Laws of Motion: JEE Mains Previous Year Questions (2021- 2024) | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Laws of Motion 
2024 
Q1: A body of mass ?? ???? experiences two forces ?? ? ? 
?? = ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 and ?? ? ? 
?? = ?? ??ˆ - ?? ??ˆ - ?? ?? ˆ
. The 
acceleration acting on the body is : 
A. ?? ??ˆ + ??ˆ + ?? ˆ
 
Correct Answer 
B. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 
C. -?? ??ˆ - ??ˆ - ?? ˆ
 
D. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
To find the acceleration acting on the body, we first need to determine the resultant force acting on the 
body by adding the two forces ?? 
1
 and ?? 
2
 vectorially. Then, we apply Newton's second law of motion, 
which states that the acceleration ??  of a body is directly proportional to the total force ?? 
 acting on it 
and inversely proportional to the mass ?? of the body: 
?? 
= ?? · ??  
or 
?? =
?? 
?? 
Let's start by adding the forces: 
?? 
1
+ ?? 
2
= (5??ˆ + 8??ˆ + 7?? ˆ
) + (3??ˆ - 4??ˆ - 3?? ˆ
) 
Performing the addition component-wise: 
?? 
total 
= (5 + 3)??ˆ + (8 - 4)??ˆ + (7 - 3)?? ˆ
?? 
total 
= 8??ˆ + 4??ˆ + 4?? ˆ
 
Now, let's use the formula for acceleration with ?? = 4 kg : 
?? =
?? 
0?????? ?? =
8??ˆ + 4??ˆ + 4?? ˆ
4 kg
 
Divide each component by the mass: 
?? = 2??ˆ + 1??ˆ + 1?? ˆ
 
So, the acceleration acting on the body is: 
?? = 2??ˆ + ??ˆ + ?? ˆ
 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Laws of Motion 
2024 
Q1: A body of mass ?? ???? experiences two forces ?? ? ? 
?? = ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 and ?? ? ? 
?? = ?? ??ˆ - ?? ??ˆ - ?? ?? ˆ
. The 
acceleration acting on the body is : 
A. ?? ??ˆ + ??ˆ + ?? ˆ
 
Correct Answer 
B. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 
C. -?? ??ˆ - ??ˆ - ?? ˆ
 
D. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
To find the acceleration acting on the body, we first need to determine the resultant force acting on the 
body by adding the two forces ?? 
1
 and ?? 
2
 vectorially. Then, we apply Newton's second law of motion, 
which states that the acceleration ??  of a body is directly proportional to the total force ?? 
 acting on it 
and inversely proportional to the mass ?? of the body: 
?? 
= ?? · ??  
or 
?? =
?? 
?? 
Let's start by adding the forces: 
?? 
1
+ ?? 
2
= (5??ˆ + 8??ˆ + 7?? ˆ
) + (3??ˆ - 4??ˆ - 3?? ˆ
) 
Performing the addition component-wise: 
?? 
total 
= (5 + 3)??ˆ + (8 - 4)??ˆ + (7 - 3)?? ˆ
?? 
total 
= 8??ˆ + 4??ˆ + 4?? ˆ
 
Now, let's use the formula for acceleration with ?? = 4 kg : 
?? =
?? 
0?????? ?? =
8??ˆ + 4??ˆ + 4?? ˆ
4 kg
 
