Motion Test

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Motion Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Directions For Questions

The graph given shows the position of two cars A and B, as a function of time. The cars move along the x-axis on parallel but separate tracks, so that they can pass each others position without colliding.

...view full instructions

Which one of the following best describes the motion of car A as shown on the graphs?

A
Speeding up

B
Constant velocity

C
Slowing down

D
First speeding up, then slowing down

Solution

Velocity is given by the slope of this graph. Since the graph of car A is having an increasing slope first and then decreasing slope throughout the time, thus car A is first speeding up and then slowing down.
Whereas car B is continuously speeding throughout the time.
Question 2
1 / -0

A body moving in circular motion with constant speed has :

A
constant velocity

B
constant acceleration

C
constant kinetic energy

D
constant displacement

Solution

A body moving along a circular path moves such that it traverses equal arc-length in equal intervals of time. Displacement, velocity and acceleration changes at each point on the trajectory.Hence, the speed is uniform and consequently the kinetic energy is also constant throughout.
Question 3
1 / -0

When the speed of a car is $$v$$, the minimum distance over which it can be stopped is $$s$$. If the speed becomes $$nv$$, what will be the minimum distance over which it can be stopped during the same time?

A
$$s/n$$

B
$$ns$$

C
$$s/ { n }^{ 2 }$$

D
$${ n }^{ 2 }s$$

Solution

$$0-{ v }^{ 2 }=2as\quad \Rightarrow \quad a=\dfrac { -{ v }^{ 2 } }{ 2s } \\ 0-{ (nv) }^{ 2 }=2ad\quad \Rightarrow \quad -{ (nv) }^{ 2 }=2\times \dfrac { -{ v }^{ 2 } }{ 2s } \times d\\ \Rightarrow d={ n }^{ 2 }s$$
Question 4
1 / -0

The muzzle velocity of a certain rifle is $$330\ m{ s }^{ -1 }$$. At the end of one second, a bullet fired straight up into the air will travel a distance of

A
$$(330-4.9)\ m$$

B
$$330\ m$$

C
$$(330+4.9)\ m$$

D
$$(330-9.8)\ m$$

Solution

Given:
The initial velocity $$u=330\ m/s$$
Time taken $$t=1\ sec$$
Acceleration = $$-g$$ (As the bullet is going up)

Using the equation of motion $$s=ut-\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$\therefore s=330(1)-\dfrac { 1 }{ 2 } 9.8{ (1) }^{ 2 }=(330-4.9)m$$
Question 5
1 / -0

Two particles are projected vertically upwards with the same velocity on two different planets with accelerations due to gravities $${ g }_{ 1 }$$ and $${ g }_{ 2 }$$ respectively. If they fall back to their initial points of projection after lapse of times $${ t }_{ 1 }$$ and $${ t }_{ 2 }$$, respectively, then

A
$${ t }_{ 1 }{ t }_{ 2 }={ g }_{ 1 }{ g }_{ 2}$$

B
$${ t }_{ 1 }{ g }_{ 1 }={ t }_{ 2 }{ g }_{ 2 }$$

C
$${ t }_{ 1 }{ g }_{ 2 }={ t }_{ 2 }{ g }_{ 1}$$

D
$${ { t }_{ 1 } }^{ 2 }+{ { t }_{ 2 } }^{ 2 }={ g }_{ 1 }+{ g }_{ 2 }$$

Solution

For both particles, net vertical displacement is zero.
Particle 1: $$ y=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\quad \Rightarrow 0=u{ t }_{ 1 }-\dfrac { 1 }{ 2 } { g }_{ 1 }{ t }_{ 1 }^{ 2 }\quad or\quad u=\dfrac { 1 }{ 2 } { g }_{ 1 }{ t }_{ 1 }$$
Particle 2: $$ 0=u{ t }_{ 2 }-\dfrac { 1 }{ 2 } { g }_{ 2 }{ t }_{ 2 }^{ 2 }\quad or\quad u={ \dfrac { 1 }{ 2 } { g }_{ 2 } }{ t }_{ 2 }$$
thus $$ u={ \dfrac { 1 }{ 2 } g }_{ 1 }{ t }_{ 1 }={ \dfrac { 1 }{ 2 } { g }_{ 2 } }{ t }_{ 2 }\quad or\quad { g }_{ 1 }{ t }_{ 1 }={ { g }_{ 2 } }{ t }_{ 2 }$$
Question 6
1 / -0

The displacement of a particle as a function of time is shown in figure. The figure indicates that

A
the particle starts with a certain velocity, but the motion is retarded and finally the particle stops.

B
the velocity of the particle is constant throughout.

C
the acceleration of the particle is positive throughout.

D
the particle starts with a constant velocity, the motion is accelerated and finally the particle moves with another constant velocity.

Solution

Initially the slope is decreasing and the slope $$=$$ 0.
Thus velocity is decreasing as time increases, i.e. there is retardation and ultimately the particle stops. It must have obviously started with some velocity.
Question 7
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The given figure shows the displacement-time curve of two particles $$P$$ and $$Q$$. Which of the following statements is correct?

