Unit 2: Force and Motion — Short Questions

Right hand rule Rotate vector A into B through smaller angle. Curl the fingers of right hand in the direction of smaller rotation, the erect thumb indicates the direction of A × B.
The direction of A × B is perpendicular to the plane containing vectors A and B as shown in Fig.

Product of force and time interval for which the force acts on a body is known as impulse.
Mathematically:
Impulse = F × t
According to 2nd law of motion in terms of momentum
F = Δp/Δt
FΔt = mvf - mvi
Impulse = I = mvf - mvi
Hence impulse is always equal to change in linear momentum.

Elastic and Inelastic Collision

Elastic Collision "When the total K.E energy and total linear momentum of the system is conserved, then the collision is known as elastic collision." Collision are considered nearly elastic under certain conditions. Examples:
(i) Collision among gas molecules.
(ii) A hard . ball rebounds from hard floor loses negligible kinetic energy.

Inelastic Collision "When the total linear momentum is conserved but K.E is not conserved the collision is known as inelastic collision." It should be noted that during inelastic collision a portion of K.E is lost partially due to friction, heat and sound energies. Examples:
(i) Car crashes
(ii) Two clay balls colliding and sticking together.

Let a body of mass m is moving with velocity vi . A force F acts on it for time t and its velocity becomes vf . Then the for applied on the body by Neutron's second law.
F = ma
F = m(vf - vi)/t
F = (mvf - mvi)/t
F = Δp/t
Hence second law of motion can be stated as the "Time rate of change in momentum is equal to the applied force."

The total linear momentum of an isolated system always remains constant.
Importance of isolated system:
Real systems are not perfectly isolated, since they involve frictional effects. Under such condition, the momentum of the system does not remain same.
Usefulness of law even if system in not isolated:
In practical cases, the frictional forces are supposed to be very small as compared to the mutually interacting forces and hence the law of conservation of momentum is found useful in such situations.

Since range of projectile is
R = v₁²Sin2θ/g
Rage of projectile will be maximum
When Sin2θ=max
Sin2θ=1
2θ = Sin⁻¹(1)
2θ=90°
θ= 90°/2
θ=45°
Hence, range will be maximum at an angle of 45°.

∴ Vy = Vi Sinθ
Vy = 0
a = - g
t = ?
Applying 1st equation of motion
Vf = Vi + at
0 = Vi Sinθ + (- g) t
0 = Vi Sinθ - gt
gt = Vi Sinθ
t = Vi Sinθ / g
This is the time to reach maximum height.

Rmax = 800m
∴ Rmax = v₁²/g
g × Rmax = v₁²
9.8 × 800 = v₁²
v₁² = 7840m²s⁻²
For height of projectile
H = V₁² Sin²θ / 2g
= (7840 × (Sin60°)²) / (2 × 9.8)
= (7840 × (0.866)²) / 19.6
= (7840 × 0.75) / 19.6
= H = 300m

If A·B = 0 and A ≠ 0 then there are two possibilities
(i) B is a null vector (B = 0)
(ii.) Both are perpendicular to each other (i.e. θ=90°)
A·B = ABCos90° = AB(0)
A·B = 0

Yes, scalar product can be negative.
A·B = ABCosθ
Scalar product depends upon Cosθ, which has negative values when 90°<θ<180°. The work done against friction is negative.
w = Fd cos180°
w = - Fd. (the angle between two vectors is 180°).

When two vectors are perpendicular to each other then, their dot product will be zero i.e. θ = 90°. The product also zero of either the vector is null vector.
A·B = AB Cos90° = 0
The product also zero of either the vector is null vector.

Such a product of two vectors which results in a scalar quantity is called scalar product. Its value is defined as:
A·B = AB cosθ
Examples are: F·d = Work
F·V = Power

The scalar product of two parallel vectors is equal to the product of their magnitudes. If vectors A and B are parallel vectors i.e. the angle between them is 0°
A.B = AB cos0° = AB
[∴cos0° = 1]
∴ A.B = AB

We know that p = mv
[p = kgm⁻¹s]
Unit of momentum
P = kg × m/s × s = kg m/s s = Ns
[∴ 1 N = kg⁻²/s]
[∴ 1N = kg⁻m/s²]

Rate of change in momentum is equal to the force applied. This form of 2nd law is more general than the form F=ma, because it involves the change in mass as the body accelerate.

Since the collision between ball and bat is inelastic but law of conservation of momentum holds. When ball hits on the bat, momentum is added up. There must be the same momentum after collision as before the collision.

A product of two vectors which results in a vector quantity is called vector product. Its value is defined as:
A× B = AB sin θ n̂
r× F = τ = torque
r× P = L = angular momentum

(i) Equations of motion are useful only for linear motion with uniform acceleration.
(ii) All the vectors are manipulated like scalars,
(iii) Direction of initial velocity is taken as positive and negative sign is assigned to the quantities whose direction is opposite to that of initial velocity.

