Force and Laws of Motion

(a)

We notice from the table above that the acceleration of the object is increasing.

(b) We know Force = mass x acceleration

As mass of the object remains same, the increasing acceleration implies that the force acting on the object is increasing as well.

Given,

Mass of the car (m) = 1200 kg

When two persons push the car, it moves with uniform velocity i.e., acceleration of the car is 0 and total force on the car is balanced.

Hence, the force that accelerates the car is given by the third person.

Acceleration when third person starts pushing the car (a) = 0.2 ms^-2

We know, Force = mass x acceleration

Substituting we get,

F = 1200 × 0.2 = 240 N

The force applied by third person = 240 N.

As all 3 people push with the same muscular effort, hence, force applied by each person = 240 N

Given,

Mass of the hammer (m) = 500 g

Convert g into kg

1000 g = 1 kg

500 g = $\dfrac{500}{1000}$ = 0.5 kg

Initial velocity of the hammer (u) = 50 ms^-1

Terminal velocity of the hammer (v) = 0

Time period (t) = 0.01 s

As per the first equation of motion,

a = $\dfrac{\text{v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{0-50}{0.01}$ = -5000 ms^-2

We know, Force = mass x acceleration

Substituting we get,

F = 0.5 x 5000 = -2500 N

According to the third law of motion, the nail exerts an equal and opposite force on the hammer and as the force exerted on the nail by the hammer is -2500 N, hence, the force exerted on the hammer by the nail will be 2500 N.

Given,

Mass of the car (m) = 1200 kg

Initial velocity (u) = 90 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

90 x $\dfrac{5}{18}$ = 25 ms^-1

Final velocity (v) = 18 kmh^-1

Convert kmh^-1 to ms^-1 : multiply by $\dfrac{5}{18}$

18 x $\dfrac{5}{18}$ = 5 ms^-1

Time period (t) = 4 sec

As per the first equation of motion,

a = $\dfrac{\text{v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{5-25}{4}$ = -5ms^-2

Hence, acceleration of the car is -5 ms^-2

Negative sign shows retardation.

Initial momentum of the car = m × u = 1200 × 25 = 30,000 kgms^-1

Final momentum of the car = m × v = 1200 x 5 = 6,000 kgms^-1

∴ Change in momentum = mv - mu = 6,000 - 30,000 = -24,000 kgms^-1

Hence, change in momentum = 24,000 kgms^-1

We know, Force = mass x acceleration

Substituting we get,

F = 1200 × -5 = -6000 N

Hence, the magnitude of force required to slow down the vehicle = 6000 N

Yes, it is possible. According to Newton's first law of motion, an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force.
A net zero external unbalanced force means that the object has balanced force. Hence, if it had an initial velocity then the object will continue travelling with a non-zero velocity.
So, the conditions for an object to be traveling with a non-zero velocity while experiencing a net zero external unbalanced force are:

1. The object must have an initial velocity.
2. The object must have experienced a force in the past that set it in motion, and the net force acting on it is now balanced, resulting in a constant velocity.

The part of the carpet where the stick strikes, comes in motion at once, while the dust particles settled on it's fur, remain in position due to inertia of rest. Thus, the part of the carpet moves ahead with the stick, leaving behind the dust particles which fall down due to the earth's pull.

When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in the forward direction will exert a force (in the backward direction) on the luggage due to luggage's inertia of rest.

On the other hand, when a moving bus suddenly stops due to the application of brakes, a force (in the forward direction) is exerted on the luggage due to luggage's inertia of motion. Hence, tying the luggage will secure its position and prevent it from falling off the bus.

there is a force on the ball opposing the motion.

Reason — When the ball rolls on a level ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball.

Given,

Distance (s) = 400 m

Time (t) = 20 sec

Initial velocity (u) = 0

According to the second equation of motion,

S = ut + $\dfrac{1}{2}$at²

Substituting we get,

400 = 0 + $\dfrac{1}{2}$ x a x 20 x 20

a = $\dfrac{400 \times 2}{20 \times 20}$ = 2 ms^-2

∴ acceleration of the car = 2 ms^-2.

