Force Test

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Force Test - 9

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Question 1
1 / -0

When a constant force is applied to a body, it moves with uniform

A
acceleration

B
velocity

C
speed

D
momentum

Solution

When a constant force is applied, the body moves with a constant acceleration.
F = ma
If F is constant, a is constant.
Question 2
1 / -0

When a force of constant magnitude always acts perpendicular to the motion of a particle, then

A
velocity is constant

B
acceleration is constant

C
K.E. is constant

D
none of these

Solution

When a force of constant magnitude acts perpendicular to the motion of a particle, it only changes the direction of the particle and hence the speed remains constant, but there is change in the velocity.
Hence, K.E. (Kinetic Energy) remains constant as it is independent of the direction.
Question 3
1 / -0

A body of mass 2 kg moving on a horizontal surface with an initial velocity of 4 m/s comes to rest after 2 seconds. If one wants to keep this body moving on the same surface with a velocity of 4 m/s, the force required is

A
8 N

B
4 N

C
Zero

D
2 N

Solution

Initial velocity = 4 m/s
Mass = 2 kg
Time after it rests = 2 s
Final velocity = 0
v = u + at
0 = 4 + a (2) a = – 2 m/s²
Hence, an opposing force of F = 2(–2) = – 4 N is acting on the body.
Therefore, to maintain its speed at 4 m/s, a force of 4 N is required.
Question 4
1 / -0

A vehicle is moving on a track as shown in figure. The weight of the vehicle is

A
maximum at A

B
maximum at B

C
maximum at C

D
same at A, B and C

Solution

The weight of the vehicle would be maximum at B as can be seen in the following figure:

The weight is Normal force between the block and earth
At B
N - mg =mv²/R
N= mg + mv²/R

At C
mg - N = mv²/ R
N = mg - mv²/ R
Question 5
1 / -0

A surface is hit elastically and normally by n balls per unit time. All the balls have the same mass m and moving with the same velocity u. The force on the surface is

A
mnu²

B
2mnu

C
(1/2) mnu²

D
2mnu²

Solution

Using the relation for one ball,
F. t =
F = m(v₁ - v₂) (as t = 1 s)
Here, v₁ = U
V₂ = - U (as balls hit elastically, therefore, there is no loss of energy)
F = m (20)
For n balls, F = 2mnu
Question 6
1 / -0

A rocket is ejecting 1 kg of gases per sec at a velocity of 60 km/s. The accelerating force on the rocket is

A
60 N

B
600 N

C
6000 N

D
60000 N

Solution

According to Newton`s third law of motion, every action has an equal and opposite reaction.
Expelled force due to gas = 60 10³ m/s 1 kg/s
= 6 10⁴ N
The same force acts on the rocket = 6 10⁴ N
Question 7
1 / -0

A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 kg/s and at a speed of 5 m/s. The initial acceleration of the block will be

A
2.5 m/s²

B
5 m/s²

C
10 m/s²

D
5 m/s

Solution

Force F on jet = ma
Here, mass/unit time = 1 kg/ s
V = 5 m/s
F = 5 1 kg m/s²
F = 5 N
Acceleration of block = = 2.5 m/s²
Question 8
1 / -0

A ball weighing 10 g hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for (0.01) sec. The average force exerted by the surface on the ball is

A
100 N

B
10 N

C
1 N

D
0.1 N

Solution

The relation

where, v is the final velocity, u is the initial velocity and m is mass of the body
Here, v = 5 and u = - 5
F =
F = 10 N
Question 9
1 / -0

A projectile is moving at 60 m/s at its highest point, where it breaks into two equal parts due to an internal explosion. One part moves vertically up at 50 m/s with respect to the ground. The other part will move at

A
110 m/s

B
120 m/s

C
130 m/s

D
10 m/s

Solution

The projectile undergoes the following case:

Let mass of each part be `m`
Let mass of projectile be M
According to the conservation of linear momentum along horizontal and vertical axis,
M (60) = mv cos M(60) = V cos = 120
and m 50 = m V sin
50 = V sin â¦â¦ (1)
120 = V cos â¦â¦ (2)
Squaring and adding, we get V² = 16900
V = 130 m/s
Question 10
1 / -0

A light particle moving horizontally with a speed of 12 m/s strikes a very heavy block moving in the same direction at 10 m/s. The collision is one-dimensional and elastic. After the collision, the particle will

