Motion in A Straight Line Test

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Motion in A Straight Line Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

A car is moving with a uniform speed on a level road. Inside the car there is a balloon filled with helium and attached to a piece of string tied to the floor. The string is observed to be vertical. The car now takes a left turn maintaining the speed on the level road. The balloon in the car will

A
Continue to remain vertical

B
Burst while taking the curve

C
Be thrown to the right side

D
Be thrown to the left side

Solution

It happens due to moment of intertia. Air outside the balloon is heavier so it will have more tendency to move towards right and will keep the balloon towards left side (Here in this question car is supposed to be air tight).
Question 2
1 / -0

Rewrite the sentence using proper alternative.
The total energy of an falling freely toward the ground ____

A
decreases

B
remains unchanged

C
increases

D
increases in the beginning and then decreases

Solution

Here potential energy is converting into kinetic energy.
Hence, The total energy of an falling freely toward the ground remains unchanged .
Question 3
1 / -0

On another planet, a marble is released from rest at the top of a high cliff. It falls $$4.00 m$$ in the first $$1 s$$ of its motion. Through what additional distance does it fall in the next $$1 s$$?

A
$$4.00m$$

B
$$8.00m$$

C
$$12.0m$$

D
$$16.0m$$

E
$$20.0m$$

Solution

We take downward as the positive direction with $$y = 0$$ and $$t = 0$$ at the top of the cliff. The freely falling marble then has $$v_{0} = 0$$ and its displacement at $$t = 1.00 s$$ is $$Δy = 4.00 m$$. To find its acceleration, we use
$$y=y_{0}+v_{0}t+\frac{1}{2}at^{2}\rightarrow (y-y_{0})=\Delta y=\frac{1}{2}at^{2}\rightarrow a=\frac{2\Delta y}{t^{2}}$$
$$a=\frac{2(4.00m)}{(1.00s)^{2}}=8.00m/s^{2}$$
The displacement of the marble (from its initial position) at $$t = 2.00 s$$ is found from
$$\Delta y=\frac{1}{2}at^{2}$$
$$\Delta y=\frac{1}{2}(8.00m/s^{2})(2.00s)^{2}=16.0m$$
The distance the marble has fallen in the $$1.00 s$$ interval from $$t = 1.00 s$$ to $$t = 2.00 s$$ is then
$$∆y = 16.0 m − 4.0 m = 12.0 m$$.
and the answer is (c).
Question 4
1 / -0

An electron with a speed of $$3.00 \times 10^6 m/s$$ moves into
a uniform electric field of magnitude $$1.00 \times10^3 \,N/C$$. The field lines are parallel to the electrons velocity and pointing in the same direction as the velocity. How far does the electron travel before it is brought to rest?

A
$$2.56 \,cm$$

B
$$5.12 \,cm$$

C
$$11.2 \,cm$$

D
$$3.34 \,m$$

E
$$4.24 \,m$$

Solution

The electric force is opposite to the field direction, so it is opposite to the velocity of the electron. From Newton's second law, the acceleration the electron will be
$$a_x=\dfrac{F_x}{m}=\dfrac{qE_x}{m}=\dfrac{(-1.60\times 10^{-19}C)(1.00\times 10^3 \,N/C}{9.11\times10^{-31} Kg}$$
$$-1.76\times 10^{14} \,m/s^2$$
The kinematics equation $$v_x^2=v_{0x}^{2}+2a_x(\Delta x)$$, with $$v_x=0$$, gives the stopping distance as
$$\Delta x=\dfrac{-v_{0x}^2}{2a_x}=\dfrac{-(3.00\times 10^6 \,m/s^2)}{2(1.76\times 10^{14} \,m/s^2)}=2.56\times 10^{-2} \,m=2.56m$$
Question 5
1 / -0

A particle moves with a non zero initial velocity $${ v }_{ 0 }$$ and retardation $$kv$$, where $$v$$ is the velocity at any time $$t$$.

