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Important Questions on Laws of Motion - CBSE Class XI
Solution:
Given:
- Mass, m = 1000 kg
- Initial velocity, u = 20 m/s
- Final velocity, v = 0 m/s
- Time, t = 5 seconds
The acceleration a can be found using:
v = u + at
0 = 20 + a × 5
a = -20 / 5 = -4 m/s²
Using F = ma:
F = 1000 × (-4) = -4000 N
The force required to stop the car is 4000 N in the direction opposite to the motion.
Solution:
Given:
- Mass, m = 2 kg
- Coefficient of friction, μ = 0.3
- Acceleration due to gravity, g = 9.8 m/s²
The maximum force of friction fmax is given by:
fmax = μ × m × g
fmax = 0.3 × 2 × 9.8
fmax = 5.88 N
The maximum force of friction acting on the block is 5.88 N.
Solution:
Given:
- Force, F = 200 N
- Mass, m = 50 kg
The acceleration a can be found using Newton’s second law:
F = ma
a = F / m
a = 200 / 50 = 4 m/s²
The acceleration of the object is 4 m/s².
Solution:
Static Friction: This is the frictional force that prevents two surfaces from sliding past each other when they are at rest. It must be overcome to start the motion. For example, the friction between a stationary car and the road that prevents the car from slipping.
Kinetic Friction: This is the frictional force that opposes the motion of two surfaces sliding past each other. It is usually less than static friction. For example, the friction between the tires of a moving car and the road.
Solution:
Given:
- Mass, m = 5 kg
- Initial velocity, u = 10 m/s
- Final velocity, v = 0 m/s
The change in momentum Δp is:
Δp = m × (v - u)
Δp = 5 × (0 - 10)
Δp = -50 kg·m/s
The change in momentum is -50 kg·m/s.
Solution:
Given:
- Mass, m = 10 kg
- Coefficient of kinetic friction, μk = 0.2
- Acceleration due to gravity, g = 9.8 m/s²
The force of friction fk is given by:
fk = μk × m × g
fk = 0.2 × 10 × 9.8
fk = 19.6 N
The force of friction acting on the block is 19.6 N.
Solution:
Given:
- Mass, m = 3 kg
- Radius, r = 2 m
- Speed, v = 6 m/s
The centripetal force Fc is given by:
Fc = (m × v²) / r
Fc = (3 × 6²) / 2
Fc = 54 N
The centripetal force acting on the object is 54 N.
Solution:
Newton's Third Law of Motion: This law states that for every action, there is an equal and opposite reaction. In other words, if one body exerts a force on a second body, the second body exerts a force of equal magnitude but in the opposite direction on the first body.
Example: When a person jumps off a boat, the person exerts a force backward on the boat, and the boat exerts an equal and opposite force on the person, causing the person to move forward and the boat to move backward.
Application to Rocket Motion: In a rocket, the engine expels gas particles downward (action). The reaction to this is that the rocket experiences an equal and opposite force pushing it upward. This is how rockets are propelled into space.
Solution:
Given:
- Initial velocity, u = 0 m/s
- Final velocity, v = 30 m/s
- Time, t = 10 seconds
The acceleration a can be calculated using:
a = (v - u) / t
a = (30 - 0) / 10
a = 3 m/s²
The acceleration of the car is 3 m/s².
Solution:
Given:
- Initial velocity, u = 40 m/s
- Angle, θ = 30°
- Acceleration due to gravity, g = 9.8 m/s²
The vertical component of the initial velocity is:
uy = u × sin(θ)
uy = 40 × sin(30°)
uy = 40 × 0.5 = 20 m/s
The maximum height H can be found using:
H = (uy²) / (2 × g)
H = (20²) / (2 × 9.8)
H = 20.41 m
The maximum height reached by the projectile is 20.41 m.
Solution:
Given:
- Mass, m = 5 kg
- Angle of incline, θ = 30°
- Acceleration due to gravity, g = 9.8 m/s²
The component of gravitational force parallel to the incline Fparallel is given by:
Fparallel = m × g × sin(θ)
Fparallel = 5 × 9.8 × sin(30°)
Fparallel = 5 × 9.8 × 0.5
Fparallel = 24.5 N
The component of gravitational force acting parallel to the incline is 24.5 N.
