Forces and Laws of Motion Test

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Forces and Laws of Motion Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

Two solid rubber balls A and B having masses 200 g and 400 g respectively are moving in opposite directions with velocity of A equal to 0.3 ms$$^{-1}$$. After collision the two balls come to rest then velocity of B is :

A
-0.15 ms$$^{-1}$$

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

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

D
none of these

Solution

Here, two solid rubber balls A and B having masses 200 g and 400 g respectively are moving in opposite directions with velocity of A equal to $$0.3 ms^{-1}$$ and after collision the two balls come to rest.
Final velocities $$v_A$$ and $$v_B$$ are zero because both come to rest.
Initial momentum=Final momentum
$$m_A u_A + m_B u_B = m_A\times0 +m_B\times0$$
$$m_A u_A + m_B u_B = 0$$
$$m_B u_B = -m_A u_A $$
$$u_B = -\dfrac{m_A u_A}{m_B}$$
$$u_B = -\dfrac{200 \times 0.3}{400} = -0.15 ms^{-1}$$
Question 2
1 / -0

The speed-times relation of a car whose weight is $$1500kg$$ as shown in the given graph. How much breaking force has been applied at the end of $$7$$sec, to stop the car in $$2$$ sec?

A
$$2000N$$

B
$$1200N$$

C
$$4800N$$

D
$$8400N$$

E
$$9000N$$

Solution

Speed - time graph is given,
Slope of the speed-time graph gives an acceleration of the moving object.

In between 7th and 9th second,
Slope of line $$ =\dfrac{0-12}{9-7}=-6m/s^{2} $$
Slope of line is negative which show deacceleration of object. Hence When driver will apply break, car will deaccelerate with $$-6m/s^{2}$$.

Breaking force, $$F=ma=1500\times 6=9000\ N$$
Question 3
1 / -0

Two bodies of masses $$m$$ and $$3m$$, moving with velocities $$3v$$ and $$v$$ respectively along same direction, collide with each other. After collision they stick together and move with a velocity $$V$$ in the same direction. Then:

A
$$\displaystyle V= v$$

B
$$\displaystyle V=\frac {3}{2}v$$

C
$$\displaystyle V=2v$$

D
$$\displaystyle V = \frac {4}{3}v$$

Solution

Here, the collision is inelastic. And law of conservation of momentum for inelastic collision is given by,

$$m_{1}u_{1} + m_{2}u_{2} = \left ( m_{1} + m_{2} \right ) V$$

$$\Rightarrow V = \dfrac{m_{1}u_{1} + m_{2}u_{2}}{m_{1} + m_{2}}$$

$$\Rightarrow V = \dfrac{m \times 3v + 3m \times v}{m + 3m}$$

$$\Rightarrow V = \dfrac{3mv + 3mv}{4m}$$

$$\Rightarrow V = \dfrac{6mv}{4m} = \dfrac{3}{2}v$$
Question 4
1 / -0

A $$20 g$$ bullet passes through a plate of mass $$1 kg$$ and finally comes to rest inside another plate of mass $$2980 g$$. It makes the plates move from rest to same velocity. The percentage loss in velocity of bullet between the plate is:

A
$$0$$

B
$$50$$%

C
$$75$$%

D
$$25$$%

Solution

Using conservation of momentum
$$0.02 u = 0.02 v + 1 V_1$$ (i)
(where $$V_1$$ be the velocity of plate of $$1 kg$$)
and $$0.02 v = (2.98 + 0.02) V_1$$ (ii)
($$\because $$ plate of $$3 kg$$ has also same velocity i.e. $$V_1$$)
or $$V_1 = 0.02 \dfrac{v}{3}$$
Substituting
$$\displaystyle \therefore 0.02 u = 0.02v + 0.02 \dfrac{v}{3}$$
or $$\displaystyle u = \dfrac{4v}{3}$$ or $$\displaystyle v = \dfrac{3}{4}u$$
percentage loss in velocity $$= \displaystyle |\dfrac{u-v}{u}| \times 100$$
$$= \displaystyle |\dfrac{u - \dfrac{3}{4} u}{u}| \times 100 = 25$$%
Question 5
1 / -0

A heavy uniform bar is being carried by two men on their shoulders. The weight of the bar is $$w$$. What is the work done by gravity?

A
$$w/4$$

B
$$w/2$$

C
$$w$$

D
$$0$$

Solution

The weight of the bar is acting downward and the men are moving forward.
Therefore, the force and the displacement are perpendicular to each other.
So, the force is not doing any work.
Question 6
1 / -0

Statement 1: When a man is standing on a stationary horizontal surface, the normal force exerted by the ground on him is equal and opposite to his weight.
Statement 2: According to Newton's $$3^{ rd }$$ law , every action has an equal and opposite reaction.