Now, mass = 7 tonnes

Converting tonne into kg

1 tonne = 1000 kg

∴ 7 tonnes = 7000 kg

Force = mass x acceleration

F = 7000 x 2 = 14000 N

Hence, Force acting = 14000 N

Given,

Mass (m) = 1 kg

Initial velocity (u) = 20 ms^-1

Final velocity (v) = 0

Distance travelled (s) = 50 m

According to the third equation of motion,

2as = v² - u²

or

a = $\dfrac{\text{v}^2 - \text{u}^2 }{2\text{s}}$

Substituting we get,

a = $\dfrac{0 - (20)^2}{2 \times 50} = -\dfrac{400}{100}$ = -4 [retardation]

Now, as Force = mass x acceleration

Substituting we get,

F = 1 × (-4) = -4 N

Negative sign indicates the opposing force which is friction

Hence, force of friction between the stone and the ice = -4 N

(a) Given,

Force exerted (F) = 40,000 N

Force of friction = -5000 N

Net accelerating force = 40,000 + (-5000) = 35,000 N

Hence, net accelerating force = 35,000 N

(b) Total mass of the train = mass of engine + mass of each wagon = 8000 + (5 × 2000) = 8000 + 10000 = 18000 kg

Now, as Force = mass x acceleration

Substituting we get,

35000 = 18000 x a

a = $\dfrac{35000}{18000}$ = 1.94 ms^-2

Hence, acceleration of the train is 1.94 ms^-2

Given,

Mass (m) = 1500 kg

Acceleration (a) = -1.7 ms^-2

Now, as Force = mass x acceleration

F = 1500 × (-1.7) = -2550 N

Hence, force required = -2550 N. (Negative sign indicates opposite direction.)

mv

Reason — The momentum of a body is defined as the product of its mass and velocity. For a body of mass m, moving with velocity v, linear momentum p is expressed as : p = mv

As the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. Hence, the magnitude of opposing frictional force is equal to the force exerted on the cabinet i.e., 200 N. So, the total friction force is 200 N.

The student's justification is incorrect since action-reaction force always apply to different bodies and hence cannot cancel each other. When two equal and opposite forces act on the same body, they cancel each other.

As, the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. Hence, the truck will only move if a force greater than the friction force is applied to it.

Given,

Mass (m) = 200 g

Initial velocity (u) = 10 ms^-1

Final velocity (v) = -5 ms^-1

Initial momentum of the ball = mu = 200 × 10 = 2000 gms^-1

Final momentum of the ball = mv = 200 × -5 = -1000 gms^-1

∴ The change in momentum (mv - mu) = -1000 - 2000 = -3000 gms^-1

Convert gms^-1 to kgms^-1 :

1000 gms^-1 = 1 kgms^-1

So, 3000 gms^-1 = $\dfrac{3000}{1000}$ = -3 kgms^-1

Hence, momentum of the ball reduces by 3 kgms^-1 after being struck by the hockey stick.

Given,

Mass of the bullet (m) = 10 g

Convert g into kg

1000 g = 1 kg

So, 10 g = $\dfrac{10}{1000}$ = 0.01 kg

Initial velocity of the bullet (u) = 150 ms^-1

Terminal velocity of the bullet (v) = 0

Time period (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated

Let the distance of penetration be s

According to the first equation of motion,

v = u + at

or

a = $\dfrac{\text {v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{0-150}{0.03} = \dfrac{150}{0.03} =\dfrac{-15000}{3}$ = -5000 ms^-2

∴ Acceleration = -5000 ms^-2.

According to the third equation of motion,

2as = v² - u²

or

a = $\dfrac{\text{v}^2 - \text{u}^2 }{2\text{s}}$

Substituting we get,

s = $\dfrac{0 - (150)^2}{2 \times -5000} = \dfrac{-22500}{-10000} =$ = 2.25 m.

Hence, the total distance travelled is 2.25 m

Now, as Force = Mass x Acceleration

Substituting we get,

F = 0.01 × -5000 = -50 N

Negative sign indicates direction of recoil of bullet is opposite to the direction of bullet.

Hence, magnitude of the force = 50 N

Given,

Mass of the object (m₁) = 1 kg

Mass of the block (m₂) = 5 kg

Initial velocity of the object (u₁) = 10 ms^-1

Initial velocity of the block (u₂) = 0

Mass of the resulting object = m₁ + m₂ = 6 kg

Velocity of the resulting object (v) = ?

