NCERT Exemplar: Laws of Motion | Physics Class 11 - NEET PDF Download

Ans. (c)
Solution.
When a body moves in such a way that the linear distance covered by each particle of the body is same during the motion, then the motion is said to be translatory or translation motion.
Translatory motion can be, again of two types viz., curvilinear (shown in fig. (a)) or rectilinear (shown in fig. (b)), accordingly as the paths of every constituent particles are similarly curved or straight line paths. Here it is important that the body does not change its orientation. Here we can also define it further in uniform and non-uniform translatory motion. Here figure
(b) is uniformly translatory motion.

Ans. (b)
Solution.
As the meter, the scale is moving with uniform velocity.
∴ No change in its velocity i.e., acceleration of it zero by Newton’s second law
∴ Hence net or resultant force must act on body zero.

So torque must be zero.
Hence, for uniform motion force and torque, both must be zero.So, It verifies the option (b).

Ans. (c)
Solution.
m = 150g = 0.15kg

Change in momentum momentum-Initial momentum



Hence, this verifies the option(c).

Ans. (c)
Solution.
By previous solution,


= 1.5 kg ms^-1
So, this verifies the option (c).

Ans. (d)
Solution.
(i) By Newton’s second law 

As in law of conservation of momentum is zero.
i.e.,  = 0
 = 0
⇒
(ii) By Newton’s third law action force is equal to reaction force in magnitude but in opposite direction.
∴ ( = 0)


So proves the law of conservation of momentum and verifies the option (d).

Ans. (c)
Solution. 
Step 1: Understand the Initial and Final Directions
The hockey player is initially moving northward. When he turns to move westward, he changes his direction but maintains the same speed.
Step 2: Analyze the Change in Momentum
Momentum is a vector quantity defined as the product of mass and velocity. The change in momentum occurs due to the change in direction.

• Initial momentum (P₁) is directed north.
• Final momentum (P₂) is directed west.

Step 3: Determine the Change in Momentum
The change in momentum (ΔP) can be calculated as:
ΔP = P₂ − P₁
This represents a vector subtraction where you find the resultant vector from the initial and final momentum vectors.
Step 4: Find the Direction of the Change in Momentum
To find the direction of ΔP, you can visualize the initial and final momentum vectors. The change in momentum vector will point from the initial direction (north) to the final direction (west).
This results in a vector that points diagonally between the north and west directions, which is southwest.
Step 5: Relate the Change in Momentum to Force
According to Newton's second law, the force acting on an object is equal to the rate of change of momentum. Thus, the direction of the force will be the same as the direction of the change in momentum.
Step 6: Identify the Type of Force
In this scenario, the force acting on the hockey player is primarily due to friction, which allows him to change direction while maintaining speed. The direction of this frictional force is southwest, as derived from the change in momentum.
Conclusion
The force that acts on the hockey player when he turns from north to west while maintaining the same speed is a frictional force directed southwest.

Ans. (a)
Solution.
To find the force acting on the body at t = 2 s, we follow these steps:
Step 1: Equation for acceleration
The position of the body as a function of time is given by: x(t) =pt + qt² + rt³ 
where p= 3m/s, q = 4m/s², r =5m/s³.
Velocity is the first derivative of position:
v(t) = dx/dt =p + 2qt + 3rt²
Acceleration is the first derivative of velocity:
a(t) = dv/dt = 2q + 6rt
Step 2: Compute acceleration at t = 2 s
Substitute q = 4m/s² and r = 5 m/s³ into a(t):
a(t) = 2q + 6rt
a(2) = 2(4) +6(5)(2) = 8 + 60 = 68m/s²
Step 3: Compute force
The force acting on the body is given by Newton's second law:
F = ma
Substitute m = 2 kg and a = 68 m/s²:
F = 2 x 68 = 136N

Ans. (b)
Solution.
To determine when the body will have a velocity only along the y-axis, we need to analyze the motion of the body under the given force.
Step 1: Force and acceleration
The force acting on the body is related to the acceleration by Newton's second law:
where m = 5kg.
Let the components of the force be F_x and F_y, and the corresponding accelerations be ax = F_x/m and a_y = F_y/m.
Step 2: Initial velocity
Let the initial velocity of the body be where and are the initial components along the x- and y-axes.
At time t, the velocity components are:

For the body to have a velocity only along the y-axis, the x-component of velocity must be zero:

Step 3: Solve for t
From we get:

Step 4: Conditions for t

• If  the body already has no x-component of velocity.
• If the x-component of velocity will never change.
• If and substitute the values of and a_x to calculate  t.
  

