Motion Test

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Motion Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

A stone is dropped from the top of a tower of height $$49$$ $$m$$. Another stone is thrown up vertically with velocity of $$24.5\ m/s$$ from the foot of the tower at the same instant. They will meet in a time of

A
1 s

B
2 s

C
0.5 s

D
0.25 s
Question 2
1 / -0

Two balls A and B are thrown simultaneously. A vertically upwards at a speed of $$15$$ m/s from the ground and B, vertically downwards from a height of $$30$$ m at the same speed along the same line of motion. They meet after a time of:

A
1 s

B
2 s

C
3 s

D
4 s
Question 3
1 / -0

A body is projected vertically upwards with a velocity $$u$$. It crosses a point in its journey at a height $$h$$ meter twice, just after $$1$$ and $$7 $$ seconds .The value of $$u$$ in $$ms^{-1}$$ is $$ (g= 10ms^{-2})$$

A
$$50$$

B
$$40$$

C
$$30$$

D
$$20$$

Solution

$$h = ut - \dfrac {1}{2}gt^2$$
$$h = u(1) - \dfrac {1}{2}g = u - \dfrac {g}{2} \ \ \ ..... (1)$$
$$h = u(7) - \dfrac {1}{2}g (7)^2 = 7 u - \dfrac {49g}{2} \ \ \ ..... (2)$$
Equating (1) & (2)
$$u - \dfrac {g}{2} = 7u - \dfrac {49g}{2}$$
$$24g = 6u$$
$$u = 4g$$
$$\Rightarrow u = 40\ m/s$$
Question 4
1 / -0

The displacement ($$x$$) versus time ($$t$$) curve for two particles is as shown in figure. Which of the following statements is correct for time interval between $$0$$ to $$10$$ s?

A
The particle $$A$$ is speeding up while $$B$$ is slowing down.

B
Both the particles are initially speeding up and then slowing down.

C
Both the particles are initially slowing down and then speeding up.

D
Particle $$A$$ is speeding up first and then slowing down while $$B$$ is slowing down first and then speeding up.

Solution

Since both A & B have slope increasing initially & decreasing afterwards to zero thus option (B) is correct
Question 5
1 / -0

A body moving with uniform acceleration in a straight line is at points A, B, C, D after successive equal intervals of time. The distance AD is equal to:

A
BC

B
2(BC)

C
3(BC)

D
4(BC)

Solution

Let the starting point is O. And the body reach,
A at time t,
B at time 2t.
C at time 3t,
D at time 4t.
Using $$s=ut+\dfrac{1}{2}at^2$$
O to A is $$ut+\dfrac{t^2}{2}$$
O to B is $$ut+\dfrac{4t^2}{2}$$
O to C is $$ut+\dfrac{9t^2}{2}$$
O to D is $$ut+\dfrac{16t^2}{2}$$
So AD is (OD-OA)= $$\dfrac{16t^2}{2}+ut$$$$-\dfrac{t^2}{2}-ut$$$$=\dfrac{15t^2}{2}$$
And BC is (OC-OB) = $$\dfrac{9t^2}{2}+ut$$$$-\dfrac{4t^2}{2}-ut$$$$=\dfrac{5t^2}{2}$$
Therefore, AD=3BC
Hence option C is correct.
Question 6
1 / -0

A ball is released from the top of a tower of height $$h\ m$$. It takes $$T$$ seconds to reach the ground. What is the position of the ball in $$\dfrac{T}{3}$$ second?

A
$$\dfrac{h}{9}$$ metres from the ground

B
$$\dfrac{7h}{9}$$ metres from the ground

C
$$\dfrac{8h}{9}$$ metres from the ground

D
$$\dfrac{17h}{18}$$ metres from the ground

Solution

Apply second equTION OF MOTION,
$$h = ut + \dfrac{1}{2} gt^{2}$$
$$h$$: height of the tower
$$u = 0$$; Initial velocity
$$h = \dfrac{1}{2} gT^{2}$$

In $$\dfrac{T}{3}\ s$$,
$$h_{1} = \dfrac{1}{2} g(\dfrac{T}{3})^{2}$$

$$\Rightarrow h_1 = \dfrac{1}{9}(\dfrac{1}{2} gT^{2})$$

$$\Rightarrow h_{1} = \dfrac{h}{9}$$ from top.
Height from ground-
$$h_g=h- h_{1} = h-\dfrac{h}{9}$$
$$\Rightarrow h_g = \dfrac{8h}{9}$$ from ground
Question 7
1 / -0

The displacement-time graph of motion of a particle is shown in the figure. The ratio of the magnitudes of the speeds during the first two seconds and the next four seconds is:

