Motion Test

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Motion Test - 8

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Question 1
1 / -0

Velocity of a body is defined as

A
rate of change of its position

B
change of its position only

C
rate of change of its direction

D
rate of change of its position in a definite direction

Solution

Velocity of a body is defined as the rate of change of its position in a definite direction. As velocity is a vector quantity, so its magnitude and direction both will be considered.
Question 2
1 / -0

A man drives 15 km to north and then 20 km to the east. His displacement from the starting point is

A
25 km

B
20 km

C
15 km

D
30 km

Solution

Let displacement = d
Therefore, by Pythagoras theorem,
d² = (15)² + (20)²
d = 25 km
Question 3
1 / -0

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

A
x

B
x²

C
ln x

D
e^x

Solution
Question 4
1 / -0

Two trains take 3 seconds to pass one another when going in opposite directions, but only 2.5 seconds if speed of one is increased by 50%. The time that one train would take to pass the other when going in the same direction at the original speed is

A
10 sec

B
12 sec

C
15 sec

D
18 sec

Solution

Let the speed of 1^st train be v.
Let the speed of 2^nd train be u.
Let their combined length (distance) be d.
When going in opposite directions with original speeds,
=> (v+u) = d/3
=> d= 3v + 3u .......... Eq (1)
When speed of one train is increased by 50 %, v = 1.5v
(1.5v+u)=d/2.5
d = 3.75v + 2.5 u............Eq (2)
Eq (2) - Eq (1)
0.75v - 0.5u = 0
1.5v = u
Time = Distance/Speed
When going in same direction, time= d/(u - v)
Since d = 3 v + 3 u = 3v + 3 x 1.5 v= 7.5 v,
u - v= 0.5v
Time = 7.5v/0.5v = 15 seconds
Question 5
1 / -0

Two bodies move in a straight line towards each other at initial velocities v₁and v₂ and with constant acceleration a₁and a₂directed against the corresponding velocities at the initial instant. The maximum initial separation l_maxbetween the bodies for which they will meet during the motion is

A

B

C

D

Solution

Relative velocity,
Relative acceleration,

Distance travelled by both bodies,
Question 6
1 / -0

A projectile weighing 50 kg is fired from a gun weighing 200 kg with a velocity of 72 km/hour. What is the velocity of recoil?

A
80 m/s

B
15 m/s

C
18 km/h

D
36 km/h

Solution

By conservation of linear momentum,
M₁V₁ = M₂V₂
Here, M₁ = mass of gun
M₂ = mass of projectile
V₁ = velocity of recoil of gun
V₂ = velocity of projectile
200 × V₁ = 50 ×
200 × V₁ = 20 × 50 V₁ = = 5 m/s
Or km/h = 18 km/h
Question 7
1 / -0

A distance travelled by a freely falling body is proportional to

A
mass of the body

B
time of fall

C
square of the time of fall

D
square of the acceleration due to gravity

Solution

The distance travelled by a freely falling body is proportional to square of the time of fall.
This can be seen from the relation,
h =
Here, h is the height from which body is falling, g is acceleration due to gravity and t is the time of fall.
Question 8
1 / -0

Water drops fall at regular intervals from a roof. At an instant, when a drop is about to leave the roof, the distances of separation between 3 successive pairs of drops (including this drop) are in the ratio

A
1 : 2 : 3

B
1 : 4 : 9

C
1 : 3 : 5

D
1 : 5 : 13

Solution

Distance travelled in the n^th second,

u = 0

The drops are moving under acceleration due to gravity.

As the time interval is regular, we can assume it to be 1 sec.
Hence, ratio of the separation between the three drops is:
Question 9
1 / -0

A particle, tossed up vertically, reaches its highest point in time t₁ and returns to the ground in a further time t₂. The air resistance exerts a constant force on the particle opposite to its direction of motion. Which of the following relations holds true?

A
t₁> t₂

B
t₁ = t₂

C
t₁ < t₂

D
Either (1) or (3), depending on the ratio of the force of air resistance to the weight of the particle.

Solution

Let us consider the particle tossed up, as shown in the figure below:

Case-1: When the particle is tossed up in a vertical direction:

Let the acceleration due to air friction be a.
As the air friction is in opposite direction particle movement, both gravity and acceleration due to air friction work downwards.
Therefore, the total acceleration will be (g + a) in downward direction.

And,

Case-2: When the particle moves in a downward direction:

As the air friction is in opposite direction of particle movement, acceleration due to air friction works upwards.
Therefore, the total acceleration will be (g - a) in downward direction.

Now, equation (2)/(3):

Hence, from equation (4), it is clear that:
t₁ < t₂
Question 10
1 / -0

With what speed a body should be thrown upwards so that the distances traversed in the 5^th second and 6^th second are equal?