A
Both $$P$$ and $$Q$$ move with uniform equal speed

B
$$P$$ is accelerated and $$Q$$ is retarded

C
Both $$P$$ and $$Q$$ move with uniform speeds but the speed of $$P$$ is more than the speed of $$Q$$

D
Both $$P$$ and $$Q$$ move with uniform speeds but the speed of $$Q$$ is more than the speed of $$P$$

Solution

As the $$x - t$$ graph is a straight line, in either case, the velocity of both the particles is uniform. As the slope of the $$x - t$$ graph for P is greater, therefore, the velocity of P is greater than that of Q. (Slope of $$x-t$$ graph represents velocity)
Question 8
1 / -0

A bullet, moving with a velocity of $$200cm/s$$ penetrates a wooden block and comes to rest after traversing $$4cm$$ inside it. What velocity is needed for traversing a distance of $$6cm$$ in the same block

A
$$104.3cm/s$$

B
$$136.2cm/s$$

C
$$244.9cm/s$$

D
$$272.7cm/s$$

Solution

$${ v }^{ 2 }-{ u }^{ 2 }=2as\\ 0-{ (200) }^{ 2 }=2a\times 4\\ \Rightarrow a=5000\quad cm/{ s }^{ 2 }$$
Now initial velocity to travel 6 cm,
$$0-{ u }^{ 2 }=-2\times 5000\times 6\\ \Rightarrow u=244.9\quad cm/{ s }$$
Question 9
1 / -0

Directions For Questions

Suppose that you were called upon to give some advice to a lawyer concerning the physics involved in one of his cases. The policeman has charged a driver for breaking the speed limit of $$60 km/hr$$ and had arrested him. The length of the skid marks is $$15\ m$$ when he made a reasonable assumption that the maximum deceleration of the car could not exceed $$g$$.

...view full instructions

Is the policeman right in arresting the driver to break speed limit?

A
Yes

B
No

C
Insufficient data to reply

D
The policeman demanded bribe

Solution

Max deceleration
$$a = -g$$
$$\Rightarrow a = -9.8ms^{-2}$$
Distance,
$$S= 15\ m$$
Using third equation of motion:
$$2as= v^2 - u^2$$
(v=0, final speed as vehicle has stopped)
$$\Rightarrow 2 \times (-9.8) \times 15 = -u^2 $$
$$\Rightarrow u = 17.1464 ms^{-1}$$
$$\Rightarrow u = \dfrac{18}{5} \times 17.1464\ kmph$$
$$\Rightarrow u = 61.7271\ kmph$$

So yes the policeman was right in arresting the driver to break the speed limit.
Question 10
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Two balls $$A$$ and $$B$$ are simultaneously thrown. $$A$$ is thrown from the ground level with a velocity of $$20\ m{ s }^{ -1 }$$ in the upward direction and $$B$$ is thrown from a height of $$40\ m$$ in the downward direction with the same velocity. Where will the two balls meet?

A
$$15\ m$$

B
$$25\ m$$

C
$$35\ m$$

D
$$45\ m$$

Solution

Let both the ball meet after time$$ t$$ and $$ x$$ be the distance covered by ball $$A$$.
So$$,\ x=20t-\dfrac { 1 }{ 2 } \left( 10 \right) { t }^{ 2 }...\left( 1 \right) $$
For ball $$B,\quad \left( 40-x \right) =20t+\dfrac { 1 }{ 2 } \left( 10 \right) { t }^{ 2 }...\left( 2 \right) $$
Adding $$ (1)\ \&\ (2) ,\quad 40=40t\quad or\quad t=1\ s$$
so $$x=20-5=15\ m$$

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Re-Attempt Weekly Quiz Competition

Motion Test - 39

Result

Motion Test - 39

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Rank

-

SHARING IS CARING

Question 1

Speeding up

Constant velocity

Slowing down

First speeding up, then slowing down

Solution

Question 2

constant velocity

constant acceleration

constant kinetic energy

constant displacement

Solution

Question 3

$$s/n$$

$$ns$$

$$s/ { n }^{ 2 }$$

$${ n }^{ 2 }s$$

Solution

Question 4

$$(330-4.9)\ m$$

$$330\ m$$

$$(330+4.9)\ m$$

$$(330-9.8)\ m$$

Solution

Question 5

$${ t }_{ 1 }{ t }_{ 2 }={ g }_{ 1 }{ g }_{ 2}$$

$${ t }_{ 1 }{ g }_{ 1 }={ t }_{ 2 }{ g }_{ 2 }$$

$${ t }_{ 1 }{ g }_{ 2 }={ t }_{ 2 }{ g }_{ 1}$$

$${ { t }_{ 1 } }^{ 2 }+{ { t }_{ 2 } }^{ 2 }={ g }_{ 1 }+{ g }_{ 2 }$$

Solution

Question 6

the particle starts with a certain velocity, but the motion is retarded and finally the particle stops.

the velocity of the particle is constant throughout.

the acceleration of the particle is positive throughout.

the particle starts with a constant velocity, the motion is accelerated and finally the particle moves with another constant velocity.

Solution

Question 7

Both $$P$$ and $$Q$$ move with uniform equal speed

$$P$$ is accelerated and $$Q$$ is retarded

Both $$P$$ and $$Q$$ move with uniform speeds but the speed of $$P$$ is more than the speed of $$Q$$

Both $$P$$ and $$Q$$ move with uniform speeds but the speed of $$Q$$ is more than the speed of $$P$$

Solution

Question 8

$$104.3cm/s$$

$$136.2cm/s$$

$$244.9cm/s$$

$$272.7cm/s$$

Solution

Question 9

Yes

No

Insufficient data to reply

The policeman demanded bribe

Solution

Question 10

$$15\ m$$

$$25\ m$$

$$35\ m$$

$$45\ m$$

Solution

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