Forces and Laws of Motion Test

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Forces and Laws of Motion Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

A man and a cart move towards each other. The man weighs 64 kg and the cart 32 kg. The velocity of the man is 5.4 km/hr and that of the cart is 1.8km/hr. When the man approaches the cart, he jumps on to it. The velocity of the cart carrying the man will be:

A
30 km/hr

B
3 km/hr

C
1.8 km/ hr

D
zero

Solution

Using conservation of linear momentum
$$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$$
Thus we get
$$64\times 1.5-32\times 0.5=96\times v$$
$$\Rightarrow v=\dfrac{80}{96}m/s$$
$$=3km/hr$$ in the direction of man.
Question 2
1 / -0

The net force on a rocket with a weight of $$1.5\times 10^{4}N$$ is $$2.4 \times 10^{4}N$$. About how much time is needed to increase the rocket's speed from $$12 m/s$$ to $$36 m/s$$ near the surface of the Earth at take off? (Take $$g=10m/s^2$$)

A
$$15 s$$

B
$$0.78 s$$

C
$$1.5 s$$

D
$$3.8 s$$

Solution

Weight of the rocket $$=1.5\times10^4 N$$
Weight = mass $$\times$$ gravitational acceleration
$$\therefore$$ Mass of the rocket $$= \dfrac{Weight}{gravitational \ acceleration}=\dfrac{1.5\times10^4}{10}=1.5\times10^3 \ kg$$

According to Newton's second law of motion,
$$F_{net}=mass \times acceleration$$
Acceleration of the rocket $$=\dfrac{Net \ force}{mass}=\dfrac{2.4\times10^4}{1.5\times10^3}=16\ ms^{-2}$$

Given:
$$v=36m/s$$
$$u=12m/s$$
$$a=16 m/s^2$$

Using the equation of motion: $$v=u+at$$
$$t=\dfrac{36-12}{16}=1.5\ sec$$
Question 3
1 / -0

A $$60 \ kg$$ boy lying on a surface of negligible friction throws a stone of mass $$1 \ kg$$ horizontally with a speed of $$12 \ m/s$$ away from him. With what kinetic energy does he move back due to result of this action?

A
$$2.4$$ $$J$$

B
$$7.2$$ $$J$$

C
$$1.2$$ $$J$$

D
$$36$$ $$J$$

Solution

The given surface is frictionless so no external force is acting on the system. Hence linear momentum of the system will remain conserved.

$$\Rightarrow $$ Initial momentum $$= $$ Final momentum
$$\Rightarrow 0 = 60\times V + 1\times 12 $$
$$ \Rightarrow v = -\dfrac{1}{5}m/s $$ (-ve sign show man moves in opposite direction of stone)

Kinetic energy of the boy after separation $$(K.E.) = \dfrac{1}{2} mv^{2}$$
$$K.E. = \dfrac{1}{2}\times 60\times \dfrac{1}{25}$$
$$K.E.= 1.2J$$
Question 4
1 / -0

When a gun of mass $$M$$ fires a bullet of mass $$m$$, the total energy released in the explosion is found to be $$E$$. The kinetic energy of the bullet is:

A
$$\dfrac{EM}{(M+m)}$$

B
$$\dfrac{Em}{(M+m)}$$

C
$$\dfrac{E(M+m)}{M}$$

D
$$\dfrac{E(M+m)}{m}$$

Solution

Using momentum conservation on bullet + gun as a system

$$\Rightarrow$$ initial momentum$$=$$ Final momentum

$$0=-MV+mv$$

$$\Rightarrow mv=MV$$..........(1)

Now total energy $$=E$$

$$\Rightarrow \dfrac{1}{2}MV^{2}+\dfrac{1}{2}mv^{2}=E$$

From (1)

$$V=\dfrac{mv}{M}$$

$$\Rightarrow\dfrac{1}{2}M\dfrac{m^{2}}{M^2}v^{2}+\dfrac{1}{2}mv^{2}=E$$

$$\dfrac{1}{2}mv^{2}\left ( \dfrac{m}{M}+1 \right )=E$$

$$\dfrac{1}{2}mv^{2}=\dfrac{EM}{M+m}$$
$$\downarrow $$
K.E bullet
Question 5
1 / -0

A man stands on the surface of the earth. The number of action reaction pairs that can be seen here is/are

A
1

B
2

C
4

D
5
Solution
The action-reaction pair seen is:
the normal reaction force by the surface of the earth on the man and the normal reaction by the man on the surface of the earth
also, earth applies gravitational force on the man and man applies gravitational force on the earth
So two action-reaction pairs can be seen.
Question 6
1 / -0

A horizontal force $$F $$ produces an acceleration of $$6 \ m/ s^{2}$$ on a block resting on a smooth horizontal surface. The same force produces an acceleration of $$3\ m/ s^{2}$$ on a second block resting on a smooth horizontal surface. If the two blocks are tied together and the same force acts, the acceleration produced will be:

A
$$9\ m/ s^{2}$$

B
$$2\ m/ s^{2}$$

C
$$4\ m/ s^{2}$$

D
$$\dfrac{1}{2}\ m/ s^{2}$$

Solution

Here, $$F=m_1a_1$$ and $$F=m_2a_2$$.
Now if we tie both of them together then $$F=ma$$ and $$m=m_1 +m_2$$

$$\displaystyle \frac{F}{a}=\frac{F}{a_{1}}+\frac{F}{a_{2}}$$

If the same force acts, the acceleration produced will be, $$a=\dfrac{(a_1a_2)}{(a_1+a_2)}=2\ m/s^2$$.
Question 7
1 / -0