Solution:
Given:
- Mass, m = 1000 kg
- Initial velocity, u = 20 m/s
- Final velocity, v = 30 m/s
- Time, t = 5 seconds
The work done W can be calculated using the change in kinetic energy:
\( W = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 \)
\( W = \frac{1}{2} \times 1000 \times 30^2 - \frac{1}{2} \times 1000 \times 20^2 \)
\( W = \frac{1}{2} \times 1000 \times 900 - \frac{1}{2} \times 1000 \times 400 \)
\( W = 450000 - 200000 \)
\( W = 250000 \, \text{J} \)
The work done by the car's engine is 250000 J.
Solution:
Given:
- Mass of the block, m = 5 kg
- Angle of the incline, \(\theta = 30^\circ\)
- Coefficient of static friction, \(\mu_s = 0.4\)
The block will start sliding when the angle of the incline exceeds the angle of friction, which can be calculated using:
\[ \theta = \arctan(\mu_s) \]
Substitute \(\mu_s = 0.4\):
\[ \theta = \arctan(0.4) \]
The value of \(\theta\) is approximately:
\[ \theta \approx 21.8^\circ \]
The maximum angle at which the block will start sliding is approximately 21.8°.
Solution:
Given:
- Mass of the car, m = 1200 kg
- Initial velocity, u = 20 m/s
- Final velocity, v = 0 m/s
- Time taken, t = 5 s
The average acceleration can be calculated using:
\[ a = \frac{v - u}{t} \]
Substitute the values:
\[ a = \frac{0 - 20}{5} = -4 \, \text{m/s}^2 \]
The negative sign indicates deceleration.
The average braking force can be calculated using Newton’s second law:
\[ F = m \cdot a \]
Substitute the values:
\[ F = 1200 \times (-4) = -4800 \, \text{N} \]
The magnitude of the average braking force applied on the car is 4800 N.
Solution:
Given:
- Initial velocity, u = 40 m/s
- Angle of projection, \(\theta = 30^\circ\)
The maximum height \( H \) can be calculated using the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
where g is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).
Substitute the values:
\[ H = \frac{40^2 \cdot \sin^2 30^\circ}{2 \cdot 9.8} \]
Since \(\sin 30^\circ = 0.5\):
\[ H = \frac{1600 \cdot (0.5)^2}{19.6} = \frac{1600 \cdot 0.25}{19.6} \approx 8.16 \, \text{m} \]
The maximum height reached by the projectile is approximately 8.16 meters.
Solution:
Given:
- Mass of the block, m = 5 kg
- Angle of the incline, \(\theta = 45^\circ\)
The acceleration a of the block down the incline can be calculated using:
\[ a = g \sin \theta \]
where g is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)).
Substitute the values:
\[ a = 9.8 \cdot \sin 45^\circ \]
Since \(\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707\):
\[ a = 9.8 \cdot 0.707 \approx 6.93 \, \text{m/s}^2 \]
The acceleration of the block down the incline is approximately 6.93 m/s².
Solution:
Given:
- Initial velocity, u = 0 m/s
- Final velocity, v = 30 m/s
- Time, t = 10 s
The acceleration a can be calculated using the formula:
\[ a = \frac{v - u}{t} \]
Substitute the values:
\[ a = \frac{30 - 0}{10} = 3 \, \text{m/s}^2 \]
The distance s covered can be calculated using:
\[ s = ut + \frac{1}{2} a t^2 \]
Since u = 0:
\[ s = \frac{1}{2} \cdot 3 \cdot 10^2 = \frac{1}{2} \cdot 3 \cdot 100 = 150 \, \text{m} \]
The distance covered by the car is 150 meters.