Ans. (b)
Solution. Here, mass of car = m
As it starts from rest, u = 0 final velocity along east υ = υî , t = 2 sec
Fromυ = u + at

force of car is mv/2 east ward
This is due to friction on the tyres exerted by the road.

Ans. (a, b, d)
Solution.
We have given mass = m, x(t) = 0 for t < 0

Now let’s look for time interval

x(t) = A sin 4πt

Also we know,

1^st derivative of x(t) w.r.t. time will give velocity

2^nd derivative of x(t) w.r.t. time will give velocity

Then,

Here, this force is obviously time dependent so force is not constant hence option (d) is correct.

Now let’s look to option (a) i.e. At

So from this we can say option (a) also correct.

Now for option (b) i.e. between t = 0 s and t = (1/4) s, as we know impulse is change in momentum or  so at t = (1/4) s impulse will be

 here we have to keep in mind that F(t) varies from t=0 to maximum 1/8 s, so F (1/4) will be replaced by F (1/8) also from above we have found

so option(b) is also correct.

Option (c) is absolutely wrong as we have above found that force is acting.

Option (e) is also wrong because we have already calculated the impulse.

Ans. (a, b, d, e)
Solution.
We have given  Let acceleration in body A and B is ‘a’.

Body A will move along with body B by force F till the force of friction between surface of A and B is larger or equal to zero.

Now taking system A+B then acceleration will be

So force on A

If F_AB is equal or smaller than f₂ then body A will move along with body B.

N = Reaction force by B on A

[N₂ = Normal reaction on B along with A by surface]

From (i)

 so adding this with equation (ii) we can say F = 0.45mg ...(iii)

F = 0.45mg Newton is the maximum force on B. so that A and B can move together. So option (e) is correct.

Both bodies can move together if F is less than or equal to 0.45mg Newton.

So options (a) and (b) are also correct and rejects the option (c) as 0.5mg>0.45mg.

For option (d): Minimum force which can move A and B together

Given force in option (d) 0.1mg Newton <0.25mg Newton. So body A and B will not move i.e.

Bodies A and B will remain in rest hence option (d) is also correct.

Ans. (b, d)
Solution.
Forces Acting on m₁  on the Slope:

• Component of weight along the slope: m₁gsinθ
• Normal force: N = m₁gcosθ 
• Frictional force: F_friction = μ_N = μm₁gcosθ
• Tension in the string: T (acts up the slope).

The net force on m₁ is given by:
Forces Acting on m₂:
The force on m₂ is:
Equation of Motion for the System:
The system will move when the forces on m₁ and m₂  are unbalanced. Considering the tension T and using Newton's second law for both masses, the condition for motion can be derived.

Case 1: Motion up the slope (m₁  moves up)
For m₁  to move up the slope, the force due to m₂  must overcome the sum of the downhill forces acting on m₁ . This gives:
m₂g > m₁g(sinθ + μcosθ)
Simplifying:
m₂ > m₁(sinθ + μcosθ)
Case 2: Motion down the slope (m₁ moves down)
For m₁  to move down the slope, the downhill forces acting on m₁  must overcome the tension caused by m₂. This gives:
m₁g(sinθ − μcosθ) > m₂g
Simplifying:
m₂ < m₁(sinθ − μcosθ)
Answer Choices:
(a) m₂  > m₁ sin θ : Incorrect, as friction is ignored in this condition.
(b) m₂ > m₁(sinθ+μcosθ): Correct, this is the condition for m₁ to move up the plane.
(c) m₂ < m₁(sin⁡θ + μcos⁡θ): Incorrect, this condition doesn’t align with motion upward.
(d) m₂ < m₁(sinθ−μcosθ): Correct, this is the condition for m₁ to move down the plane.

Ans. (b, c)

Solution.

Condition	Free body diagrams
Equation	Tension and acceleration
T - m1g sin α = m1a
m2a = m2g sin β - T

In this problem first we have to decide the direction of motion. Let block A moves up the plane friction force on A will be downward (along the plane) as shown.

Ans. (c, d)

According to the problem, balls are identical.

Initial velocity (u) = u₁ = u₂ = 5 m/s
Final velocity (v) = v₁ = v₂ = -5 m/s
Time duration of collision = 10^-3s
Change in linear momentum = m(v - u)



Impulse and force are opposite in directions.

Ans. (a, c)
Solution.
Recall the concept of resultant of two vectors, when they are perpendicular

As they are perpendicular, cos 90° = 0
So resultant will be
As shown in the diagram
According to the problem, mass = m = 10 kg
F₁ = 6 N, F₂ = 8 N


Let θ₁ be the angle between F and F₁.