A trolley of mass $$60 kg$$ moves on a smooth horizontal surface and has kinetic energy $$120 J$$.A mass of $$40 kg$$ is lowered vertically on to the trolley. The total kinetic energy of the system after lowering the mass is:

A
$$60 J$$

B
$$72 J$$

C
$$120 J$$

D
$$144 J$$

Solution

KE of trolley $$=120=0.5mV^2$$
$$\therefore 120=0.5\times 60\times V^2$$
$$V=2\ m/s$$

Using momentum conservation:
$$60\times 2=100\times v$$
$$1.2\ m/s=v$$

$$(K.E)total=\dfrac{1}{2}\times 100\times (1.2)^{2}$$ $$=72J$$
Question 8
1 / -0

A boy weighing 50 kg throws a stone of mass 10 kg horizontally with a velocity of $$8 ms^{-1}$$. With what velocity does he move after throwing?

A
$$ -1.6 ms^{-1}$$

B
$$-16 ms^{-1}$$

C
$$ -160 ms^{-1}$$

D
$$-1.9 ms^{-1}$$

Solution

Let mass of boy: $$m_1=50kg$$
mass of stone: $$m_2=10 kg$$
Velocity of boy after throwing stone:$$v_1$$
Velocity of stone:$$v_2 = 8m/s$$
Using conservation of momentum :
$$m_1v_1+m_2v_2=0$$ (initial momentum of the system = 0)
$$50 \times v_1 + 10 \times 8 = 0$$
$$v_1 = \dfrac{-80}{50}$$
$$v_1 = -1.6m/s$$
The boy moves with a speed of $$1.6 m/s$$ in the opposite direction of the stone.
Question 9
1 / -0

A gun fires a bullet of mass 50 g with a velocity of mass $$130$$ ms$$^{-1}$$. Because of this, the gun is pushed back with a velocity of $$1$$ ms$$^{-1}$$. The mass of the gun is :

A
15 kg

B
30 kg

C
6.5 kg

D
20 kg

Solution

Mass of the gun = $$m_1=?$$
Mass of the bullet $$m_2=50\ g= 0.050\ kg$$
Final velocity of the gun $$v_1=-1\ m/s$$ (The minus sign indicates that the gun is pushed back)
Final velocity of the bullet $$v_2=130\ m/s$$

Let the system consists of the gun and the bullet. Now the initial momentum is zero.
By using momentum conservation, we get,
$$m_1v_1 + m_2 v_2=0$$
But $$v_1=-1$$ ms$$^{-1},v_2=130$$ms$$^{-1}, m_2=\dfrac{50}{1000}$$kg
$$\Rightarrow m_1(-1)+\dfrac{50}{1000}\times 130= 0$$
$$ \Rightarrow m_1 = 6.5 $$kg
Question 10
1 / -0

A bullet of mass 20 gm is fired from a rifle of mass 20 kg with a muzzle velocity of $$200\ ms^{-1}$$. Find the velocity of recoil of the rifle.

A
$$0.4ms^{-1}$$

B
$$0.5ms^{-1}$$

C
$$0.2ms^{-1}$$

D
$$0.3ms^{-1}$$

Solution

By the law of conservation of momentum,
Initial momentum $$ = $$ find momentum
$$ \therefore P_{1} = P_{2} $$
$$ \therefore m_{1}v_{1} = m_{2}v_{2} $$
here, $$ m_{1} = 20\,gm. v_{1} = 200\,m\,s^{-1},m_{2} = 20kg = 20,000\,gm $$

$$ \therefore v_{2} = \dfrac{m_{1}v_{1}}{m_{2}} $$

$$ = \dfrac{20\times 200}{20,000} $$

$$ \therefore \boxed{v_{2} = 0.2}m/s $$

$$ \therefore $$ option (c) is correct.

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Re-Attempt Weekly Quiz Competition

Forces and Laws of Motion Test - 38

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Forces and Laws of Motion Test - 38

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SHARING IS CARING

Question 1

30 km/hr

3 km/hr

1.8 km/ hr

zero

Solution

Question 2

$$15 s$$

$$0.78 s$$

$$1.5 s$$

$$3.8 s$$

Solution

Question 3

$$2.4$$ $$J$$

$$7.2$$ $$J$$

$$1.2$$ $$J$$

$$36$$ $$J$$

Solution

Question 4

$$\dfrac{EM}{(M+m)}$$

$$\dfrac{Em}{(M+m)}$$

$$\dfrac{E(M+m)}{M}$$

$$\dfrac{E(M+m)}{m}$$

Solution

Question 5

1

2

4

5

Solution

Question 6

$$9\ m/ s^{2}$$

$$2\ m/ s^{2}$$

$$4\ m/ s^{2}$$

$$\dfrac{1}{2}\ m/ s^{2}$$

Solution

Question 7

$$60 J$$

$$72 J$$

$$120 J$$

$$144 J$$

Solution

Question 8

$$ -1.6 ms^{-1}$$

$$-16 ms^{-1}$$

$$ -160 ms^{-1}$$

$$-1.9 ms^{-1}$$

Solution

Question 9

15 kg

30 kg

6.5 kg

20 kg

Solution

Question 10

$$0.4ms^{-1}$$

$$0.5ms^{-1}$$

$$0.2ms^{-1}$$

$$0.3ms^{-1}$$

Solution

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