Forces and Laws of Motion Test

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Forces and Laws of Motion Test - 44

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Question 1
1 / -0

Student A and student B sit in identical office chairs facing each other, as shown in figure. Student A is heavier than student B. Student A suddenly pushes student B with his feet. Suppose $$v_A$$ and $$v_B$$ are the velocities of student A and student B respectively, then

A
$$v_A > v_B, v_A$$ is leftward and $$v_B$$ is rightward

B
$$v_A < v_B, v_A$$ is leftward and $$v_B$$ is rightward

C
$$v_A = v_B, v_A$$ is leftward and $$v_B$$ is rightward

D
$$v_A > v_B, v_A$$ and $$v_B$$ both are rightward
Question 2
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 $$0.3\:m/s$$. After collision the two balls come to rest, then the velocity of $$B$$ is-

A
$$0.15\:m/s$$

B
$$1.5\:m/s$$

C
$$-0.15\:m/s$$

D
None of the above

Solution

Given, mass of ball A $$m_A = 200 \ g = 0.2 \ kg$$
mass of ball B $$m_B = 400 \ g = 0.4 \ kg$$

initial velocity of ball A $$u_A = 0.3 \ m/s$$
initial velocity of ball B $$u_B = ?$$

final velocity of ball A $$v_A = 0 \ m/s$$
final velocity of ball B $$v_B = 0\ m/s$$

By law of conservation of momentum
$$m_Au_A+m_Bu_B = m_Av_A+m_Bv_B$$

$$0.2\times 0.3+0.4\:u_B=0$$ or $$u_B=-0.15\:m/s$$
Question 3
1 / -0

The 35 kilogram girl is standing on a 20 kilogram wagon and jumps off, giving the wagon a kick that sends it off at 3.8 meters per second. How fast is the girl moving?

A
1.2 m/s

B
3 m/s

C
4 m/s

D
2.2 m/s

Solution

Answer is D.

The total momentum is initially zero, and must remain zero. Therefore, the momentum acquired by the wagon in one direction equals the momentum acquired by the girl in the other case.
That is, $${ (mv) }_{ girl }={ (mv) }_{ wagon }$$.
$${ 35\quad kg\quad \times \quad (v) }_{ girl }={ \quad 20\quad kg\quad \times \quad }3.8\quad m/s$$.
Therefore, $${ (v) }_{ girl }={ \quad 2.2\quad m/s }$$.
Hence, the girl is moving at a velocity of 2.2 m/s.
Question 4
1 / -0

A space-craft of mass $$M$$ moves with a velocity $$V$$ and suddenly explodes into two pieces. If a part of mass $$m$$ becomes stationary, then the velocity of another part will be :

A
$$\dfrac{MV}{(M - m)}$$

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

C
$$\dfrac{mV}{M - m)}$$

D
$$\dfrac{mV}{M + m)}$$

Solution

Given:
Mass of the spacecraft = $$M$$
Velocity of the spacecraft = $$V$$
Mass of one part = $$m$$, velocity of this part = $$0$$
Mass of the other part = $$M-m$$, velocity of this part = $$v=?$$

No external force is acting on the spacecraft in the explosion. Therefore, the total linear momentum must be conserved.

$$\underline{\text{Applying conservation of momentum:}}$$
$$\text{Total initial linear momentum of the spacecraft}=\text{Total final momentum of the two parts}$$
$$\Rightarrow MV=m\times0+(M-m)v$$
$$\Rightarrow v=\dfrac{MV}{M-m}$$

Hence, option A is correct.
Question 5
1 / -0

If a constant external force starts acting on a moving particle, then:

A
The line of motion of the particle will keep changing

B
The speed of the particle will keep changing

C
The particle will keep moving

D
All of the above are correct.

Solution

If a constant force is applied to a moving object then it will produce acceleration or deceleration.
It depends on the direction of force applied to the moving object If there is acceleration or deceleration then there will always be a change in the speed of the object.
Speed may decrease or increase, it depends on the direction of the force.
So, the correct answer is B. Speed of the particle will keep changing.
Question 6
1 / -0

A bullet of mass m = 50 gm strikes a bag of mass M = 5 kg hanging from a fixed point, with a horizontal velocity $$\bar{V}_p$$. If bullet sticks to the sand bag then just after collision the ratio of final & initial kinetic energy of the bullet is approximately :

A
$$10^{-2}$$

B
$$10^{-3}$$

C
$$10^{-6}$$

D
$$10^{-4}$$

Solution

By momentum conservation we have
$$mV_p=(m+M)V$$ ,m=0.05 kg
$$0.05\times V_p=(5+0.05)V\Rightarrow V_p=101V $$
initial KE of bullet=$$\dfrac{mV_p^2}{2}$$
Final KE=$$\dfrac{mV^2}{2}$$
So ratio =$$\dfrac{mV^2}{mV_p^2}=10^{-4}$$
Question 7
1 / -0

Four forces are acting on a body. If the body doesn't change its position or shape, the forces :

A
Must be of equal magnitude

B
Must be parallel & opposite

C
Must add up to zero when taken as vectors

D
Must be in a single line

Solution

Answer is C.
Force is needed for a body to change its position and/or shape. Since the there is no change in the position or shape of the body, the net force acting on it should be zero.
Question 8
1 / -0

A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist the downward acceleration if the shoots 40 bullets a second at the speed of 500 m/s. If the mass of a bullet is 49 gm, what is the mass of the man with the gun ? Ignore resistance due to air and assume the acceleration due to gravity g = 9.8 $$m/s^2$$.

