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JEE Main 2024 Physics Kinematics Previous Year Questions with Solutions
For JEE Main other Engineering Entrance Exam Preparation, JEE Main Physics Kinematics Previous Year Questions with Solutions is given below.
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Q1. A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 Kg is fired vertically upward, with a velocity 100 m/s, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is (g = 10 m/s)
- 10 m
- 30 m
- 20 m
- 40 m
Solution: Suppose both collide at the point P after time t. Time taken for the particles to collide,
t = d/vrel = 100/100 = 1s
Speed of wood just before collision =gt = 10m/s
Speed of bullet just before collision
v -gt = 100 -10 = 90 m/s
Before
0.03 kg ↓ 10 m/s
0.02 kg ↑ 90 m/s
After
↑ v
0.05 kg
Now, the conservation of linear momentum just before and after the collision
-(0.03)(10) + (0.02)(90) = (0.05)v ⇒ v = 30 m/s
The maximum height reached by the body a = v2/2g
= (30)2/2(10)
= 45 m
(100 -h) = ½ gt2 = ½ x 10 x1 ⇒h = 95 m
Height above tower = 40 m
Answer: (d)
Q2. A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite direction, is
(a) 25/11
(b) 3/2
(c) 5/2
(d) 11/5
Solution:
The total distance to be travelled by train is 60 + 120 = 180 m.
When the trains are moving in the same direction, the relative velocity is v1 – v2 = 80 – 30 = 50 km hr–1.
So time taken to cross each other, t1 = 180/(50 x 103/3600) = [(18 x 18)/25] s
When the trains are moving in opposite direction, relative velocity is |v1 – (–v2 )| = 80 + 30 = 110 km hr–1
So time taken to cross each other
t2= 180/(110 x 103/3600) = [(18 x 36)/110] s
t1/t2= [(18 x 18)/25] / [(18 x 36)/110] = 11/5
Answers: (d)
Q3. An automobile travelling at 40 km/h can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding)
(a) 100 m
(b) 75 m
(c) 160 m
(d) 150 m
Solution :
Using v2 = u2 – 2as
0 = u2 – 2as
S = u2 /2a
S1/S2 = u12/u22
S2 = (u12/u22)S1 = (2)2(40) = 160 m
Answer: (c)
Q4. A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bailout?
(a) 293 m
(b) 111 m
(c) 91 m
(d) 182 m
Solution:
Initially, the parachutist falls under gravity u 2 = 2ah = 2 × 9.8 × 50 = 980 m2s –2
He reaches the ground with speed = 3 m/s,
a = –2 m s–2 ⇒ (3)2 = u 2 – 2 × 2 × h1
9 = 980 – 4 h1
h1 = 971/4
h1 = 242.75 m
Total height = 50 + 242.75 = 292.75 = 293 m.
Q5. A car, starting from rest, accelerates at the rate f through a distance s, then continues at a constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed in 15 s, then
(a) s= ½ft2
(b) s= (¼)ft2
(c) s = ft
(d) s= (1/72)ft2
Solution: For the first part of the journey, s = s1,
s1 = ½ ft12………………………(1)
v = f t1 …………………………(2)
For second part of journey,
s2 = vt or s2 = f t1 t ……………(3)
For the third part of the journey,
s3 = ½(f/2)(2t1)2= ½ x ft12
s3 = 2s1 = 2s ………………….(4)
s1 + s2 + s3 =15s
Or s + ft1t + 2s = 15s
ft1t = 12 s
From (1) and (5) we get
(s/12 s )= ft12/(2 x ft1t)
Or t1 = t/6
Or s= ½ ft12
s= ½ f(t/6)2
s= ft2/72
Answer: (d) s= (1/72)ft2
Q6. An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e., 120 km/h, the stopping distance will be
(a) 20 m
(b) 40 m
(c) 60 m
(d) 80 m
Solution:
Let a be the retardation for both the vehicles.
For automobile, v 2 = u 2 – 2as
u1 2 – 2as1 = 0
u1 2 = 2as1
Similarly for car, u2 2 = 2as2
(u2/u1)2 = s2/s1 = (120/60)2 = s2/20
S2 = 80 m
Answer: (d)
Q7. A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in T/3 second?