Divide each component by the mass: 
?? = 2??ˆ + 1??ˆ + 1?? ˆ
 
So, the acceleration acting on the body is: 
?? = 2??ˆ + ??ˆ + ?? ˆ
 
Thus, the correct option is: 
Option A : 
2??ˆ + ??ˆ + ?? ˆ
 
Q2: A cricket player catches a ball of mass ?????? ?? moving with ???? ?? /?? speed. If the catching process is 
completed in ?? .?? ?? then the magnitude of force exerted by the ball on the hand of player will be (in SI 
unit) : 
A. 30 
B. 24 
C. 12 
D. 25 
Ans: (a) 
The first step in solving this problem is to calculate the change in momentum of the ball when it is 
caught. The change in momentum, or impulse, is the product of the mass of the ball and the change in 
velocity (as momentum is mass times velocity). 
The ball is initially moving with a velocity of ?? ?? = 25 m/s before the catch and finally comes to rest with 
a velocity of ?? ?? = 0 m/s after the catch. Since the ball is caught, the final velocity is zero. The change in 
velocity ??? = ?? ?? - ?? ?? = 0 - 25 = -25 m/s. Remember that the direction of the force exerted by the 
ball on the hand will be opposite to the direction of the ball's initial motion. 
The mass of the ball ?? is given as 120 g which needs to be converted into kilograms to maintain Sl 
units: 
?? = 120 g = 120× 10
-3
 kg = 0.12 kg.  
Now we can calculate the change in momentum (impulse): ??? = ?? ??? = 0.12 kg × (-25 m/s). 
Substituting the values we get: ??? = 0.12× -25 = -3 kg · m/s. 
The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial 
motion, which makes sense because the ball's velocity is reduced to zero. 
The magnitude of the impulse is independent of the sign and is 3 kg · m/s. 
Impulse is also equal to the average force exerted on the ball times the time interval during which the 
force is exerted. We can use the formula: ??? = ?? ??????
??? 
Where ?? ??????
 is the average force and ??? is the time interval of 0.1 s. Re-arranging the formula to solve 
for ?? ??????
 gives us: ?? ??????
=
??? ??? . Substituting the known values we have: ?? avg 
=
3
0.1
= 30 N. 
The magnitude of the average force exerted by the hand of the player to catch the ball is 30 N. 
Q3: A light string passing over a smooth light fixed pulley connects two blocks of masses ?? ?? and ?? ?? . 
If the acceleration of the system is ?? /?? , then the ratio of masses is: 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Laws of Motion 
2024 
Q1: A body of mass ?? ???? experiences two forces ?? ? ? 
?? = ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 and ?? ? ? 
?? = ?? ??ˆ - ?? ??ˆ - ?? ?? ˆ
. The 
acceleration acting on the body is : 
A. ?? ??ˆ + ??ˆ + ?? ˆ
 
Correct Answer 
B. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 
C. -?? ??ˆ - ??ˆ - ?? ˆ
 
D. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
To find the acceleration acting on the body, we first need to determine the resultant force acting on the 
body by adding the two forces ?? 
1
 and ?? 
2
 vectorially. Then, we apply Newton's second law of motion, 
which states that the acceleration ??  of a body is directly proportional to the total force ?? 
 acting on it 
and inversely proportional to the mass ?? of the body: 
?? 
= ?? · ??  
or 
?? =
?? 
?? 
Let's start by adding the forces: 
?? 
1
+ ?? 
2
= (5??ˆ + 8??ˆ + 7?? ˆ
) + (3??ˆ - 4??ˆ - 3?? ˆ
) 
Performing the addition component-wise: 
?? 
total 
= (5 + 3)??ˆ + (8 - 4)??ˆ + (7 - 3)?? ˆ
?? 
total 
= 8??ˆ + 4??ˆ + 4?? ˆ
 
Now, let's use the formula for acceleration with ?? = 4 kg : 
?? =
?? 
0?????? ?? =
8??ˆ + 4??ˆ + 4?? ˆ
4 kg
 
