Newton’s Second Law for a System of Particles MCQ Quiz - Objective Question with Answer for Newton’s Second Law for a System of Particles - Download Free PDF

$B_{in}={\displaystyle \frac{{\mu }_{0}Jr}{2}}$

$B_{out}={\displaystyle \frac{{\mu }_{0}J{R}^2}{2r}}$

$r=$ distance from the axis of the cylinder.

$R=$ Radius of the cylinder.

Assuming the bigger cylinder to carry a positive current density and the smaller cylinder carry a negative current density of magnitude J each.

$\therefore$ Magnetic field at point P = $B=B_1+B_2$

$B_1={\displaystyle \frac{{\mu }_{0}Ja}{2}}$

$B_2={\displaystyle \frac{-{\mu }_{0}J{({\displaystyle \frac{a}{2}})}^2}{2{\displaystyle \frac{3a}{2}}}}$

$\therefore B_2={\displaystyle \frac{-{\mu }_{0}Ja}{12}}$

$\therefore B={\displaystyle \frac{{5\mu }_{0}Ja}{12}}$

$\therefore N=5$

Concept:

Impulse and Momentum

• Momentum: The product of the mass and velocity of an object in motion is called momentum.
  • It is a vector quantity.
  • The unit of momentum is kg m s -1. 
• Impulse: The change of momentum is called impulse. 
  • It has the same unit as momentum. 
• The rate of change of momentum is called Force by the second law of motion. 

$F=\frac{\Delta P}{t}$

Δ P is a change in momentum or impulse, t is time.

• Impulse can also be defined as the product of force and time. 

Δ P = F × t

• The Force is also defined as the product of mass and acceleration. 

Calculation:

ΔP = 20 - 5 = 15 kgm/s, time t = 3 s

$F=\frac{\Delta P}{t}$

$F=\frac{15}{3}=5kgm/s$

Concept:

Linear Momentum:

• It is a product of the mass and velocity of the body.
• The formula of the momentum, P = m v, where m= mass of the body, v= velocity of the body
• In terms of force, $F_{ext}=\frac{dP}{dt}$
• For an isolated system, the external force is zero.

Torque:

• It is the measure of the force that can cause an object to rotate about an axis.
• Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes angular acceleration.
• Formula, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$, where, F = force vector, r = displacement vector, τ = torque

Explanation:

consider an external force acting on the center of mass of a body placed on a smooth surface.
Net torque about point O is zero but there is a net force acting on the body.

F = ma

$a=\frac{F}{m}$

Thus the velocity of COM will continuously change because of the acceleration a.

For an isolated system, F_ext = 0

$F_{ext}=\frac{dP}{dt}=0$

⇒P = constant

Hence linear momentum of an isolated system is conserved.

CONCEPT:

Centre of mass:

• The centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particle appeared to be concentrated.

The motion of the centre of mass:

• Let there are n particles of masses m1, m2,..., mn.
• If all the masses are moving then,

⇒ Mv = m1v1 + m2v2 + ... + mnvn

⇒ Ma = m1a1 + m2a2 + ... + mnan

$\Rightarrow M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$

⇒ M = m1 + m2 + ... + mn

• Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.
• The internal forces contribute nothing to the motion of the centre of mass.

Given:

Given m₁ = 1 kg, m2 = 2 kg, m3 = 2 kg $\overrightarrow{F_1}=(3\hat{i}+6\hat{j})$, $\overrightarrow{F_2}=(5\hat{i}-8\hat{j})$, and $\overrightarrow{F_3}=(-4\hat{i}+5\hat{j})$

• We know that the total mass of a system of particles times the acceleration of its centre of mass is equal to the vector sum of all the forces acting on the system of particles.

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$     -----(1)

By equation 1,

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}$

⇒$(1+2+2)\overrightarrow{a}=(3\hat{i}+6\hat{j})+(5\hat{i}-8\hat{j})+(-4\hat{i}+5\hat{j})$

⇒ $5\overrightarrow{a}=4\hat{i}+3\hat{j}$     -----(2)

By equation 2 the magnitude of the vector $4\hat{i}+3\hat{j}$ is given as,

⇒ $5a=\sqrt{4^2+3^2}$

⇒ a = 1 m/sec²

• Hence, option 1 is correct.

CONCEPT:

Centre of mass:

• The centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particle appeared to be concentrated.

The motion of the centre of mass:

• Let there are n particles of masses m1, m2,..., mn.
• If all the masses are moving then,

⇒ Mv = m1v1 + m2v2 + ... + mnvn

⇒ Ma = m1a1 + m2a2 + ... + mnan

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$

⇒ M = m1 + m2 + ... + mn

• Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.
• The internal forces contribute nothing to the motion of the centre of mass.