Total momentum before the collision = m₁u₁ + m₂u₂ = (1 × 10) + 0 = 10 kgms^-1

According to the law of conservation of momentum,

total momentum before the collision = total momentum after the collision.

∴ the total momentum post the collision is also 10 kgms^-1

Hence, total momentum just before the impact = 10 kgms^-1 and just after the impact = 10 kgms^-1

Now, (m₁ + m₂) × v = 10 kgms^-1

6 x v = 10

v = $\dfrac{10}{6}$ = $\dfrac{5}{3}$ ms^-1

Hence, velocity of the combined object = $\dfrac{5}{3}$ ms^-1

Given,

Mass of the object (m) = 100 kg

Initial velocity (u) = 5 ms^-1

Final velocity (v) = 8 ms^-1

Time period (t) = 6 s

Now, initial momentum (m × u) = 100 × 5 = 500 kgms^-1

Final momentum (m × v) = 100 × 8 = 800 kgms^-1

Hence, Initial momentum = 500 kgms^-1 and final momentum = 800 kgms^-1

As per the first equation of motion,

a = $\dfrac{\text{v - u}}{\text{t}}$

Substituting we get,

a = $\dfrac{8-5}{6}$ = 0.5 ms^-2

We know, Force = mass x acceleration

Substituting we get,

F = 100 x 0.5 = 50 N

Hence, magnitude of the force = 50 N

Kiran's statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum.

Akhtar's statement is false. Force exerted is directly proportional to the mass and change in momentum. It does not depend on velocity. Secondly, according to Newton's third law of motion car and insect would apply equal and opposite force on each other.

Rahul's statement is correct. As per the Newton's third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. Hence, insect and car experience same force.

The change in momentum, however, depends on the mass and resulting acceleration. The motorcar, with its larger mass, experiences less acceleration and, therefore, less change in velocity. On the other hand, the mass of the insect is very small compared to the motorcar, it suffers a huge change of velocity as compared to the motorcar. Due to this, insect dies.

Given,

Mass of the dumb-bell (m) = 10 kg

Distance covered (s) = 80 cm

Convert cm to m

100 cm = 1 m

So, 80 cm = $\dfrac{80}{100}$ = 0.8 m

Initial velocity (u) = 0

Acceleration (a) = 10 ms^-2

Terminal velocity (v) = ?

Momentum of the dumb-bell when it hits the ground = mv

As per the third law of motion,

v² - u² = 2as

Substituting we get,

v² - 0 = 2 x 10 x 0.8

v² = 16

v = $\sqrt{16}$ ms^-1

v = 4 ms^-1

Hence, dumb-bell strikes the ground with a velocity of 4 ms^-1

The momentum transferred by the dumb-bell to the floor = mv = 10 × 4 = 40 kgms^-1

Hence, momentum transferred = 40 kgms^-1

As inertia of a body is directly proportional to its mass, hence, more the mass more is the inertia. The following objects therefore, hold greater inertia because of their higher mass.

(a) Stone

(b) Train

(c) Five-Rupee coin

The velocity of the ball changes :

1. When the football player kicks a football to another player,
2. When that player kicks the football towards the goal.
3. When the goalkeeper stops the football.
4. When the goalkeeper kicks the football towards his team player.

Hence, the velocity of the football changes four times.

Agent supplying the force:

(a) The First case is the First player

(b) The Second case is the Second player

(c) The Third case is Goalkeeper

(d) The Fourth case is Goalkeeper

When the tree is at rest, the leaves are also at rest. Now when the tree is vigorously shaken, the tree is in motion while the leaves are still at rest, due to inertia of rest. The force acts on the leaves in changing directions. Therefore, the leaves that are weakly attached to the branch fall off whereas the leaves that are firmly attached to the branch remain attached.

With the application of brakes, the bus stops but our body tends to continue in the same state of motion because of its inertia. Hence, we fall forward.

An opposite experience is encountered when we are sitting in a bus and the bus begins to move suddenly. Now, we tend to fall backwards. This is because the sudden start of the bus brings motion to the bus. But, our body opposes this motion because of its inertia. Hence, we fall backwards.