Divide each component by the mass: 
?? = 2??ˆ + 1??ˆ + 1?? ˆ
 
So, the acceleration acting on the body is: 
?? = 2??ˆ + ??ˆ + ?? ˆ
 
Thus, the correct option is: 
Option A : 
2??ˆ + ??ˆ + ?? ˆ
 
Q2: A cricket player catches a ball of mass ?????? ?? moving with ???? ?? /?? speed. If the catching process is 
completed in ?? .?? ?? then the magnitude of force exerted by the ball on the hand of player will be (in SI 
unit) : 
A. 30 
B. 24 
C. 12 
D. 25 
Ans: (a) 
The first step in solving this problem is to calculate the change in momentum of the ball when it is 
caught. The change in momentum, or impulse, is the product of the mass of the ball and the change in 
velocity (as momentum is mass times velocity). 
The ball is initially moving with a velocity of ?? ?? = 25 m/s before the catch and finally comes to rest with 
a velocity of ?? ?? = 0 m/s after the catch. Since the ball is caught, the final velocity is zero. The change in 
velocity ??? = ?? ?? - ?? ?? = 0 - 25 = -25 m/s. Remember that the direction of the force exerted by the 
ball on the hand will be opposite to the direction of the ball's initial motion. 
The mass of the ball ?? is given as 120 g which needs to be converted into kilograms to maintain Sl 
units: 
?? = 120 g = 120× 10
-3
 kg = 0.12 kg.  
Now we can calculate the change in momentum (impulse): ??? = ?? ??? = 0.12 kg × (-25 m/s). 
Substituting the values we get: ??? = 0.12× -25 = -3 kg · m/s. 
The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial 
motion, which makes sense because the ball's velocity is reduced to zero. 
The magnitude of the impulse is independent of the sign and is 3 kg · m/s. 
Impulse is also equal to the average force exerted on the ball times the time interval during which the 
force is exerted. We can use the formula: ??? = ?? ??????
??? 
Where ?? ??????
 is the average force and ??? is the time interval of 0.1 s. Re-arranging the formula to solve 
for ?? ??????
 gives us: ?? ??????
=
??? ??? . Substituting the known values we have: ?? avg 
=
3
0.1
= 30 N. 
The magnitude of the average force exerted by the hand of the player to catch the ball is 30 N. 
Q3: A light string passing over a smooth light fixed pulley connects two blocks of masses ?? ?? and ?? ?? . 
If the acceleration of the system is ?? /?? , then the ratio of masses is: 
 
 
A. 
?? ?? 
B. 
?? ?? 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
?? =
(?? 1
- ?? 2
)?? (?? 1
+ ?? 2
)
=
?? 8
8?? 1
- 8?? 2
= ?? 1
+ ?? 2
7?? 1
= 9?? 2
?? 1
?? 2
=
9
7
 
Q4: Three blocks ?? ,?? and ?? are pulled on a horizontal smooth surface by a force of ???? ?? as shown in 
figure 
 
The tensions ?? ?? and ?? ?? in the string are respectively: 
A. ???? ?? ,???? ?? 
B. ???? ?? ,???? ?? 
C. ???? ?? ,?????? ?? 
D. ???? ?? ,???? ??   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (a) 
 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Laws of Motion 
2024 
Q1: A body of mass ?? ???? experiences two forces ?? ? ? 
?? = ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 and ?? ? ? 
?? = ?? ??ˆ - ?? ??ˆ - ?? ?? ˆ
. The 
acceleration acting on the body is : 
A. ?? ??ˆ + ??ˆ + ?? ˆ
 
Correct Answer 
B. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 
C. -?? ??ˆ - ??ˆ - ?? ˆ
 
D. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
To find the acceleration acting on the body, we first need to determine the resultant force acting on the 
body by adding the two forces ?? 
1
 and ?? 
2
 vectorially. Then, we apply Newton's second law of motion, 
which states that the acceleration ??  of a body is directly proportional to the total force ?? 
 acting on it 
and inversely proportional to the mass ?? of the body: 
?? 
= ?? · ??  
or 
?? =
?? 
?? 
Let's start by adding the forces: 
?? 
1
+ ?? 
2
= (5??ˆ + 8??ˆ + 7?? ˆ
) + (3??ˆ - 4??ˆ - 3?? ˆ
) 
Performing the addition component-wise: 
?? 
total 
= (5 + 3)??ˆ + (8 - 4)??ˆ + (7 - 3)?? ˆ
?? 
total 
= 8??ˆ + 4??ˆ + 4?? ˆ
 
Now, let's use the formula for acceleration with ?? = 4 kg : 
?? =
?? 
0?????? ?? =
8??ˆ + 4??ˆ + 4?? ˆ
4 kg
 