A
60 kg

B
75 kg

C
100 kg

D
125 kg

Solution

We know
$$F=\dfrac{\Delta P}{\Delta t}$$
$$\Rightarrow \Delta P=F. \Delta t $$ . . . (i)
where $$\Delta P$$ is change in momentum and $$\Delta t$$ is the time interval
Let the mass of man with gun be $$M$$.
The force acting on the man alongwith the gun is $$F=Mg$$ downwards
Now this force is balanced by the change in momentum in a given time say $$\Delta t= 1 \ sec$$
The mass of one bullet $$m=49 \ gm=0.049 \ kg$$
Velocity of each bullet $$v= 500 \ m/s$$
The momentum change due to one bullet $$\Delta p=p_f-p_i=mv-0=mv$$
If $$n$$ bullets are fired per sec,
then the total momentum change $$\Delta P= P_f-P_i= (n\times mv) -(n \times 0)=nmv $$
Given $$n=40$$
We consider a time interval of $$\Delta t=1 \ sec$$
So using (i),
$$ \Delta P=F. \Delta t $$
$$nmv=Mg \times (1)$$
$$M \times 9.8 = 40 \times 0.049 \times 500$$
$$M= \dfrac{40\times49\times 500}{9.8\times 1000}$$
$$M = 100 \ kg$$
So the mass of the man along with the gun is 100 kg.
Question 9
1 / -0

A massive ball moving with speed v collides head-on with a tiny ball at rest having a very small mass as compared to the first ball. If the collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to

A
v

B
2v

C
v/2

D
$$\infty$$

Solution

$$m_1>>m_2$$ ; velocity df massive ball$$=$$v
In a head-on elastic collision where the projectile is much more massive than the target, the velocity of the target particle after the collision will be about twice that of the projectile and the projectile velocity will be essentially unchanged. So, the second ball will move with a speed approximately equal to 2v.
Question 10
1 / -0

A body $$A$$ of mass $$150 \ kg$$, travelling at $$20 \ m/s$$ collides with a body $$B$$ of mass $$100 \ kg$$ travelling at $$10 \ m/s$$ in the same direction. After collision the velocity of $$A$$ becomes $$15 \ m/s$$. What is the magnitude of velocity of $$B$$ after the collision?

A
$$22.5 \ m/s$$

B
$$17.5 \ m/s$$

C
$$5 \ m/s$$

D
$$12.5 \ m/s$$

Solution

Given : $$m_A =150$$ kg $$u_A = 20$$ m/s   $$m_B =100$$ kg $$u_B = 10$$ m/s $$v_A =15$$ m/s

Let the speed of B after the collision be $$v_B$$.

Using conservation of linear momentum before and after the collision :
$$P_i = P_f$$

$$\therefore$$ $$m_A u_A + m_B u_B = m_Av_A + m_B v_A$$

OR $$150 \times 20 + 100\times 10 = 150\times 15 + 100v_B$$

   $$\implies v_B = 17.5$$ $$m/s$$

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Forces and Laws of Motion Test - 44

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Forces and Laws of Motion Test - 44

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Question 1

$$v_A > v_B, v_A$$ is leftward and $$v_B$$ is rightward

$$v_A < v_B, v_A$$ is leftward and $$v_B$$ is rightward

$$v_A = v_B, v_A$$ is leftward and $$v_B$$ is rightward

$$v_A > v_B, v_A$$ and $$v_B$$ both are rightward

Question 2

$$0.15\:m/s$$

$$1.5\:m/s$$

$$-0.15\:m/s$$

None of the above

Solution

Question 3

1.2 m/s

3 m/s

4 m/s

2.2 m/s

Solution

Question 4

$$\dfrac{MV}{(M - m)}$$

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

$$\dfrac{mV}{M - m)}$$

$$\dfrac{mV}{M + m)}$$

Solution

Question 5

The line of motion of the particle will keep changing

The speed of the particle will keep changing

The particle will keep moving

All of the above are correct.

Solution

Question 6

$$10^{-2}$$

$$10^{-3}$$

$$10^{-6}$$

$$10^{-4}$$

Solution

Question 7

Must be of equal magnitude

Must be parallel & opposite

Must add up to zero when taken as vectors

Must be in a single line

Solution

Question 8

60 kg

75 kg

100 kg

125 kg

Solution

Question 9

v

2v

v/2

$$\infty$$

Solution

Question 10

$$22.5 \ m/s$$

$$17.5 \ m/s$$

$$5 \ m/s$$

$$12.5 \ m/s$$

Solution

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