Let θ₂ be angle between F and F₂.

Ans. Let m₁ and m₂ be the mass of girl with bicycle and stone respectively i.e. m₁ = 50kg & m₂ = 0.5kg
Also we have given u₁ = 5m/s, u₂ = 5m/s and v₂ = 15 m/s and we have to find v₁
As we know by law of conservation of momentum initial momentum (girl, bicycle, stone) = final momentum (cycle and girl) + stone





Therefore, the speed of cycle and girl decreased by

Ans. Let the acceleration be ‘a’ when the lift is descending, then the apparent weight decreases on weighing scale
So W' = R
= mg - ma
= m(g - a)
Apparent weight due to reaction force by the lift on weighing scale will be



Reading of weighing scale = R/g = 50/10 = 5 kg

Ans. Here, m=2kg
As is clear from position time graph in
At t = 0 , x = 0 body is at rest . Therefore impulse of body (at t = 0) = 0
From t = 0 to t = 4 s position time graph of body is a st line with positive slope. The body moves with a unifrom velocity.
Beyoud t = 4 s position time graph is parallel to time axis Therefore, body comes to rest
Now initial vel of body (u) = tanθ = 3/4 m/s final vel of body
(v) = 0
∴ Impulse (at t = 4 s) = Change in momentum

Ans. The phenomenon occurs due to inertia of motion, as described by Newton's First Law of Motion. Here's the explanation:

• When the car is moving, the person inside it is also moving at the same velocity as the car.
• When the brakes are suddenly applied, the car decelerates rapidly, coming to a stop.
• However, the person's body tends to maintain its state of motion (inertia) and continues to move forward with the original velocity of the car.
• Without a seatbelt to restrain him, the person moves forward and hits their head against the steering wheel.

This demonstrates how inertia resists changes in the state of motion, emphasizing the importance of wearing seatbelts to prevent such injuries.

Ans. we have mass m = 2kg and

So momentum at 2 sec p (2) = mv (2)





Hence momentum and force at 2 sec will be  and  respectively.

Ans. Let f (frictional force) at y-axis and F at x-axis
Now when a small force F₁ is applied on a heavier box, this box does not move, at this state, force of friction f₁ is equal to F₁. Increasing force on box does not move till F = f_s (the maximum static frictional or limiting force).
After force f_s, the frictional force decrease i.e. less kinetic force F_k < f_s is applied on body and it starts to move with less friction
In diagram A is equivalent to limiting frictional force and a B is equivalent to kinetic frictional force.

Ans. Porcelain (or glass) objects are brittle in nature and can crack even small jerk on it. During transportation sudden jerks or even fall takes place.
When objects are packed in paper or straw etc., the objects takes more time to stop or change velocity during jerks (due to breaks, or uneven road) so acceleration (v - u)/t decreased. So the force on objects will be smaller and objects become safer.

Ans. The effect of force F = ma. i.e., if the mass is constant for a system to decrease force, the ‘a’ should be decreased a = (v - u)/t initial and final velocity of falling body on a surface are u and zero. so it cannot be changed. If time during hitting is increased, the acceleration decreased and force will decrease.
On cemented hard floor the time to stop after fall on it is very-very small. But when she/he falls on soft ground of garden she/he sinks in ground and takes more time to stop hence smaller force or pain acts on her/him.

Ans. (a) Mass of object m = 500 g = 0.5 kg
u = 0, v = 25 m/s

Hence, the or force is opposite to the initial velocity of ball.

Ans. On an inclined plane force of friction on a body going upward is f_s = μN cos θ where θ is angle of inclination of a plane with horizontal if θ is small, the force of friction is high and there is a less chance of skidding. The road straight up would have a larger angle and smaller would be the value of friction, hence more are the chances of skidding.

Ans. Forces Acting on Thread AB:
1. AB supports: 

• The weight of the 2 kg mass (W = mg = 2kg . 9.8m/s² = 19.6N).
• The additional tension due to the force applied by pulling thread CD downward.

Thus, the tension in thread AB increases as the force applied on CD increases.
Forces Acting on Thread CD:
CD experiences only the force applied downward, increasing gradually.

Breaking Scenario:

• If the applied force on CD is increased very slowly, the tension in AB will increase gradually because AB must support both the weight of the mass and the additional force applied through CD. In this case, AB will break first, as it experiences a greater total force.
• If the force on CD is applied suddenly and very hard, thread CD will likely break first due to the sudden jerk. This happens because the tension in CD can momentarily exceed its breaking point before AB has time to transmit the full force.