Divide each component by the mass: 
?? = 2??ˆ + 1??ˆ + 1?? ˆ
 
So, the acceleration acting on the body is: 
?? = 2??ˆ + ??ˆ + ?? ˆ
 
Thus, the correct option is: 
Option A : 
2??ˆ + ??ˆ + ?? ˆ
 
Q2: A cricket player catches a ball of mass ?????? ?? moving with ???? ?? /?? speed. If the catching process is 
completed in ?? .?? ?? then the magnitude of force exerted by the ball on the hand of player will be (in SI 
unit) : 
A. 30 
B. 24 
C. 12 
D. 25 
Ans: (a) 
The first step in solving this problem is to calculate the change in momentum of the ball when it is 
caught. The change in momentum, or impulse, is the product of the mass of the ball and the change in 
velocity (as momentum is mass times velocity). 
The ball is initially moving with a velocity of ?? ?? = 25 m/s before the catch and finally comes to rest with 
a velocity of ?? ?? = 0 m/s after the catch. Since the ball is caught, the final velocity is zero. The change in 
velocity ??? = ?? ?? - ?? ?? = 0 - 25 = -25 m/s. Remember that the direction of the force exerted by the 
ball on the hand will be opposite to the direction of the ball's initial motion. 
The mass of the ball ?? is given as 120 g which needs to be converted into kilograms to maintain Sl 
units: 
?? = 120 g = 120× 10
-3
 kg = 0.12 kg.  
Now we can calculate the change in momentum (impulse): ??? = ?? ??? = 0.12 kg × (-25 m/s). 
Substituting the values we get: ??? = 0.12× -25 = -3 kg · m/s. 
The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial 
motion, which makes sense because the ball's velocity is reduced to zero. 
The magnitude of the impulse is independent of the sign and is 3 kg · m/s. 
Impulse is also equal to the average force exerted on the ball times the time interval during which the 
force is exerted. We can use the formula: ??? = ?? ??????
??? 
Where ?? ??????
 is the average force and ??? is the time interval of 0.1 s. Re-arranging the formula to solve 
for ?? ??????
 gives us: ?? ??????
=
??? ??? . Substituting the known values we have: ?? avg 
=
3
0.1
= 30 N. 
The magnitude of the average force exerted by the hand of the player to catch the ball is 30 N. 
Q3: A light string passing over a smooth light fixed pulley connects two blocks of masses ?? ?? and ?? ?? . 
If the acceleration of the system is ?? /?? , then the ratio of masses is: 
 
 
A. 
?? ?? 
B. 
?? ?? 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
?? =
(?? 1
- ?? 2
)?? (?? 1
+ ?? 2
)
=
?? 8
8?? 1
- 8?? 2
= ?? 1
+ ?? 2
7?? 1
= 9?? 2
?? 1
?? 2
=
9
7
 
Q4: Three blocks ?? ,?? and ?? are pulled on a horizontal smooth surface by a force of ???? ?? as shown in 
figure 
 
The tensions ?? ?? and ?? ?? in the string are respectively: 
A. ???? ?? ,???? ?? 
B. ???? ?? ,???? ?? 
C. ???? ?? ,?????? ?? 
D. ???? ?? ,???? ??   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (a) 
 
 
 
 
 
Q5: All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. 
The acceleration of the block of mass ?? ???? is : 
 
A. 
?? ?? 
B. 
?? ?? 
C. ?? 
D. 
?? ??     [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: (d) 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Laws of Motion 
2024 
Q1: A body of mass ?? ???? experiences two forces ?? ? ? 
?? = ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 and ?? ? ? 
?? = ?? ??ˆ - ?? ??ˆ - ?? ?? ˆ
. The 
acceleration acting on the body is : 
A. ?? ??ˆ + ??ˆ + ?? ˆ
 
Correct Answer 
B. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 
C. -?? ??ˆ - ??ˆ - ?? ˆ
 
D. ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
To find the acceleration acting on the body, we first need to determine the resultant force acting on the 
body by adding the two forces ?? 
1
 and ?? 
2
 vectorially. Then, we apply Newton's second law of motion, 
which states that the acceleration ??  of a body is directly proportional to the total force ?? 
 acting on it 
and inversely proportional to the mass ?? of the body: 
?? 
= ?? · ??  
or 
?? =
?? 
?? 
Let's start by adding the forces: 
?? 
1
+ ?? 
2
= (5??ˆ + 8??ˆ + 7?? ˆ
) + (3??ˆ - 4??ˆ - 3?? ˆ
) 
Performing the addition component-wise: 
?? 
total 
= (5 + 3)??ˆ + (8 - 4)??ˆ + (7 - 3)?? ˆ
?? 
total 
= 8??ˆ + 4??ˆ + 4?? ˆ
 