EXPLANATION:

Given $\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}=0$

• We know that the total mass of a system of particles times the acceleration of its centre of mass is equal to the vector sum of all the forces acting on the system of particles.

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$     -----(1)

∴ $M\overrightarrow{a}=0$     -----(2)

• We know that the total mass of the system of particles can't be zero.
• Therefore in this case acceleration of the centre of mass must be zero. The position of the centre of mass may change. Hence, option 2 is correct.

CONCEPT:

• Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.

P = mv

where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.

• Newton's Second Law of Motion: It says that the net external force on a system or body is equal to the change in momentum of the system or body divided by the time over which it changes. 

Mathematically:

$F_{ext}=\frac{\Delta p}{\Delta t}$

where F_ext is the external force on the system, Δp is the change in momentum, and Δt is the change in time.

CALCULATION:

Given that m = 10g = 10 × 10^-3 Kg; Δt = 0.01 sec; initial velocity u = 500 m/s; final velocity v = 0.

$F_{ext}=\frac{\Delta p}{\Delta t}$

$F_{ext}=\frac{mv-mu}{\Delta t}$

$F_{ext}=\frac{10\times {10}^{-3}\times 500-0}{0.01}$

$F_{ext}=500N$

So the correct answer is option 2.

The correct answer is option 1) i.e. increase the contact time and avoid getting hurt

Concept:

Linear momentum:

Linear momentum is the impact caused due to a body in translational motion.

The greater the mass and velocity of the moving body the more will be the impact and thus greater the momentum.

The linear momentum of a moving object is defined as:

p = mv

​Where,

p is the linear momentum, m is the mass of the object, and v is the velocity of the object.

According to Newton's second law of motion, the rate of change of momentum is directly proportional to the force applied by the moving body. 

$\Rightarrow \frac{dp}{dt}\propto F$

$\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}=ma=F$

Explanation:

• The boxer experiences momentum due to the punch on the head.
• On moving the head backward, he increases the time of contact.

Since $\frac{dp}{dt}\propto F$, increasing the time of contact for the same momentum decreases the force acting on his head.

• This will help in reducing the force acting on his head and helps avoid injury or getting hurt.

CONCEPT:

• Impulse: A large amount of force acting on an object for a short interval of time is called impulse or impulsive force.

Numerically impulse is the product of force F and time dt. 

Impulse = F. dt

EXPLANATION:

Impulse is calculated by doing the product of force F and time dt. 

Impulse = F. dt

• So the correct answer is option 2.

Additional Information

The impulse of an object is equal to the change in momentum of the object.

Impulse = Δp =mΔv

F. dt = mΔv

$F_{ext}=\frac{\Delta p}{\Delta t}$

This is Newton's Second Law of motion in momentum form.

CONCEPT:

Centre of mass:

• The centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particle appeared to be concentrated.

The motion of the centre of mass:

• Let there are n particles of masses m1, m2,..., mn.
• If all the masses are moving then,

⇒ Mv = m1v1 + m2v2 + ... + mnvn

⇒ Ma = m1a1 + m2a2 + ... + mnan

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$

⇒ M = m1 + m2 + ... + mn

• Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.
• The internal forces contribute nothing to the motion of the centre of mass.

EXPLANATION:

Given $\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}=0$

• We know that the total mass of a system of particles times the acceleration of its centre of mass is equal to the vector sum of all the forces acting on the system of particles.

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$     -----(1)

∴ $M\overrightarrow{a}=0$     -----(2)

• We know that the total mass of the system of particles can't be zero.
• Therefore in this case acceleration of the centre of mass must be zero. The position of the centre of mass may change. Hence, option 2 is correct.

Concept:

Impulse and Momentum

• Momentum: The product of the mass and velocity of an object in motion is called momentum.
  • It is a vector quantity.
  • The unit of momentum is kg m s -1. 
• Impulse: The change of momentum is called impulse. 
  • It has the same unit as momentum. 
• The rate of change of momentum is called Force by the second law of motion. 

$F=\frac{\Delta P}{t}$

Δ P is a change in momentum or impulse, t is time.

• Impulse can also be defined as the product of force and time. 

Δ P = F × t

• The Force is also defined as the product of mass and acceleration. 

Calculation:

ΔP = 20 - 5 = 15 kgm/s, time t = 3 s

$F=\frac{\Delta P}{t}$

$F=\frac{15}{3}=5kgm/s$

CONCEPT:

Centre of mass:

• The centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particle appeared to be concentrated.