A
Statement 1 - is true, statement- 2 is true and statement-2 is correct explanation for statement-1

B
Statement 1 - is true, statement- 2 is true and statement-2 is NOT the correct explanation for statement-1

C
Statement 1 - is true, statement- 2 is false

D
Statement 1 - is false, statement- 2 is true

Solution

The gravitational force and the normal force both act on the man and in Newton's third law the force pair of action-reaction is acting on different bodies. However, statement one is true, as the forces acting on the man balances itself. However, the reason is not the correct explanation.
Ans: (B)
Question 7
1 / -0

A bullet of mass $$20g$$ and moving with a velocity of $$200m/s$$ strikes a heap of sand and comes to rest after penetrating $$3cm$$ inside it. The force exerted by the sand on the bullet will be:

A
$$11.2\times { 10 }^{ 8 }dyne$$

B
$$15.7\times { 10 }^{ 8 }dyne$$

C
$$13.3\times { 10 }^{ 8 }dyne$$

D
$$18.6\times{ 10 }^{ 8 }dyne$$
Solution
Given:

mass of the bullet (m) = 0.02 kg
initial velocity of the bullet (u) = 200 m/s
final velocity of the bullet (v) = 0 m/s
distance traversed by the bullet in the sand before coming to rest (S) = 0.03 m

From the equation of motion,

$$v^2 – u^2 = 2 a S$$

$$\Rightarrow0 - 200^2 = 2 a S$$

$$\Rightarrow a = - \dfrac{2}{3}\times 10^6$$ (Here a is the deceleration)

Hence the force exerted by the sand on the bullet is

$$F = m (-a)$$

$$\Rightarrow F = 0.02 \times \dfrac{2}{3} \times 10^6$$

$$\Rightarrow F = 13.3 \times 10^3 N$$

Note: $$1 N = 10^5 dyne$$

$$\Rightarrow F = 13.3 \times 10^8 dyne$$

Hence option C is correct.
Question 8
1 / -0

A rocket is moving at a constant speed in space by burning its fuel and ejecting out the burnt gases through a nozzle. Is there any force acting on the rocket? If yes, how much?

A
Yes, equal to rate of change in momentum

B
No

C
Yes, equal to rate of change in velocity

D
Yes, equal to change in mass

Solution

As it is moving upward by ejecting gases its mass is getting decreased gradually. As momentum is the product of the mass and velocity that is p=mv. Hence momentum decreases as mass is decreasing and nothing with speed as it is constant. As we know force is defined as rate of change of momentum. Here momentum is changing hence a force will be acting on rocket.
Question 9
1 / -0

A force of $$5N$$ acts horizontally on a body of weight $$9.8N$$. What is the acceleration produced (in $${ ms }^{ -2 }$$)?

A
$$0.51$$

B
$$1.46$$

C
$$49.00$$

D
$$5.00$$

Solution

Suppose the mass of the body is $$m$$.
Weight of body $$= mg$$
Here,
$$mg=9.8$$
$$ \Rightarrow m=1 kg $$
So for acceleration,
$$a=\dfrac{F}{m}$$
$$a=\dfrac{5}{1}$$
$$\Rightarrow a =5 ms^{-2}$$
Question 10
1 / -0

A body acted upon by a force $$F$$ has an acceleration $$A$$. When it is acted upon by two forces each of magnitude $$F$$ perpendicular to each other, its acceleration will be:

A
$$2A$$

B
$$A$$

C
$$A \sqrt 2$$

D
$$\sqrt 2$$

Solution

As the acceleration of force $$F$$ is $$A$$,
so the resultant acceleration ,$$ A_R=\sqrt{A^2+A^2}=A\sqrt 2$$

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Re-Attempt Weekly Quiz Competition

Forces and Laws of Motion Test - 42

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Forces and Laws of Motion Test - 42

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Question 1

-0.15 ms$$^{-1}$$

0.15 ms$$^{-1}$$

1.5 ms$$^{-1}$$

none of these

Solution

Question 2

$$2000N$$

$$1200N$$

$$4800N$$

$$8400N$$

$$9000N$$

Solution

Question 3

$$\displaystyle V= v$$

$$\displaystyle V=\frac {3}{2}v$$

$$\displaystyle V=2v$$

$$\displaystyle V = \frac {4}{3}v$$

Solution

Question 4

$$0$$

$$50$$%

$$75$$%

$$25$$%

Solution

Question 5

$$w/4$$

$$w/2$$

$$w$$

$$0$$

Solution

Question 6

Statement 1 - is true, statement- 2 is true and statement-2 is correct explanation for statement-1

Statement 1 - is true, statement- 2 is true and statement-2 is NOT the correct explanation for statement-1

Statement 1 - is true, statement- 2 is false

Statement 1 - is false, statement- 2 is true

Solution

Question 7

$$11.2\times { 10 }^{ 8 }dyne$$

$$15.7\times { 10 }^{ 8 }dyne$$

$$13.3\times { 10 }^{ 8 }dyne$$

$$18.6\times{ 10 }^{ 8 }dyne$$

Solution

Question 8

Yes, equal to rate of change in momentum

No

Yes, equal to rate of change in velocity

Yes, equal to change in mass

Solution

Question 9

$$0.51$$

$$1.46$$

$$49.00$$

$$5.00$$

Solution

Question 10

$$2A$$

$$A$$

$$A \sqrt 2$$

$$\sqrt 2$$

Solution

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