(a) (h/9)metre from the ground
(b) (7h/9) metre from the ground
(c) (8h/9) metre from the ground
(d) (17h/18) metre from the ground
Solution: Equation of motion
s= ut + gt2
h = 0 + ½ gT2
Or 2h = gT2………(1)
After T/3 sec, s = 0 +½ x g(T/3)2= gT2/18
18 s = gT2 …………(2)
From (1) and (2), 18 s = 2h
S = (h/9) m from top.
Height from ground = h – h/9 = (8h/9) m
Answer: (c)
Q8. Which of the following statements is false for a particle moving in a circle with a constant angular speed?
(a) The velocity vector is tangent to the circle
(b) The acceleration vector is tangent to the circle
(c) The acceleration vector points to the centre of the circle
(d) The velocity and acceleration vectors are perpendicular to each other
Answer: (b) The acceleration vector acts along the radius of the circle. The given statement is false.
Q9. From a building, two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If vA and vB are their respective velocities on reaching the ground, then
(a) vB > vA
(b) vA = vB
(c) vA > vB
(d) their velocities depend on their masses
Solution
Ball A projected upwards with velocity u, falls back with velocity u downwards. It completes its journey to the ground under gravity.
vA 2 = u2 + 2gh …(1)
Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h vB 2 = u2 + 2gh …(2)
From (1) and (2)
vA = vB
Answer: (b)
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Q10: If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Solution
For first part of penetration, by equation of motion,
(u/2)2 = u2 -2a(3)
3u2 = 24a ⇒ u2 = 8a …(1)
For latter part of penetration,
0= (u/2)2 -2ax
or u2 = 8ax……………(2)
From (1) and (2)
8ax = 8a x = 1 cm
Answer: (a)
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Q11. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is
(a) gH = (n – 2)u2
(b) 2gH = n 2u2
(c) gH = (n – 2)2u2
(d) 2gH = nu2(n – 2)
Solution
Time taken by the particle to reach the topmost point is,
t =u/g … (1)
Time taken by the particle to reach the ground = nt
Using, s = ut + ½ at2
-H = u(nt) – ½ g(nt)2
-H = u x n (u/g) = ½ g2(u/g)2 [using (1)]
⇒-2gH = 2nu2 – n2u2
⇒ 2gH = nu2(n -2)
Answer: (d) 2gH = nu2(n – 2)
Q12: For a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H,u and n is
- 2gH = nu2 (n-2)
- gH = (n-2)u2
- 2gH = n2u2
- gH = (n – 2)2u2
Solution:
Time to reach highest point = t = u/g
Time to reach ground = nt
S = ut + ½ at2
-H = u(nt) – ½ g (nt)2
2gH = nu2 (n-2)
Answer: (a)
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Multiple Choice with ONE or More Than ONE correct answers
A particle is moving eastwards with a velocity of 5m/s. In 10s the velocity changes to 5m/s northwards. The average acceleration in this time is [1982-3 marks]
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12. The co-ordinates of a particle moving in a plane are given by x (t) = a cos (pt) and y (t) = b sin (pt) where a, b (< a) & p are positive constants of appropriate dimensions. Then [1999-3 marks]
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13. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of V J contact, B is the centre of a the sphere and C is topmost point, Then,[2009]
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14. Column-I describes some situations in which a small object moves. Colymn-H describes some characteristics of these motions. Match the situations in column-I with the characteristics in column-II. [2007-6 marks]
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15. Column-I gives a list of possible set of parameters measured in some experiments.The variations of the parameters in the form of graphs are shown in Column-II. Match the set of parameters given in Column-I with the graphs given in Column-II.
Assertion & Reasoning type
Instructions : The following question contains statement-I (assertion) and statement -2 (reason) of these statements, mark correct choice if
a) Statement-1 and 2 are true and statement-2 is a correct explanation for statement-1
b) Statement-l and 2 are true and statement-2 is not a correct explanation for statement-1
c) Statement-1 is true, statement-2 is false
d) Statement-1 is false, statement-2 is true.