In the absence of air resistance, 'all objects in freefall motion near the surface of the earth, moves towards the earth with a uniform acceleration. This acceleration is known as acceleration due to gravity. It is denoted by g. Its value is 9.8 m s⁻².

R = v₁² sin2θ / g
h = v₁² sin²θ / 2g
As R = 4h
v₁² sin2θ / g = 4 × v₁² sin²θ / 2g
sin2θ = 4v₁² sin²θ / 2g × g / v₁²
sin2θ = 2sin²θ
2sinθ cosθ = 2sin²θ
Or
l = sinθ / cosθ
Tan θ=1
θ = tan⁻¹1 = 45°

The force responsible for projectile motion is the force of gravity.

The horizontal distance covered by the projectile during its motion is known as range and its formula is given below:
R = v₁² sin 2θ / g
Where v₁ and θ are initial velocity and angle of projection respectively.

As we know R = v₁² sin 2θ / g
And h = v₁² sin²θ / 2g
∴ R / h = (v₁² sin 2θ / g) × (2g / v₁² sin²θ)
= R / h = 2sin 2θ / sin²θ
= 2(2cosθ sinθ) / sin²θ
= 4cosθ / sinθ = 4cotθ

Yes, the range is same for both angles
θ₁ = 30° and θ₂ = 60°
R₁ = v₁² sin2θ₁ / g = v₁² sin(30°)² / g
R₁ = v₁² sin 60° / g
R₁ = v₁² sin(0.866) / g .... (1)
For θ = 60°
R₂ = v₁² sin(2 × 60°) = v₁² sin120° / g
R₂ = v₁² × (0.866) / g
R₂ = 0.866v₁² / g .... (2)
Hence, from Eq. (1) and (2)
R₁ = R₂

We know that the range of projectile is written as:
R = v₁² sin 2θ / g
Where v₁ / g = R_max
R = R_max sin 2θ
R / R_max = sin 2θ

The mass of the fuel bum per unit time multiplied by change in velocity of the rocket is equal to the force exerted on the rocket due to which it is propelled in space. This is the principal of rocket propulsion.

Let m = Mass of the gases ejected per second, v = Velocity of gases ejected relative to rocket.
Then mv = change in momentum per second of the ejecting gases.
Thus thrust produced by the engine on the body of the rocket is:
F = mv
Ma = mv
a = mv / M

It is a system on which no external agency exerts any force, e.g. the molecules of a gas enclosed in a gas vessel at constant temperature.

If the moving objects are moving with variable velocity, then they will have impulse. But if they are moving with uniform velocity, then the impulse will be zero.

Torque = τ× F
∴ S.I units are (Nm)
Torque = m kg × m / sec²
Torque = kg × m² / sec²
So dimensions of torque are [ML²T⁻²] – (1)
Work = F.d
∴ S.I units are N m
Torque = kg × m / sec² × m
Torque = Kgm²s⁻²
So dimensions of work are [ML²T⁻²] – (2)
From the equation (1) and (2) we see that dimensions of torque and work are same.

The hunter misses the target due to gravity. When the hunter fires, the bullet does not travel in a straight line, it follows a curved path due to gravity pulling it down. So, even the hunter aim exactly at the bird, the bullet misses the bird. As a result, the bullet passes beneath the bird causing the hunter to miss the target. To hit the bird, the hunter should aim slightly ahead of the bird.

A person falling on sand does not hurt more because sand compresses and increases the time of impact so force on body decreases by relation F = Δp / Δt while concrete stops the body suddenly. Δt is shorter, resulting in large force on person and more injury occurs.

The birds fly in air by obeying the principles of aerodynamics and Newton's laws of motion.
Lift is upward force generated by the wings as air flows over them (the shape of a bird's is wing). Air foil can causes air to move faster over the top than the bottom, creating low pressure on top.

• This lift must be greater than the weight of the bird.
• Thrust is forward motion produced by wing flapping must be greater than drag force opposing the motion.
• When a bird pushes air downward and backward with its wings, the air pushes the bird upward and forward.

(i) When brakes are applied and vehicle come to stop then v is zero but a is not zero.
(ii) When vehicle is moving with constant velocity then it's a is zero but v is not zero.
(iii) When vehicle is moving in a circular path then v and a are perpendicular to each other at every point.

Air resistance will slow down projectile forward motion, reducing its velocity vi . The reduction in vi will result in a decrease in the range of projectile (because R = v₁² sin 2θ / g).
Furthermore, air resistance is not constant throughout the flight of object. As the object slows down, the air resistance experienced by it also decreases. This means that the object retards more slowly and accelerates more slowly as it falls down.
This results in a trajectory that is not perfectly parabolic, but is skewed, with steeper descent than ascent.

See Q.5 of theory.

See Q.6 of theory.

See Q.8 to 10 of theory.

See Q.12 of theory.

See Q.12 of theory.

See Q.18 of theory.

See Q.19 of theory.

See Q.20 and 21 of theory.