Solution:
Given:
- Velocity function, v(t) = 4t^2 - 2t + 1 m/s
- Time interval: t = 2 s to t = 4 s
The displacement s is the integral of the velocity function:
\[ s = \int_{t_1}^{t_2} v(t) \, dt \]
Substitute v(t) = 4t^2 - 2t + 1:
\[ s = \int_{2}^{4} (4t^2 - 2t + 1) \, dt \]
Integrate:
\[ \int (4t^2 - 2t + 1) \, dt = \frac{4t^3}{3} - t^2 + t + C \]
Evaluate from t = 2 to t = 4:
\[ s = \left(\frac{4 \cdot 4^3}{3} - 4^2 + 4\right) - \left(\frac{4 \cdot 2^3}{3} - 2^2 + 2\right) \]
\[ s = \left(\frac{256}{3} - 16 + 4\right) - \left(\frac{32}{3} - 4 + 2\right) \]
\[ s = \left(\frac{256}{3} - 12\right) - \left(\frac{32}{3} - 2\right) \]
\[ s = \frac{256 - 36}{3} - \frac{32 - 6}{3} = \frac{220}{3} - \frac{26}{3} = \frac{194}{3} \approx 64.67 \, \text{m} \]
The displacement of the particle over the time interval from t = 2 s to t = 4 s is approximately 64.67 meters.
Solution:
Given:
- Initial velocity, v_0 = 40 m/s
- Angle, \theta = 30^\circ
The maximum height H can be calculated using:
\[ H = \frac{v_0^2 \sin^2 \theta}{2g} \]
where g is the acceleration due to gravity, g = 9.8 \, m/s^2.
Substitute the values:
\[ H = \frac{40^2 \sin^2 30^\circ}{2 \cdot 9.8} \]
\[ \sin 30^\circ = 0.5 \]
\[ H = \frac{1600 \cdot (0.5)^2}{2 \cdot 9.8} = \frac{1600 \cdot 0.25}{19.6} = \frac{400}{19.6} \approx 20.41 \, \text{m} \]
The maximum height reached by the projectile is approximately 20.41 meters.
Solution:
Given:
- Mass of the block, m = 5 kg
- Coefficient of friction, \mu_s = 0.4
- Acceleration due to gravity, g = 9.8 \, m/s^2
The maximum force of static friction F_s can be calculated using:
\[ F_s = \mu_s \cdot m \cdot g \]
Substitute the values:
\[ F_s = 0.4 \cdot 5 \cdot 9.8 = 19.6 \, \text{N} \]
The maximum force of static friction acting on the block is 19.6 N.
Solution:
Inertia is a fundamental property of matter that describes the tendency of an object to resist changes in its state of motion. According to Newton's First Law of Motion, also known as the Law of Inertia, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This can be expressed mathematically as:
\[ \text{If } \vec{F} = 0, \text{ then } \vec{v} = \text{constant} \]
Inertia is directly related to the mass of an object. The more massive an object is, the greater its inertia and, therefore, the more force is needed to alter its velocity. For instance, a heavy truck has more inertia than a small car, making it harder to accelerate or decelerate.
Solution:
Momentum is a vector quantity defined as the product of an object's mass and its velocity. Mathematically, it is given by:
\[ \vec{p} = m \vec{v} \]
The principle of conservation of momentum states that in an isolated system (where no external forces act), the total momentum before and after a collision remains constant. This principle can be expressed as:
\[ \text{Total Momentum Before Collision} = \text{Total Momentum After Collision} \]
In collisions, there are two main types:
- Elastic Collisions: In an elastic collision, both momentum and kinetic energy are conserved. This means that the total kinetic energy of the system remains the same before and after the collision. The conservation equations are:
- Inelastic Collisions: In an inelastic collision, while momentum is conserved, kinetic energy is not conserved. The colliding objects may stick together or deform, resulting in some of the kinetic energy being converted into other forms of energy such as heat or sound. The conservation equation for momentum remains:
\[ m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i} = m_1 \vec{v}_{1f} + m_2 \vec{v}_{2f} \]
\[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \]
\[ m_1 \vec{v}_{1i} + m_2 \vec{v}_{2i} = m_1 \vec{v}_{1f} + m_2 \vec{v}_{2f} \]
However, the kinetic energy is not conserved:
\[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 \ne \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \]
Solution:
Newton's Third Law of Motion states: "For every action, there is an equal and opposite reaction." This means that if one object exerts a force on a second object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.
