Motion Test

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Motion Test - 41

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Question 1
1 / -0

A steel ball is bouncing up and down on a steel plate with a period of oscillation 1 second. If $$g=10 { m }{ { s }^{-2 } }$$, then it bounces up to a height of

A
$$5$$ $$m$$

B
$$10$$ $$m$$

C
$$2.5$$ $$m$$

D
$$1.25$$ $$m$$

Solution

The ball takes 1 second to bounce up and down through the same height; so, the time of fall is equal to time of rise i.e. $$\displaystyle\frac { 1 }{ 2 }$$ second.
Using second law of motion, $$s= ut + \cfrac{1}{2}at^2$$
Height, $$s=h, $$ initial velocity, $$ u=0$$
$$h=0 + \displaystyle\frac { 1 }{ 2 } g{ \left( \displaystyle\frac { 1 }{ 2 } \right) }^{ 2 }$$
$$h=\displaystyle\frac { 10 }{ 8 }$$
$$h=1.25 m$$
So, the ball bounces up to a height of $$1.25m$$.
Question 2
1 / -0

Two objects of mass ratio $$1 :4$$ are dropped from the same height. Then

A
both objects will have the same velocity

B
the velocity of the first object is twice that of the second one

C
the velocity of the $$2^{nd}$$ object is one fourth of that of the $$1^{st}$$ object

D
the velocity of the $$2^{nd}$$ object is $$4$$ times that of the $$1^{st}$$ one

Solution

Let the masses of the two objects be $$m_1$$ and $$m_2$$, $$u_1, u_2$$ be their initial velocities respectively. Let $$h$$ be the height from where the objects are dropped. Let $$v_1, v_2$$ be the their final velocities and $$t_1, t_2$$ be the time taken to strike the ground respectively.
$$u_1=u_2=0, h_1=h_2=h$$
$$h=\dfrac{1}{2}gt^2_1$$ .........$$(1)$$
$$h=\dfrac{1}{2}gt^2_2$$ .........$$(2)$$
$$\therefore t_1=t_2=t=\sqrt{\dfrac{2h}{g}}$$ ..........$$(3)$$
now $$v_1=gt_1 v_2=gt_2$$
From equation $$(3), v_1=v_2=gt$$
$$\therefore \dfrac{v_1}{v_2}=\frac{1}{1}$$
Question 3
1 / -0

A body is thrown vertically upwards with an initial velocity $$u$$. The expression for the maximum height attained by the body is

A
$$\displaystyle\frac{{u}}{2g}$$

B
$$\displaystyle\frac{{g}^{2}}{2u}$$

C
$$\displaystyle\frac{{u}^{2}}{g}$$

D
$$\displaystyle\frac{{u}^{2}}{2g}$$

Solution

From third equation of motion,
$$v^{2} = u^{2} - 2gS$$
where$$ \text{v = final velocity = 0 , u = initial velocity , S = distance}$$
$$ 0 = u^{2} - 2gS$$
$$u^{2} = 2gS$$
$$ S = \dfrac{u^{2}}{2g}$$
Question 4
1 / -0

Two stones are thrown from the top of a tower, one straight down with an initial speed $$u$$ and the second straight up with the same speed $$u$$. When the two stones hit the ground, they will have speeds in the ratio

A
$$2:3$$

B
$$2:1$$

C
$$1:2$$

D
$$1:1$$

Solution

In case of stone thrown downwards, initial velocity,$$ u = u$$
So, using equation $$ v^{2} - u^{2} = 2aS $$,
here, $$u = u, v = 0, a = a, S = Distance $$
So, final equation$$ = u^{2} = 2aS ...... (1)$$
Again, for stone thrown upwards, initial velocity,$$ u = -u$$
So, using equation $$ v^{2} - u^{2} = 2aS $$,
here, $$u = -u, v = 0, a = a, S = Distance = S $$
So, final equation$$ = u^{2} = 2aS .......(2)$$
Dividing $$(1)$$ by $$2,$$ we get ratio of $$1 : 1$$
In both the cases$$, ( u$$ is positive or negative$$) u^{2}$$ is positive.
Question 5
1 / -0

Statement 1:Two balls of different masses are thrown vertically upward with same speed.They will pass through their point of projection in downward direction with the same speed.
Statement 2:The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball

A
Statement -1 is false , Statement- 2 is true.

B
Statement-1 is true, Statement-2 is true , Statement-2 is a correct explanation for statement-1

C
Statement-1 is true , Statement-2 is true; Statement-2 is not a correct explanation for statement-1

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

Solution

Take upward direction as positive
Height is given by
$$ h=ut-\frac{1}{2}gt^2 $$
Velocity is given by
$$ v^2=u^2-2gh $$
These equation are independent of mass.
Question 6
1 / -0

A body dropped from a height $$h$$ with an initial speed zero, strikes the ground with a velocity $$3\ { km }/{ h }$$. Another body of same mass dropped from the same height $$h$$ with an initial speed $$u=4\ { km }/{ h}$$. Find the final velocity of second mass, with which it strikes the ground.