Now, let's use the formula for acceleration with ?? = 4 kg : 
?? =
?? 
0?????? ?? =
8??ˆ + 4??ˆ + 4?? ˆ
4 kg
 
Divide each component by the mass: 
?? = 2??ˆ + 1??ˆ + 1?? ˆ
 
So, the acceleration acting on the body is: 
?? = 2??ˆ + ??ˆ + ?? ˆ
 
Thus, the correct option is: 
Option A : 
2??ˆ + ??ˆ + ?? ˆ
 
Q2: A cricket player catches a ball of mass ?????? ?? moving with ???? ?? /?? speed. If the catching process is 
completed in ?? .?? ?? then the magnitude of force exerted by the ball on the hand of player will be (in SI 
unit) : 
A. 30 
B. 24 
C. 12 
D. 25 
Ans: (a) 
The first step in solving this problem is to calculate the change in momentum of the ball when it is 
caught. The change in momentum, or impulse, is the product of the mass of the ball and the change in 
velocity (as momentum is mass times velocity). 
The ball is initially moving with a velocity of ?? ?? = 25 m/s before the catch and finally comes to rest with 
a velocity of ?? ?? = 0 m/s after the catch. Since the ball is caught, the final velocity is zero. The change in 
velocity ??? = ?? ?? - ?? ?? = 0 - 25 = -25 m/s. Remember that the direction of the force exerted by the 
ball on the hand will be opposite to the direction of the ball's initial motion. 
The mass of the ball ?? is given as 120 g which needs to be converted into kilograms to maintain Sl 
units: 
?? = 120 g = 120× 10
-3
 kg = 0.12 kg.  
Now we can calculate the change in momentum (impulse): ??? = ?? ??? = 0.12 kg × (-25 m/s). 
Substituting the values we get: ??? = 0.12× -25 = -3 kg · m/s. 
The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial 
motion, which makes sense because the ball's velocity is reduced to zero. 
The magnitude of the impulse is independent of the sign and is 3 kg · m/s. 
Impulse is also equal to the average force exerted on the ball times the time interval during which the 
force is exerted. We can use the formula: ??? = ?? ??????
??? 
Where ?? ??????
 is the average force and ??? is the time interval of 0.1 s. Re-arranging the formula to solve 
for ?? ??????
 gives us: ?? ??????
=
??? ??? . Substituting the known values we have: ?? avg 
=
3
0.1
= 30 N. 
The magnitude of the average force exerted by the hand of the player to catch the ball is 30 N. 
Q3: A light string passing over a smooth light fixed pulley connects two blocks of masses ?? ?? and ?? ?? . 
If the acceleration of the system is ?? /?? , then the ratio of masses is: 
 
 
A. 
?? ?? 
B. 
?? ?? 
C. 
?? ?? 
D. 
?? ??              [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (b) 
?? =
(?? 1
- ?? 2
)?? (?? 1
+ ?? 2
)
=
?? 8
8?? 1
- 8?? 2
= ?? 1
+ ?? 2
7?? 1
= 9?? 2
?? 1
?? 2
=
9
7
 
Q4: Three blocks ?? ,?? and ?? are pulled on a horizontal smooth surface by a force of ???? ?? as shown in 
figure 
 
The tensions ?? ?? and ?? ?? in the string are respectively: 
A. ???? ?? ,???? ?? 
B. ???? ?? ,???? ?? 
C. ???? ?? ,?????? ?? 
D. ???? ?? ,???? ??   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (a) 
 
 
 
 
 
Q5: All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. 
The acceleration of the block of mass ?? ???? is : 
 
A. 
?? ?? 
B. 
?? ?? 
C. ?? 
D. 
?? ??     [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: (d) 
 
40 - 2 T = 4a
T - 10 = 4a ? 20 = 12a
 ? a = 5/3 ? 2a =
?? 3