A
$$1:1$$

B
$$1:2$$

C
$$2:1$$

D
$$1:\sqrt{2}$$

Solution

Slope of displacement-time curve gives the speed.
For first 2 seconds, speed $$v= \cfrac {40}{2} =20\ m/s$$
For the last 4 seconds, speed $$v= \cfrac {40}{4} =10\ m/s$$
Hence, ratio $$= 20/10=2:1$$
Question 8
1 / -0

A ball is thrown vertically upwards with a speed of $$10\ ms^{-1}$$ from the ground at the bottom of a tower $$200$$ $$m$$ high. Another is dropped vertically downward simultaneously, from the top of a tower. If $$g=10\ ms^{-2}$$ the time interval after which the projected body will be at the same level as the dropped body is:

A
$$20\ s$$

B
$$25\ s$$

C
$$2\sqrt{10}\;s$$

D
$$5\ s$$
Question 9
1 / -0

A particle starts with a velocity $$200\ cm/s$$ and moves in a straight line with a retardation of $$10\ cm/s^{2}$$. Its displacement will be $$1500\ cm$$:

A
Only once at $$30\ s$$ from start

B
Only once at $$10\ s$$

C
Twice at $$10\ s$$ and $$30\ s$$

D
Always

Solution

Given,
Initial velocity, $$u = 200\ cm/s$$
Acceleration, $$a = - 10\ cm/s^{2}$$
Dispplacement, $$s = 1500\ cm$$
Using Newton's second equation of motion,
$$s=ut+\dfrac{1}{2}at^2$$

$$1500 = 200t - \dfrac 12 \times 10 t^{2}$$

$$1500 = 200t - 5t^{2}$$

$$300 = 40t - t^{2}$$

$$t^{2}-40t +300 =0$$

$$t^{2}-10t-30t +300 =0$$

$$t(t-10)-30(t -10) =0$$

$$(t-10)(t-30)=0$$

$$\implies t =10\ s\ \text{and}\ 30\ s$$
Question 10
1 / -0

A ball is released from the top of a tower of height $$h$$ m. It takes $$T$$ s to reach the ground. What is the position of the ball in $$\dfrac{T}{3}$$ s?

A
$$\dfrac{h}{9}$$ m from the ground

B
$$\dfrac{7h}{9}$$ m from the ground

C
$$\dfrac{8h}{9}$$ m from the ground

D
$$\dfrac{17h}{18}$$ m from the ground

Solution

Lets consider the equation $$S=ut+\dfrac{1}{2}at^2$$
Here $$u=0 , a=g $$
So $$S=\dfrac{1}{2}gt^2$$
Now $$h=\dfrac{1}{2}gT^2$$
The distance traveled by ball in $$\dfrac{T}{3}$$ s will be $$S=\dfrac{1}{2}g \times {(\dfrac{T}{3})}^2=\dfrac{h}{9}$$
So height of ball above ground = (Total Height) $$-$$ (Distance Traveled in $$\dfrac{T}{3}$$ s)
$$=h-\dfrac{h}{9}=\dfrac{8h}{9}$$ m above ground

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Motion Test - 32

Result

Motion Test - 32

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SHARING IS CARING

Question 1

1 s

2 s

0.5 s

0.25 s

Question 2

1 s

2 s

3 s

4 s

Question 3

$$50$$

$$40$$

$$30$$

$$20$$

Solution

Question 4

The particle $$A$$ is speeding up while $$B$$ is slowing down.

Both the particles are initially speeding up and then slowing down.

Both the particles are initially slowing down and then speeding up.

Particle $$A$$ is speeding up first and then slowing down while $$B$$ is slowing down first and then speeding up.

Solution

Question 5

BC

2(BC)

3(BC)

4(BC)

Solution

Question 6

$$\dfrac{h}{9}$$ metres from the ground

$$\dfrac{7h}{9}$$ metres from the ground

$$\dfrac{8h}{9}$$ metres from the ground

$$\dfrac{17h}{18}$$ metres from the ground

Solution

Question 7

$$1:1$$

$$1:2$$

$$2:1$$

$$1:\sqrt{2}$$

Solution

Question 8

$$20\ s$$

$$25\ s$$

$$2\sqrt{10}\;s$$

$$5\ s$$

Question 9

Only once at $$30\ s$$ from start

Only once at $$10\ s$$

Twice at $$10\ s$$ and $$30\ s$$

Always

Solution

Question 10

$$\dfrac{h}{9}$$ m from the ground

$$\dfrac{7h}{9}$$ m from the ground

$$\dfrac{8h}{9}$$ m from the ground

$$\dfrac{17h}{18}$$ m from the ground

Solution

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