The motion of the centre of mass:

• Let there are n particles of masses m1, m2,..., mn.
• If all the masses are moving then,

⇒ Mv = m1v1 + m2v2 + ... + mnvn

⇒ Ma = m1a1 + m2a2 + ... + mnan

$\Rightarrow M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$

⇒ M = m1 + m2 + ... + mn

• Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.
• The internal forces contribute nothing to the motion of the centre of mass.

Given:

Given m₁ = 1 kg, m2 = 2 kg, m3 = 2 kg $\overrightarrow{F_1}=(3\hat{i}+6\hat{j})$, $\overrightarrow{F_2}=(5\hat{i}-8\hat{j})$, and $\overrightarrow{F_3}=(-4\hat{i}+5\hat{j})$

• We know that the total mass of a system of particles times the acceleration of its centre of mass is equal to the vector sum of all the forces acting on the system of particles.

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$     -----(1)

By equation 1,

⇒ $M\overrightarrow{a}=\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}$

⇒$(1+2+2)\overrightarrow{a}=(3\hat{i}+6\hat{j})+(5\hat{i}-8\hat{j})+(-4\hat{i}+5\hat{j})$

⇒ $5\overrightarrow{a}=4\hat{i}+3\hat{j}$     -----(2)

By equation 2 the magnitude of the vector $4\hat{i}+3\hat{j}$ is given as,

⇒ $5a=\sqrt{4^2+3^2}$

⇒ a = 1 m/sec²

• Hence, option 1 is correct.

CONCEPT:

• Newton's Second Law of Motion: It says that the net external force on a system or body is equal to the change in momentum of the system or body divided by the time over which it changes. 

Mathematically:

$F_{ext}=\frac{\Delta p}{\Delta t}$

where Fext is the external force on the system, Δp is the change in momentum, and Δt is the change in time.

CALCULATION:

Given that F = 120 N, and Δp = 240 Kg-m/s

$F_{ext}=\frac{\Delta p}{\Delta t}$

$120=\frac{240}{\Delta t}$

Δt = 2 sec

So the correct answer is option 4.

CONCEPT:

• Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.

P = mv

where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.

• Newton's Second Law of Motion: It says that the net external force on a system or body is equal to the change in momentum of the system or body divided by the time over which it changes. 

Mathematically:

$F_{ext}=\frac{\Delta p}{\Delta t}$

where Fext is the external force on the system, Δp is the change in momentum, and Δt is the change in time.

EXPLANATION:

Newton's second law of motion in momentum form can be written as:

$F_{ext}=\frac{\Delta p}{\Delta t}$

which says the rate of change of momentum $\frac{\Delta p}{\Delta t}$ is equal to the external force on a system.

So the correct answer is option 1.

CONCEPT:

• Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.

P = mv

where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.

• Newton's Second Law of Motion: It says that the net external force on a system or body is equal to the change in momentum of the system or body divided by the time over which it changes. 

Mathematically:

$F_{ext}=\frac{\Delta p}{\Delta t}$

where F_ext is the external force on the system, Δp is the change in momentum, and Δt is the change in time.

CALCULATION:

Given that m = 10g = 10 × 10^-3 Kg; Δt = 0.01 sec; initial velocity u = 500 m/s; final velocity v = 0.

$F_{ext}=\frac{\Delta p}{\Delta t}$

$F_{ext}=\frac{mv-mu}{\Delta t}$

$F_{ext}=\frac{10\times {10}^{-3}\times 500-0}{0.01}$

$F_{ext}=500N$

So the correct answer is option 2.

The correct answer is option 1) i.e. increase the contact time and avoid getting hurt

Concept:

Linear momentum:

Linear momentum is the impact caused due to a body in translational motion.

The greater the mass and velocity of the moving body the more will be the impact and thus greater the momentum.

The linear momentum of a moving object is defined as:

p = mv

​Where,

p is the linear momentum, m is the mass of the object, and v is the velocity of the object.

According to Newton's second law of motion, the rate of change of momentum is directly proportional to the force applied by the moving body. 

$\Rightarrow \frac{dp}{dt}\propto F$

$\frac{dp}{dt}=\frac{d(mv)}{dt}=m\frac{dv}{dt}=ma=F$

Explanation:

• The boxer experiences momentum due to the punch on the head.
• On moving the head backward, he increases the time of contact.

Since $\frac{dp}{dt}\propto F$, increasing the time of contact for the same momentum decreases the force acting on his head.

• This will help in reducing the force acting on his head and helps avoid injury or getting hurt.