16. Statement -1 : For an observer looking out through the window of a fast moving train the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.
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18. Is the time variation of position shown in the figure observed in nature ? [1979]
Ans. If we draw a straight line parallel to x-axis, it cuts the given figure at 2 points indicating that at one time, the,particle is at two different positions which is absurd. Thus such a motion as shown in the figures is never observed in nature.
19. Particles P and Q of mass 20 g and 40g respectively are simultaneously projected from points A and B on the ground.The initial velocities of P and Q make 45° and 135° angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49m/s. The separation AB is 245m. Both particle travel in the same vertical plane and undergo a collision. After the collision, P retraces its path. Determine the position of Q when it hits the ground. How much time after the collision does the particle Q take to reach the ground ? Take g = 9.8m / s2 [1982-8 marks]
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20. A body falling freely from a given height H hits an inclined plane in its path at a height h. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of (h/H) the body will take maximum time to reach the ground?
Ans. PQR is an inclined plane. The body falls under gravity vertically from A to B till it strikes the inclined plane. At B, its velocity is horizontal. From B to C, the body follows a parabolic path. At C, it reaches the ground.
21. Two towers AB and CD are situated a distance d apart as shown in figure. AB is 20 m high and CD is 30m high from the ground. An object of mass m is thrown from the top of AB horizontally with a velocity of lOm/s towards CD. Simultaneously another object of mass 2m is thrown from the top of CD at an angle of 60°t0 the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick to each other [1994-4 marks]
i) Calculate the distance d between the towers and,
ii) Find the position where the objects hit the ground.
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23. A cart is moving along x-direction with a velocity of 4m/s. A person on the cart, throws a stone with a velocity of 6m/s, relative to himself. In the frame of reference of the cart, the stone is thrown in y-z plane making an angle of 30° with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass, hung vertically from branch of a tree, by means of a string of length L. A completely inelastic collision occurs in which the stone gets embedded in the object. Determine:[1997-5 marks]
i) the speed of the combined mass immediately after the collision with respect to an observer on the ground
ii)the length L of the string such that tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.
Ans.i) The cart is moving in x-y plane
The stone, thrown from cart, travels in y-z plane while z-axis is vertical axis. The stone makes an angle of 30° with z-axis. Its path is parabolic. At the highest point of its trajectory, the vertical velocity of the stone will be zero. The velocity of the stone is thus confined to (x, y) plane at the highest point. Velocity of cart is along x-axis
ii) It is given that the tension in the string becomes zero at horizontal position. The combined mass therefore is at rest in this position. During the subsequent motion of the combined mass, the energy is conserved.
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26. A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical line AB is at a distance L/8 from O as shown in figure. The object is given a horizontal velocity u. At some point, its motion ceases to be circular and eventually the object passes through the line AB. At the instant of crossing AB, its velocity is horizontal. Find u. [1999-10 marks]
Ans. OP denotes string of length L. The particle starts from P with a horizontal velocity u. It travels along a circular path till Q. Then it passes the line AB. At Q, motion ceases to be circular. At C, its velocity is horizontal, where the particle crosses AB. After Q, the particle performs a projectile motion. At C, the velocity becomes horizontal. Thus C is at the highest point of projectile motion
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29.
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True / False Type
30. Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance).[1983-2 marks]
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31.’A projectile fired from the ground follows a parabolic path. The speed of the projectile, is minimum at the top of its path. [1984-2 marks]
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32.Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails [1985-2 marks]
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Fill in the blanks
33. A particle moves in a circle of radius R. In half the period of revolution its displacement is………. and distance covered is……… ……….. [1983-2 marks]
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34. Four persons K,L,M,N are initially at the four corners of a square of side d. Each person now moves with a uniform speed vin such a way that K always moves directly towards L, L directly towards M,M directly towards N, and N directly towards K. The four persons will meet at a time ………….. [1984-2 marks]
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36. The trajectory of a projectile in a vertical plane is y = ax-bx2,where a, b are constants, and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. The maximum height attained is……….and the angle of projection from the horizontal is [1997-2 marks]
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