A
58.4 m/s

B
49 m/s

C
98 m/s

D
m/s

Solution

This can be achieved only if the body moves up till 5^th second and moves down 6^th second. Since it is a free fall case, so a = g = 9.8ms^-2
Let initial speed be = u m/s
S_nth=
S_5th= = u + 44.1
At this point, the body will move down. So, velocity at the highest point is zero.
So, initial velocity u in this case is 0.
So, distance covered in 6^th second is,
S_6th= = 53.9
Since, S_5th= S_6th
So, u + 44.1= 53.9
Hence, u = 98 ms^-1
Question 11
1 / -0

How long will it take to stop a car travelling at a speed of 20 m/sec if the uniform acceleration during applying brakes is - 5 m/s²?

A
100 s

B
4 s

C
(1/4) s

D
(1/100) s

Solution

a = - 5 m/s²
u = 20 m/s
Final velocity, v = 0
Using v = u + at, we get
0 = 20 - 5 (t)
t = 4 seconds
Question 12
1 / -0

A body starting from rest, undergoes a uniform acceleration and covers a distance of 10 m in the fifth second. The acceleration of the body is

A
2 ms/²

B
0.2 m/s²

C
2.2 m/s²

D
4 m/s²

Solution

Distance travelled by the body in the n^t^h second is given by:

u = 0, n = 5 and S_n = 10
Question 13
1 / -0

A grasshopper finds that he can jump a maximum horizontal distance of 0.8 m. If he spends a negligible time on the ground, with what speed can he travel along the road?

A
2 m/s

B
2.8 m/s

C
104 m/s

D
1 m/s

Solution

When a grasshopper jumps, he travels in a projectile motion.

Range of the jump, R = 0.8 m
For maximum range in projectile motion, the angle of projection,

The range of the projectile motion is given by,

The speed with which he can travel along the road,
Question 14
1 / -0

Boat A and boat B are moving at 40 km h^-1and 20 km h^-1 respectively. Which of the following options is not a possible value for their relative velocity?

A
40 kmh^-1

B
30 km h^-1

C
20 km h^-1

D
10 km h^-1

Solution

Relative velocities of two bodies range from v_a - v_b to v_a + v_b
Here, 10 km/h^-1 < v_a - v_b
Therefore, this cannot be the relative velocity.
Question 15
1 / -0

A body is thrown up in a lift, with a velocity u relative to the lift and the time of flight is found to be t. The acceleration with which the lift is moving up is

A

B

C

D

Solution

Suppose the lift is moving up with the acceleration of a.
Consider the lift as the frame of reference.
Hence, total acceleration of body will be (g + a) in a downward direction.
Displacement of the body w.r.t. lift is 0.
So, we have:
Question 16
1 / -0

A body is released from a great height and falls freely towards the earth. Exactly one second later, another body is released. What is the distance between the two bodies 2 seconds after the release of the second body?

A
4.9 m

B
9.8 m

C
24.5 m

D
50 m

Solution

Distance travelled by the first body in 3 seconds =
= =
Distance travelled by the second body in 2 seconds =
=
Distance of their separation = - = = 2.5g
= 2.5 9.8 m
= 24.5 m
Question 17
1 / -0

A body is dropped from a plane moving with constant horizontal velocity. The path of the body as seen by a person on the plane will be

A
a straight line

B
parabolic

C
hyperbolic

D
None of these

Solution

The plane is flying at a constant horizontal velocity.

From the plane, the body will look as if it is travelling in a straight line.
Question 18
1 / -0

The figure shows the velocity (v) of a particle plotted against time (t).

Which of the following statements is correct?

A
The particle does not change its direction of motion at any point.

B
The acceleration of the particle remains constant.

C
The displacement of the particle and the sum of areas of the triangles both are taken as positive.

D
The initial and final velocities of the particle are the same.

Solution

Let us consider the given graph of velocity (v) of a particle plotted against time (t):

As the graph of the velocity (v) of a particle plotted against time (t) is linear with constant slope, the slope of the v-t graph is the acceleration of the particle. If the slope of the v-t graph is constant, the acceleration of the particle is constant.
Question 19
1 / -0

A tennis ball is dropped so that it falls vertically to the floor and bounces back again. Taking velocity upwards as positive, which of the following graphs best represents the variation of its velocity v with time t?

A

B

C

D

Solution

Option (3) is the best possible representation of the velocity of the ball w.r.t. time. Initially, the magnitude of velocity is increasing, but in downward direction. So, the graph starts increasing and just after the collision it becomes positive and starts decreasing.
Question 20
1 / -0

A couple is said to be acting when

A
two equal forces act on a body in same direction.

B
two parallel forces act on a body in opposite direction.

C
two opposite forces act on a body

D
two equal and parallel forces act on a body in opposite direction.

Solution

A couple is said to be acting when two equal and parallel forces are acting on a body.

Forces F_A and F_B are acting on a bar and they are parallel.