \right)$$

Solution

Kinetic energy of any moving particle is $$\frac { 1 }{ 2 } m{ v }^{ 2 }$$
Impulse of any particle is change in momentum.
Kinetic energy gained by particle is,
$$\frac { 1 }{ 2 } m{ { v }_{ f } }^{ 2 }-\frac { 1 }{ 2 } m{ { v }_{ i } }^{ 2 }\\ =\frac { 1 }{ 2 } m\left( { { v }_{ f } }^{ 2 }-{ { v }_{ i } }^{ 2 } \right) \\ =\frac { 1 }{ 2 } m\left( { v }_{ f }-{ v }_{ i } \right) \left( { v }_{ f }+{ v }_{ i } \right) \\ =\frac { 1 }{ 2 } \overrightarrow { I } \left( { v }_{ 2 }+{ v }_{ 1 } \right) $$
Hence, correct option is $$(A)$$
Question 7
1 / -0

Find the tangent of angle of banking for a curved smooth road of radius $$40 m$$ for a speed of $$72 km / h$$ $$\left( { g }=10{ m }/{ s }^{ { 2 } } \right) .$$

A
$$45 ^ { \circ }$$

B
$$37 ^ { \circ }$$

C
$$42 ^ { \circ }$$

D
$$30 ^ { \circ }$$

Solution

Given that radius of the smooth curved road $$= 40 m$$
Speed that is required for smooth turning= $$72km/h= 20m/sec$$
Let the angle of banking be $$\theta $$
$$\implies tan\theta = \dfrac{v^2}{rg} = \dfrac{20^2}{40*10}= \dfrac{400}{400} =1$$
$$\implies \theta = 45^o$$
Question 8
1 / -0

Who asserted that the natural state of body is the state of rest

A
Newton

B
Galileo

C
Aristotle

D
Einstein

Solution
Question 9
1 / -0

A ball of mass $$1$$ $$ \mathrm{kg} $$ drops vertically on to the floor with a speed of $$ 25$$ $$ \mathrm{m} / \mathrm{s} $$ . It rebounds with an initial velocity of $$10$$ $$ \mathrm{m} / \mathrm{s} $$ . What impulse acts on the ball during contact?

A
$$35$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ downwards

B
$$35$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ upwards

C
$$30$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ downwards

D
$$30$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ upwards

Solution

Given,
mass of ball is $$1kg$$
Initial velocity is $$10m/s$$
Final velocity is $$-25m/s$$ (Rebounds)
We know that impulse is change in momentum therefore,
$$\triangle P=m\left( { v }_{ f }-{ v }_{ i } \right) $$
Substitute all value in above equation we will get
$$\triangle P=1\times \left( -25-10 \right) \\ \triangle P=-35m/s$$
Here, we have taken in upward direction as positive velocity and in downward direction is as negative velocity therefore, Impulse will act in downward direction with magnitude is $$35kgm/s$$ hence, correct option is $$(A)$$
Question 10
1 / -0

A rubber band of length $$\lambda$$ has a stone of mass $$m$$ tied to its one end. It is whirled with speed $$v second$$ that the stone described a horizontal circular path. The tension $$T$$ in the rubber band is -

A
zero

B
$$mv^2/\lambda $$

C
$$>(mv^2)/\ell $$

D
none

Solution

As there is only Tension T is theonly force. So it is the necessary centripetal force
$$T= \dfrac{mv^2}{\lambda}$$

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 59

Result

Laws of Motion Test - 59

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SHARING IS CARING

Question 1

10 N-s

5.3 N-s

40 N-s

16 Ns

Solution

Question 2

$$ \dfrac{2 \mathrm{Mmv}}{\mathrm{M}+\mathrm{m}} $$

$$ \dfrac{M m v}{2(M+m)} $$

$$ \dfrac{3 \mathrm{Mmv}}{2(\mathrm{M}+\mathrm{m})} $$

$$ \dfrac{\mathrm{Mmv}}{(\mathrm{M}+\mathrm{m})} $$

Solution

Question 3

$$87.5 N$$

$$12.5 N$$

$$-12.5 N$$

$$-87.5 N$$

Solution

Question 4

$$10 Ns$$

$$5.3 Ns$$

$$40 Ns$$

$$16 Ns$$

Question 5

10000 Ns

20000 Ns

30000 Ns

40000 Ns

Solution

Question 6

$$\frac { 1 } { 2 } \vec { I } \left( \vec { v } _ { 1 } + \vec { v } _ { 2 } \right)$$

$$\frac { 1 } { 2 } \vec { I } \left( \vec { v } _ { 1 } - \vec { v } _ { 2 } \right)$$

$$\vec { 1 } \cdot \left( \vec { v } _ { 1 } - \vec { v } _ { 2 } \right)$$

$$\vec { I } \left( \vec { v } _ { 1 } + \vec { v } _ { 2 } \right)$$

Solution

Question 7

$$45 ^ { \circ }$$

$$37 ^ { \circ }$$

$$42 ^ { \circ }$$

$$30 ^ { \circ }$$

Solution

Question 8

Newton

Galileo

Aristotle

Einstein

Solution

Question 9

$$35$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ downwards

$$35$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ upwards

$$30$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ downwards

$$30$$ $$ \mathrm{kg} \mathrm{m} / \mathrm{s} $$ upwards

Solution

Question 10

zero

$$mv^2/\lambda $$

$$>(mv^2)/\ell $$

none

Solution

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$$

\dfrac{2 \mathrm{Mmv}}{\mathrm{M}+\mathrm{m}}

$$

$$

\dfrac{M m v}{2(M+m)}

$$

$$

\dfrac{3 \mathrm{Mmv}}{2(\mathrm{M}+\mathrm{m})}

$$

$$

\dfrac{\mathrm{Mmv}}{(\mathrm{M}+\mathrm{m})}

$$