Conclusion

• If the force is applied gradually, thread AB will break because it bears the combined weight and applied force.
• If the force is applied suddenly, thread CD will break first due to the jerk effect.

Ans. Thread CD will break up if CD is pulled with the jerk because pull on thread CD is not transmitted from CD to AB by mass 2 kg due to its inertia of rest.

Ans. As the whole system is going up with acceleration = a= 2ms^-2

Tension in a string is equal and opposite in all parts of a string.
Forces on mass m₁


Forces on mass m₂

Ans. Step 1: Forces on Block A
Block A rests on a frictionless incline at an angle θ = 30^∘. The forces acting on it are:
1. Its weight, W_A = 100 N, which can be resolved into:

• Component along the plane: W_A sin θ
• Component perpendicular to the plane: W_A cos θ

2. Tension in the cord, T, which acts up the plane.
For equilibrium, the tension T must balance the component of the weight along the plane:
T = W_Asinθ
Substitute W_A = 100 N and sin 30° = 0.5:
T = 100⋅0.5 = 50N
Step 2: Forces on Block B
Block B is hanging vertically. The forces acting on it are:

• Its weight, W_B  =W, which acts downward.
• The tension in the cord, T, which acts upward.

For equilibrium:
T = W
Thus:
W = 50N

Ans. Forces Acting on the Block:

Weight of the block: W=Mg, acting vertically downward due to gravity.

Frictional force: f_friction, acting upward and opposing the weight. The maximum frictional force is given by:
f_friction = μN 
where N is the normal force exerted by the finger.
Normal force: N = F, the force applied by the finger perpendicular to the wall.
Condition for Equilibrium:
For the block to remain stationary, the upward frictional force must balance the downward weight:
f_friction = Mg
Substitute f_friction =μN and N = F:
μF=Mg
Rearranging for F:

Ans. Given, the height of the cliff is,
Height of the cliff, h=500 m
Mass of the ball, m= 1 kg
Mass of the gun, M=100 kg
Range of the ball, R= 400 m
Acceleration due to gravity, g= 10 m s^-2
Let the final velocity of the ball be u and the of the gun be v.
Now we know, time of flight is


And range is given by:
R = ut
So, u=400/10 = 40 m s^-2
Therefore, the velocity is:



The recoil velocity of the gun is 0.4 m s^-1.

Ans. From graph (a)

From figure (b) y = t²





F = 1 N toward Y-axis.

Ans. Here, the initial velocity of the coin is, u = 20 m s¹
And the acceleration of the elevator is, a = 2 m s^-2
Here, let’s think this problem from the elevator reference frame. For the person inside the elevator going up with acceleration 2 m s^-2 will experience a net acceleration of (g+a) which is 12 m s^-2.
And as the coin return back to its original position (because we are in the elevator reference frame), the net displacement (s) is zero.
So we can use the laws of motion to solve the problem.

Here, a' = a + g = 12 m s^-2


So the time of flight of the coin is 3.33 seconds.

Ans. (a) We know force is a vector quantity. From the theory of planes, we know that two intersecting lines form a plane.
Let these lines be F₁ and F₂. This means that the vector sum, F₁ + F₂ is also on this plane.
As the body is moving with uniform speed, the acceleration is zero. So the net of all forces is zero.
F₁ + F₂ + F₃ = 0
So, F₃ = - (F₁ + F₂)
This means even F₃ lies on the same plane as F₁ and F₂.
Thus the forces are coplanar.
(b) As the sum of all forces are zero, the net torque in any direction is zero. For example, torque about “O” is
Τ = OP x (F₁ + F₂ + F₃)
Τ = 0 as (F₁ + F₂ + F₃) = 0
Hence, torque acting about any point is also zero.

Ans. Step 1: Motion on a Smooth Inclined Plane (No Friction)

For a smooth inclined plane with angle $\backslash theta\; =\; 45^\backslash circ$θ=45^∘:

• The acceleration of the body, $smootha\_\{\backslash text\{smooth\}\}$a_smooth, is due to gravity along the incline.
• Since there is no friction, $a\_\{\backslash text\{smooth\}\}\; =\; g\; \backslash sin\; \backslash theta$a_smooth = gsinθ.

