Force Test

Result

Force Test - 7

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Question 1
1 / -0

A system consists of masses M and m (< M). The centre of mass of the system is

A
at the middle

B
nearer to M

C
nearer to m

D
at the position of large mass

Solution

∵ Distribution of mass is more towards M(m << M)
⇒ Centre of mass will be farther from m
⇒ r₁ > r₂
So, Centre of mass will be nearer to M.
Question 2
1 / -0

A body of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The force required to keep the body moving with the same velocity is

A
8 N

B
zero

C
2 ×4² N

D
N

Solution

When a body is moving with a constant velocity on a frictionless surface, acceleration is zero. Hence, no force is being applied on it.
This is because f = ma and if a is zero, then F is also zero.
Question 3
1 / -0

Drums of oil are carried in a truck. If a constant acceleration is applied on the truck, the surface of the oil in the drum will

A
remain unaffected

B
rise towards forward direction

C
rise towards backward direction

D
nothing is certain

Solution

When a constant acceleration is applied on the truck, the oil in the drums will rise in the backward direction because of inertia of rest. This principle states that a body has a tendency to not to leave its resting place. Hence, oil moves and rises backwards.
Question 4
1 / -0

A 20 kg monkey slides down a vertical rope with a constant acceleration of 7.0 ms^-2. If g = 10 Nkg^-1, the tension in the rope will be

A
60 N

B
70 N

C
140 N

D
200 N

Solution

T = tension in the string.
The weight of monkey acting downwards, W = mg = 20 ×10 = 200 N
The acceleration of monkey (a) = 7 m/s²
Net force on monkey = W - T
W - T = m a
200 - T = 20(7)
T = 200 - 140 = 60 N
Question 5
1 / -0

A parachutist of weight W strikes the ground with his legs fixed and comes to rest with an upward acceleration of 3g. The force exerted by him on the ground during landing is

A
W

B
2W

C
3W

D
4W

Solution

A parachutist of weight W strikes the ground with his legs fixed and comes to rest with an upward acceleration of 3g.
Let F is the upward force
ma = F - mg
m(3g) = F - mg
F = 3mg + mg = 4 mg
F = 4W
Question 6
1 / -0

A leather ball strikes a wall and falls down. A tennis ball having the same mass and same velocity strikes the same wall and bounces back. Which of the following statements is correct?

A
The tennis ball suffers a greater change in momentum.

B
The leather ball suffers a greater change in momentum.

C
Both the balls suffer equal change in momentum.

D
The momentum of leather ball is greater than that of tennis ball.

Solution

let m be mass of both the balls and initial velocity is v, final velocity of leather ball is zero while that of tennis ball is -v
Change in momentum of leather ball = mv - (0) = mv
Change in momentum of tennis ball = mv - (-mv) = 2mv
Hence the change of momentum of tennis ball is more.
Question 7
1 / -0

A cracker explodes into two fragments of exactly equal masses. The two fragments will move

A
with the same speed, but at right angle to each other

B
in the same direction with different speeds

C
in opposite directions, but with the same speed

D
in opposite directions with different speeds

Solution

When it explodes into 2 fragments of equal masses, the momentum has to be conserved. The initial momentum is zero. Hence, final momentum should also be zero. This can be explained as below:
Here, V₁ and V₂ are velocities of two fragments and M is the initial mass.
M (0) = mV₁ + mV₂
(From the law of conservation of momentum)
V₁ = – V₂
Question 8
1 / -0

A body of mass M collides against a wall with velocity V and rebounds with the same speed. Its change of momentum is

A
zero

B
2 MV

C
- 2 MV

D
- MV

Solution

Mass of body = M
Initial velocity = V
Initial momentum = MV
Final velocity = - V
Final momentum = - MV
Change = Final momentum - Initial momentum
Change = - MV - (MV)
= - 2MV
Question 9
1 / -0

A ball drops vertically onto a floor with momentum p and then bounces repeatedly. The coefficient of restitution is e. The total momentum imparted by the ball to the floor is

A
p (1 + e)

B

C
p

D
p

Solution

When a particle undergoes normal collision with floor , with coefficient of restitution e,.
If the particle strikes the floor with velocity v .The speed of particle after collision is ev ,
Taking upward direction as positive.
Hence change in momentum after first collision = emv - (-mv) = p(1 + e)
Change in momentum after second collision =e²p - (-ep ) = ep(1 + e)
Similarly change in momentum after third collision =e³p - (-e²p ) = e²p(1 + e)
Total change in momentum
Question 10
1 / -0

A bomb of mass 9 kg explodes into two pieces of masses 3 kg and 6 kg. The velocity of mass 3 kg is 16 m/s. The KE of mass 6 kg (in joule) is

A
96

B
384

C
192

D
768

Solution

Mass = M = 9 kg
Mass M₁ = 3 kg , Mass M₂ = 6 kg
Initial velocity = 0 , Final velocity of 3 kg = 16 m/s
Let velocity of other = x m/s
By the law of conservation of momentum,
(M) (0) = 3(16) + 6(x)
0 = 48 + 6x
x = - 8 m/s
KE =
= = 192 kg
Question 11
1 / -0