Mathematically, this can be expressed as:
\[ \vec{F}_{12} = -\vec{F}_{21} \]
Implications and Real-World Applications:
- Walking: When you walk, your foot pushes backward against the ground. According to Newton's Third Law, the ground pushes your foot forward with an equal and opposite force. This reaction force propels you forward.
- Recoil of a Gun: When a gun is fired, the bullet is pushed forward out of the barrel with a certain force. Simultaneously, the gun is pushed backward with an equal force, causing the recoil.
- Swimming: A swimmer pushes water backward with their hands and feet. The water pushes the swimmer forward with an equal and opposite force, allowing the swimmer to move through the water.
- Rocket Propulsion: In a rocket, fuel is expelled backward at high speed. According to Newton's Third Law, the rocket experiences an equal and opposite force, propelling it forward.
This law is fundamental to understanding various physical interactions and is crucial in engineering, mechanics, and many scientific applications.
Solution:
Inertial Reference Frames: An inertial reference frame is one in which an object not subject to any net external force moves with a constant velocity, or remains at rest. In such frames, Newton's First Law holds true, stating that a body will remain at rest or in uniform motion unless acted upon by an external force.
Non-Inertial Reference Frames: A non-inertial reference frame is one that is accelerating, either linearly or rotationally. In these frames, objects appear to experience fictitious forces (also known as pseudo-forces) because the frame itself is accelerating. Newton's Laws need to be modified to account for these fictitious forces.
Differences in Laws of Motion:
- Inertial Frames:
- Newton’s Laws of Motion apply directly.
- Example: A car moving with a constant velocity on a straight road is in an inertial frame of reference. If you throw a ball inside the car, it will continue to move with the same velocity as the car (neglecting air resistance).
- Non-Inertial Frames:
- Fictitious forces, such as centrifugal force or Coriolis force, need to be introduced to account for observed accelerations.
- Example: In a rotating reference frame (like a spinning carousel), you might feel as though a force is pushing you outward (centrifugal force). This force is not real but arises due to the rotation of the reference frame.
Mathematically, the fictitious forces in non-inertial frames can be expressed as:
Centrifugal Force: \[ \vec{F}_{\text{cent}} = -m \omega^2 \vec{r} \]
Coriolis Force: \[ \vec{F}_{\text{cor}} = -2m (\vec{v} \times \vec{\omega}) \]
Solution:
Frictional Force:
1. Static Friction:
- Definition: Static friction is the force that opposes the initiation of motion between two surfaces that are in contact and at rest relative to each other.
- Characteristics:
- It acts to prevent relative motion.
- The maximum static friction force is given by \( F_s \leq \mu_s N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force.
- It varies depending on the applied force up to a maximum value, beyond which motion will start.
2. Kinetic Friction:
- Definition: Kinetic (or dynamic) friction is the force that opposes the motion of two surfaces sliding against each other.
- Characteristics:
- It acts to resist relative motion.
- The kinetic friction force is given by \( F_k = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.
- It remains constant regardless of the speed of the relative motion, unlike static friction which varies.
Factors Influencing Friction:
- Nature of Surfaces: Rougher surfaces have higher coefficients of friction. For instance, rubber on concrete has higher friction than ice on steel.
- Normal Force: The frictional force is directly proportional to the normal force acting between the surfaces. If the normal force increases, the frictional force also increases.
- Material Properties: Different materials have different coefficients of friction. For example, wood on wood has different friction compared to steel on steel.
- Presence of Lubricants: Lubricants like oil or grease reduce friction by creating a film between the surfaces, thus lowering the coefficient of friction.
Mathematically:
Static Friction: \[ F_s \leq \mu_s N \]
Kinetic Friction: \[ F_k = \mu_k N \]
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