For $\backslash theta\; =\; 45^\backslash circ$θ=45^∘: 

Using the second equation of motion, $\mathrm{<br\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">}\mathrm{<span\; style="background-color:\; transparent;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit;\; word-break:\; break-word;\; line-height:\; 1.8;\; font-size:\; 20px;\; color:\; black;">,\; where </span>}$s is the distance traveled, $a$a is the acceleration, and t is the time taken: 

Step 2: Motion on a Rough Inclined Plane (With Friction)

Now, the same body slides down a rough inclined plane with the same angle θ=45^∘ and the same distance s, but it takes a time pT where p>1.

Let μ be the coefficient of friction between the body and the rough plane. The frictional force f acting opposite to the motion is f=μN, where N is the normal force on the incline.

The normal force N on an incline is given by N=mgcosθ.

The net force acting on the body along the incline is:

The net acceleration a_rough along the incline is then:

a_rough= g sin⁡θ −  μ_gcos⁡θ

Using the same equation of motion $\mathrm{<img\; src="https://cn.edurev.in/ApplicationImages/Temp/24733558\_e9e3c80e-e759-45d8-a2a7-0bc30ea95001\_lg.png"\; style="display:\; inline-block;\; margin:\; 5px\; 0px;\; text-align:\; left;\; max-width:\; calc(100\%\; -\; 5px);\; float:\; left;"\; fr-original-class="fr-draggable"\; alt="NCERT\; Exemplar:\; Laws\; of\; Motion\; |\; Physics\; Class\; 11\; -\; NEET">}\mathrm{<span\; style="background-color:\; transparent;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit;\; word-break:\; break-word;\; line-height:\; 1.8;\; font-size:\; 20px;\; color:\; black;">we\; can\; set\; up\; the\; equation\; for </span>}$$ss$s when the body takes time $pT$pT on the rough incline:

Step 3: Equate Distances for Both Cases

Since the distance s is the same in both cases, we can equate the expressions for s obtained from both scenarios:

Ans. Here, as the velocity shows different behavior at different times, we have to divide time into intervals to specify velocity.


Now, we can differentiate velocity to calculate Forces(F).


Hence, the total force is

Ans. To find the total time, we find the time of each segments.
Segment: ABC
Here, the frictional force is
f = μmg
And the centripetal force is

Therefore,


Segment:DEF
Again, the centripetal force is equal to friction.


Now, the time taken is distance/speed.
For ABC, the distance is

And for DEF we have

Therefore,
= 86.3
Hence the time taken is 86.3 seconds.

Ans. (a) Here, the x and y coordinates are
x = A cos ωt
y = B sin ωt
The equation of an ellipse is:

Substituting x and y, we find that LHS = RHS.
So the trajectory is an ellipse.
(b) To find force, we need to find acceleration.
This can be found by differentiating r two times with respect to t.

And


We know,
F = ma
⇒ F = -mrω²

Ans. Case (a): Ball Released with Only Horizontal Velocity

In this case, the ball is released with only a horizontal velocity v_s. There is no initial vertical velocity component. Let’s break down the motion into horizontal and vertical components:

1. Horizontal Motion:

   • Horizontal velocity v_x = v_s (constant, as there’s no air resistance).

2. Vertical Motion:

   • Initial vertical velocity v_y = 0.

   • Acceleration a=g (due to gravity).

   • Using the second equation of motion to find the time taken t to reach the ground:
     

Case (b): Ball Released with Horizontal Velocity and Small Downward Velocity

In this case, the ball is released with a horizontal velocity v_s and a small initial downward vertical velocity v_y0.

• Horizontal Motion:

  • Horizontal velocity v_x = v_s (constant, as there’s no air resistance).

• Vertical Motion:

  • Initial vertical velocity v_y0 (downward).
  • Using the second equation of motion to find the time t it takes to reach the ground:
    
    Solving for t is more complex in this case, but t will be slightly shorter than in Case (a) because of the initial downward velocity v_y0.

Final Vertical Velocity:

Final Speed at the Ground:

Conclusion

The ball in Case (b) (with both horizontal velocity and a small downward velocity) will have a greater speed when it hits the ground compared to the ball in Case (a). This is because the additional initial downward velocity increases the final vertical component of velocity, resulting in a higher overall speed upon impact.

Ans. As the particle is rest or a = 0. So resultant force due to all forces will be zero.
∴ Net components along X and Y-axis will be zero.
Resolving all forces along X-axis

Resolving all forces along Y-axis

Ans. (a) From the diagram, the force on the floor(F) is given by:
F - mg = ma
Where m is the mass of the crew and a is the upwards acceleration.
(b) Here, we have to consider all the weight if the system. The force due to rotor F_r is:


(c) This force is just the reaction force of Fr. So the force due to surrounding air is 6.25 x 10⁴ N.