A
move at 2 m/s in its original direction

B
move at 8 m/s in its original direction

C
move at 8 m/s opposite to its original direction

D
move at 12 m/s opposite to its original direction

Solution

Initial speed of the block, u₁ = 12 m
Initial speed of the block, u₂ = 10 m/s
Let the speed of the particle after strike be v₁.
As the block is very heavy, the speed of the bock will not change after strike.
Hence, speed of the block after strike, v₂ = u₂ = 10 m/s

As the strike is elastic,

Hence, after the collision, the particle will move at 8 m/s in its original direction.
Question 11
1 / -0

A bullet of mass `a` and velocity `b` is fired into a large block of wood of mass `c`. The final velocity of the system is

A

B

C

D

Solution

Mass of bullet = a
Velocity = b
Mass of block of wood = c
Initial momentum = final momentum
m₁v₁ = m₂v₂
ab = (a + c) v₂
Question 12
1 / -0

Two skaters A and B of masses 50 kg and 70 kg, respectively stand facing each other 6 m apart. Then, they pull on a rope stretched between them. How far has each moved when they meet?

A
Both have moved 3 m

B
A moves 2.5 m and B 2.5 m

C
A moves 3.5 m and B 2.5 m

D
A moves 2 m and B 4 m

Solution

They will meet in such a way that product of their mass with the distance travelled will be the same for both.
M_Ad_A = M_B d_B
Let d_A = a
Then, d_B = 6 - a
50 (a) = 70 (6 - a)
a = 3.5 m
Or, d - a = 2.5 m
Hence, a travels 3.5 m and B travels 2.5 m.
Question 13
1 / -0

Two blocks of masses 2 kg and 1 kg are in contact with each other on a frictionless table. When a horizontal force of 3.0 N is applied to the block of mass 2 kg, the value of the force of contact between the two blocks is

A
4 N

B
3 N

C
2 N

D
1 N

Solution

Now, force F = 3 N is acting on the block of 2 kg
Now, 2 kg and 1 kg are in contact, so they move together.
Therefore, common acceleration a =
Force on block of mass 1 kg = m a
= 1 kg 1 m/s²
= 1 N
Question 14
1 / -0

A string of length L and mass M is lying on a horizontal table. A force F is applied at one of its ends. Tension in the string at a distance x from the end at which force is applied is

A
Zero

B
F

C
F(L - x)/ L

D
F(L - x)/ M

Solution

Let the tension be T
Mass is M
Length is L

F = Ma a =
Then, mass balance on part at a distance x
=
Mass after a distance x
T =
=
T =
Question 15
1 / -0

Two bodies of masses 5 kg and 4 kg are arranged in two positions as shown in figures (1) and (2). If the pulleys and the table are perfectly smooth and pulleys are massless, what will be the acceleration of the 5 kg body in both cases (1) and (2)?

A
g and (5/9)g

B
(4/9)g and (1/9)g

C
g/5 and g/5

D
(5/9)g and (1/9)g

Solution

Case (1):

Let the acceleration be a₁.

For the above system,

Case (2):

Let the acceleration be a₂.
For the above system,
Question 16
1 / -0

A spring obeys Hooke's law. When loaded with 12 g, its extension is 2 cm. Which of the following will produce a 3 cm extension? (g = 10 m/s²)

A
18 g mass

B
8 g mass

C
a force of 1.8 N

D
a force of 0.12 N

Solution

F -
= k
600 N/m² = k
F =
= = kg
= 270 g = = 18 g
Question 17
1 / -0

A block A of mass 2 kg rests on another block B of mass 8 kg, which rests on a horizontal floor. The coefficient of friction between A and B is 0.2, while that between B and floor is 0.5. When a horizontal force of 25 N is applied on the block B, the force of friction between A and B is

A
zero

B
3.9 N

C
5.0 N

D
49 N

Solution
Question 18
1 / -0

A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is

A
2.5 N

B
0.98 N

C
4.9 N

D
0.49 N

Solution

f_s = mg

f_s = f_n
f_s = mg = 0.1 9.8
= 0.98
f_n =
= 1.96 N
Question 19
1 / -0

A cart of mass M has a block of mass m attached to it as shown in the figure. The coefficient of friction between the block and the cart is . The minimum acceleration of the cart so that the block m does not fall is

A

B

C

D

Solution

Let the minimum acceleration of the cart so that the block m does not fall be a.
Using cart as a frame of reference,

The block is at rest with respect to the cart.
So,
μma = mg
Question 20
1 / -0

A boy of mass 40 kg is hanging from the horizontal branch of a tree. The tension in his arms is the least, when the angle between the arms is

A
0°

B
90°

C
120°

D
180°

Solution

The tension will be least when,
2 T cos = mg
T will be minimum when is 0 or cos is maximum.
Hence, T =