A
$$3\ { km }/{ h }$$

B
$$4\ { km }/{ h }$$

C
$$5\ { km }/{ h }$$

D
$$6\ { km }/{ h }$$
Question 7
1 / -0

Two balls of masses in ratio $$1 : 2$$ are dropped from the same height. Find the ratio between their velocities when they strike the ground:

A
$$1 : 2$$

B
$$2 : 1$$

C
$$1 : 1$$

D
$$1 : 4$$

Solution

We know from kinematics equations of motion $$v=\sqrt{2gh}$$.From this it is very clear that velocity of any body depends on only height it is falling but not the mass.Hence two bodies will have in the ratio of 1 is to 1.
Question 8
1 / -0

If a particle moves in a circle with constant speed, its velocity :

A
remains constant

B
changes in magnitude

C
changes direction

D
changes both in magnitude and directions

Solution

When a particle is under uniform circular motion, its direction of motion continuously changes, hence the magnitude of its velocity remains constant while its direction changes.
Question 9
1 / -0

A rubber ball is dropped from a height of $$5 m$$ on a planet where the acceleration due to gravity is same as that on the surface of the earth. On bouncing, it rises to a height of $$1.8 m$$. the ball loses its velocity by a factor of

A
$$\displaystyle\frac { 3 }{ 5 }$$

B
$$\displaystyle\frac { 9 }{ 25 }$$

C
$$\displaystyle\frac { 2 }{ 5 }$$

D
$$\displaystyle\frac { 16 }{ 25 }$$

Solution

Given,
Height from which ball is dropped $$s=5\ m$$
Height attend by ball after bouncing $$s'=1.8\ m$$
Gravitational acceleration $$g=9.8\ m/s^2$$
We know by third equation of kinetic motion,
$$v^{2} - u^{2} = 2gs$$
where $$v =$$ final velocity , $$u =$$ initial velocity ,$$S =$$ distance
Now, for downward motion, $$u = 0$$
$$v^{2} - 0 = 2 \times 9.8 \times 5$$
$$\Rightarrow v = \sqrt{98} = 9.9$$

Also for upward motion, $$v=0$$
$$0^{2} - u^{2} = 2 \times (-9.8) \times 1.8$$
$$\Rightarrow u = \sqrt{3528} = 5.94$$

Fractional loss $$ =\dfrac{9.9-5.94}{9.9} = 0.4 $$ or $$\dfrac25$$

Option C
Question 10
1 / -0

A truck of mass $$5\times 10^3\ kg$$ starting from rest travels a distance of $$0.5\ km$$ in $$10\ s$$ when a force is applied on it. Calculate the acceleration acquired by the truck

A
$$25ms^{-2}$$

B
$$1ms^{-2}$$

C
$$10cms^{-2}$$

D
$$10ms^{-2}$$

Solution

Here truck has initial velocity of 0 m/sec and travelled a distance of 0.5 Km which is also equal to 500 m in 10 seconds.
$$s = ut+0.5at^{2}$$
$$s =0\times 10 + 0.5\times a\times 10^{2}$$
$$500=0.5\times a\times 100$$
$$a=10m/s^2$$
Hence acceleration is $$10m/s^2$$.

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Re-Attempt Weekly Quiz Competition

Motion Test - 41

Result

Motion Test - 41

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SHARING IS CARING

Question 1

$$5$$ $$m$$

$$10$$ $$m$$

$$2.5$$ $$m$$

$$1.25$$ $$m$$

Solution

Question 2

both objects will have the same velocity

the velocity of the first object is twice that of the second one

the velocity of the $$2^{nd}$$ object is one fourth of that of the $$1^{st}$$ object

the velocity of the $$2^{nd}$$ object is $$4$$ times that of the $$1^{st}$$ one

Solution

Question 3

$$\displaystyle\frac{{u}}{2g}$$

$$\displaystyle\frac{{g}^{2}}{2u}$$

$$\displaystyle\frac{{u}^{2}}{g}$$

$$\displaystyle\frac{{u}^{2}}{2g}$$

Solution

Question 4

$$2:3$$

$$2:1$$

$$1:2$$

$$1:1$$

Solution

Question 5

Statement -1 is false , Statement- 2 is true.

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

Statement-1 is true , Statement-2 is true; Statement-2 is not a correct explanation for statement-1

Statement -1 is true, Statement-2 is false.

Solution

Question 6

$$3\ { km }/{ h }$$

$$4\ { km }/{ h }$$

$$5\ { km }/{ h }$$

$$6\ { km }/{ h }$$

Question 7

$$1 : 2$$

$$2 : 1$$

$$1 : 1$$

$$1 : 4$$

Solution

Question 8

remains constant

changes in magnitude

changes direction

changes both in magnitude and directions

Solution

Question 9

$$\displaystyle\frac { 3 }{ 5 }$$

$$\displaystyle\frac { 9 }{ 25 }$$

$$\displaystyle\frac { 2 }{ 5 }$$

$$\displaystyle\frac { 16 }{ 25 }$$

Solution

Question 10

$$25ms^{-2}$$

$$1ms^{-2}$$

$$10cms^{-2}$$

$$10ms^{-2}$$

Solution

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