A
The particle will cover a total distance $$\cfrac {{ v }_{ 0 }}{k}$$

B
The particle comes to rest at $$t=\cfrac{1}{k}$$

C
Particle continues to move for long time

D
at time $$\cfrac{1}{\alpha}, v=\cfrac {{ v }_{ 0 }}{2}$$

Solution

$$u=v_0, $$ retardation $$a=kv$$ and $$v$$ is the velocity at time $$t$$.
using $$v=u-at,$$ we will get $$v=v_0-kvt \Rightarrow t=\dfrac{v_0}{kv}-\frac{1}{k}$$
if $$v=0, t =\infty$$. So the particle will be moved for a long time.
Question 6
1 / -0

A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at height $$\dfrac {h}{2}$$. If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be-

A
$$\sqrt {\dfrac {5gh}{4}}$$

B
$$\sqrt {gh}$$

C
$$\sqrt {\dfrac {gh}{4}}$$

D
None of these

Solution

m mass falls a distance $$\dfrac {h}{2}$$ in time, say t. Then,
$$\dfrac {h}{2}=\frac {1}{2}gt^2\therefore t=\sqrt {\dfrac {h}{g}}$$
For mass 2m,
$$\dfrac {h}{2}=vt-\dfrac {1}{2}gt^2=v\sqrt {\dfrac {h}{g}}-\dfrac {h}{2}$$
$$\therefore v=\sqrt {gh}$$
Now, let u is the velocity with which combined mass collides with ground, then
$$u=\sqrt {(\dfrac {2}{3}v)^2+2g(\dfrac {h}{3})}$$
$$=\sqrt {\dfrac {4}{9}gh+\dfrac {2}{3} gh}$$
$$=\dfrac {\sqrt {10}}{3}\sqrt {gh}$$

Since the collision takes place in vertical direction and external force is not zero in this direction, momentum conservation cannot be applied.
If the gravitational force( weight) is much less than the impulse force, it may be ignored and conservation of momentum can be applied.
Question 7
1 / -0

A particle is projected with a velocity $$u$$ in horizontal direction as shown in the figure. Find $$u$$(approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.

A
$$10 \ m/s$$

B
$$20 \ m/s$$

C
$$10\sqrt{2}\ m/s$$

D
None of these

Solution

Considering plane along Wedge axis,
when particle collides with the plane, its component of velocity parallel to inclined plane should be zero.
$$\Rightarrow u \cos\theta - g \sin\theta \times t =0$$ {Equation along the plane}
$$t=\displaystyle \dfrac { u }{ g \tan \theta } $$
$$u=\displaystyle gt\times \tan\theta = \dfrac{gt}{\sqrt{3}}$$
Now,
Consider the plane along original x-y axis,
Thus,
Horizontal displacement before it hits the plane $$=ut$$
vertical displacement during the time t $$=\dfrac{1}{2} g t^2 $$
Now , $$ \tan 30^\circ{} =\displaystyle \dfrac{ h- \frac{1}{2} g t^2 } { ut }$$
$$ \dfrac{1}{\sqrt{3}} = \displaystyle \dfrac{ h- \dfrac{1}{2} g t^2 } { ut }$$
Hence , $$ \dfrac{1}{3} = \displaystyle ( \dfrac{h} {gt^2} - \dfrac{1}{2} ) $$

$$t=\displaystyle \sqrt{ \dfrac{6h}{5g}} $$
Hence , $$t = 2.5938 $$ s
So,
$$u = gt \, \tan 30^\circ{} = \displaystyle \dfrac{9.8 \times 2.5938}{\sqrt 3} = 14.69 \simeq 10 \sqrt{2}$$ m/s
Question 8
1 / -0

Mass $$A$$ is released from rest at the top of a frictionless inclined plane $$18 m$$ long and reaches the bottom $$3 s$$ later. At the instant when $$A$$ is released, a second mass $$B$$ is projected upwards along the plate from the bottom with a certain initial velocity. Mass $$B$$ travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with $$A$$. The two masses do not collide. Initial velocity of $$B$$ is

A
$$4\:ms^{-1}$$

B
$$5\:ms^{-1}$$

C
$$6\:ms^{-1}$$

D
$$7\:ms^{-1}$$

Solution

Here, for $$A, 18=0 \times 3+\displaystyle \frac{1}{2}a \times 3^{2}$$ or $$a=4 ms^{-2}$$
For $$B$$, time taken to move up is given by , $$t_1=u/a$$ ($$\therefore$$ the relation $$v=u+at$$ here becomes $$0=u-at_{1}$$).
Distance moved up is given by the relation $$0=u^{2}-2a S$$
i.e $$S=u^{2}/2a$$