There are two blocks A and B of masses 2 kg and 3 kg, respectively. The ground is smooth. P is an external force of 10 N. The force exerted by B on A is

A
4 N

B
6 N

C
8 N

D
10 N

Solution

Force = P = 10 N
Common acceleration (a) = m/s² = 2 m/s²
Acceleration of B = 2 m/s²
Force on B = m_B(a)
= 3× 2 N = 6 N
Question 12
1 / -0

In the given arrangement, the pulleys are fixed and ideal, the strings are light, m₁ > m₂ and S is a spring balance which itself is massless. The reading of S (in units of mass) is

A
m₁ - m₂

B
(m₁ + m₂)

C

D

Solution

The pulleys are fixed and ideal, the strings are light, m₁ > m₂ and S is a spring balance which itself is massless.
The tension in the string is:

So, the reading of S (in units of mass) is .
Question 13
1 / -0

A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force F is applied at one end of the rope, the force which the rope exerts on the block is

A
Fm(M + m)

B
F

C
FM/(m + M)

D
zero

Solution

Let the common acceleration of the block and rope be 'a' and the force exerted by the rope on the block be 'f'.
We know that equal and opposite force will be exerted by the block on the rope.

So,

Hence,

Thus, the force exerted by the rope on the block is:
Question 14
1 / -0

A spring obeying Hooke's law has a force constant k. Now, if the spring is cut in two equal parts, the force constant of each part is

A
k

B
k/2

C
2k

D
zero

Solution

When a spring is broken into two equal parts, the spring constant of each part is twice the spring constant of the original spring.
Question 15
1 / -0

A block of mass 20 kg is suspended through two light spring balances as shown in the figure. What are the readings shown by both the scales?

A
Both the scales show 10 kg reading.

B
Both the scales show 20 kg reading.

C
The upper scale will read 20 kg, while the lower one will read zero.

D
Their readings are in between 0 and 20 kg with their sum being equal to 20 kg.

Solution

When two spring balances are attached to a single load, then the tension in the two springs is the same because they are connected through a single line. Therefore, the readings shown by both the spring balances would be 20 kg.
Question 16
1 / -0

A bicycle moves on a horizontal road with some acceleration. The forces of friction between the road and the front and rear wheels are F₁ and F₂, respectively. Which of the following relations holds true?

A
Both F₁ and F₂ act in the forward direction.

B
Both F₁ and F₂ act in the reverse direction.

C
F₁ acts in the forward direction, whereas F₂acts in the reverse direction.

D
F₂ acts in the forward direction, whereas F₁acts in the reverse direction.

Solution

When a bicycle moves forward, the torque is applied to the rear wheel of the bicycle by chain gear system. Because of this, the slipping tendency of the point of contact of the rear wheel is backwards. Hence, the force of friction which opposes the motion acts in the forward direction. The opposite is the case with the front wheel. It tends to move forward. Hence, the friction F₁ acts in the reverse direction.
Question 17
1 / -0

A body of mass 60 kg is dragged with just enough force to start moving on a rough surface with coefficients of static and kinetic frictions 0.5 and 0.4, respectively. On applying the same force, what is the acceleration?

A
0.98 m/s²

B
9.8 m/s²

C
0.54 m/s²

D
5.292 m/s²

Solution

Mass = 60 kg
μ_s= 0.5
μ_k = 0.4

When body is at rest the minimum force required if equal to static friction force
F = μ_smg = (0.5 × 60 × 9.8) N
Now the body is moving and friction force applied is kinetic friction

F - μ_kmg = ma
(0.5) (60) (9.8) - (0.4) (60) (9.8) = (60) (a)
6 × 9.8 = 60 (a)
a = 0.98 m/s²
Question 18
1 / -0

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab. The static coefficient of friction between the block and the slab is 0.60, while the kinetic coefficient is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s², the resulting acceleration of the slab will be

A
0.98 m/s²

B
1.47 m/s²

C
1.52 m/s²

D
6.1 m/s²

Solution

Force acting on the block is 100 N. It is much greater than μ_s mg (= 0.6 × 10 × 9.8 = 58.8).

Hence, μ_k will prevail a force balance on the 10 kg block. Here, m is the mass of the block.

mass of slab = M = 40 Kg
The force with which the slab will move will be μ_k mg.
Hence, μ_k mg = Ma
0.4 × 10 × 9.8 = 40 × a
a = 0.98 m/s²
Question 19
1 / -0

A man tries to remain in equilibrium by pushing his hands and feet against two parallel walls. For equilibrium,

A
he may exert unequal forces on the two walls

B
the forces of friction at the two walls must be equal

C
friction must be present on both the walls

D
the coefficients of friction must be the same between the walls and the man

Solution

For equilibrium, friction must be present on both the walls. Otherwise, his hands or his feet will slip and he will not be able to maintain the equilibrium.
Question 20
1 / -0

Which of the following is an essential characteristic of equilibrium?

A
Momentum equals zero

B
Acceleration equals zero

C
K.E. equals zero

D
Velocity equals zero

Solution

For a body to be in equilibrium, the sum total of all the forces acting on the body should be zero. If ∑F is zero then, acceleration will also be zero.