For coming down the inclined plane for $$S=\displaystyle \frac{1}{2}at_{2}^{2}$$

and $$t_{2}=\sqrt{\displaystyle \frac{2S}{a}}$$

or $$t_{2}=\sqrt{\displaystyle \frac{2u^{2}}{a.2a}} =\displaystyle \frac{u}{a}$$

But $$\displaystyle \frac{u}{a}=t_{1}$$, then

$$t_1+t_2=3$$ or $$\displaystyle \frac{2u}{a}=3$$ or

$$u=\displaystyle \frac{3a}{2}$$ or

$$u=\displaystyle \frac{3}{2} \times 4=6 \: ms^{-1}$$
Question 9
1 / -0

The displacement of a particle moving in a straight line is given by $$x=16t-2t^{2}$$ (where, $$x$$ is in meters and $$t$$ is in second). The distance traveled by the particle in $$8$$ seconds [starting from $$t$$ $$=$$ 0] is

A
$$24 m$$

B
$$40 m$$

C
$$64 m$$

D
$$80 m$$

Solution

Given:
$$x=16 t - 2t^{2}$$
Comparing with second equation of motion:
$$s=ut+\cfrac{1}{2}at^2$$
It can be concluded that:
$$u=16\ m/s,\quad a=4\ m/s^2$$

Now, using first equation of motion:
$$v = u+at$$
$$v=16 - 4t$$
For $$v = 0 $$ we get, $$t= 4 s$$
Here, direction of velocity will reverse. So total distance travelled will be the sum of displacement in first four seconds and displacement in next four seconds.

Now,
Displacement at
$$t=0,\quad x=0$$
$$t=4,\quad x=32$$
$$t=8,\quad x=0$$
$$\therefore$$ Distance $$=$$ (distance from $$x=0$$ to $$x=32$$) + (distance from $$x=32$$ to $$x=0$$)
$$\therefore$$ Distance $$= 32+32 =64\ m$$
Question 10
1 / -0

A water tap leaks such that water drops fall at regular intervals. Tap is fixed $$5\ m$$ above the ground. First drop reaches the ground and at that very instant third drop leaves the tap. At this instant the second drop is at a height of

A
$$3\ m$$

B
$$4.5\ m$$

C
$$3.75\ m$$

D
$$2.5\ m$$

Solution

Velocity of drop 5m below the tap can be obtained by energy conservation.
Assume that potential energy is zero 5m below the tap.
Now, $$v=\sqrt { 2gh } =10\quad m/s$$
Now, time taken to achieve this velocity$$=\dfrac { 10 }{ 10 } =1s$$
it is given that drops leave the tap at a regular time interval. Thus, second drop will leave at 0.5s as the third drop is leaving at 1s.
Now, total time of motion of second drop is 0.5s
Thus, distance moved by second drop in 0.5s is
$$s=\dfrac { 1 }{ 2 } \times 10\times { (0.5) }^{ 2 }=1.25$$
Thus height from ground level is $$5-1.25=3.75m$$

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Motion in A Straight Line Test - 62

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Motion in A Straight Line Test - 62

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Question 1

Continue to remain vertical

Burst while taking the curve

Be thrown to the right side

Be thrown to the left side

Solution

Question 2

decreases

remains unchanged

increases

increases in the beginning and then decreases

Solution

Question 3

$$4.00m$$

$$8.00m$$

$$12.0m$$

$$16.0m$$

$$20.0m$$

Solution

Question 4

$$2.56 \,cm$$

$$5.12 \,cm$$

$$11.2 \,cm$$

$$3.34 \,m$$

$$4.24 \,m$$

Solution

Question 5

The particle will cover a total distance $$\cfrac {{ v }_{ 0 }}{k}$$

The particle comes to rest at $$t=\cfrac{1}{k}$$

Particle continues to move for long time

at time $$\cfrac{1}{\alpha}, v=\cfrac {{ v }_{ 0 }}{2}$$

Solution

Question 6

$$\sqrt {\dfrac {5gh}{4}}$$

$$\sqrt {gh}$$

$$\sqrt {\dfrac {gh}{4}}$$

None of these

Solution

Question 7

$$10 \ m/s$$

$$20 \ m/s$$

$$10\sqrt{2}\ m/s$$

None of these

Solution

Question 8

$$4\:ms^{-1}$$

$$5\:ms^{-1}$$

$$6\:ms^{-1}$$

$$7\:ms^{-1}$$

Solution

Question 9

$$24 m$$

$$40 m$$

$$64 m$$

$$80 m$$

Solution

Question 10

$$3\ m$$

$$4.5\ m$$

$$3.75\ m$$

$$2.5\ m$$

Solution

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