JEE Main Physics Kinematics Previous Year Questions

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)

1. 10 m
2. 30 m
3. 20 m
4. 40 m

Solution: Suppose both collide at the point P after time t. Time taken for the particles to collide,

t = d/v_rel = 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)

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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

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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.

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Q9. From a building, two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If v_A and v_B are their respective velocities on reaching the ground, then

(a) v_B > v_A

(b) v_A = v_B

(c) v_A > v_B

(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.

v_A ² = u² + 2gh …(1)

Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h v_B 2 = u² + 2gh …(2)

From (1) and (2)

v_A = v_B

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

1. 2gH = nu² (n-2)
2. gH = (n-2)u²
3. 2gH = n²u²
4. 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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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.

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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.

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]

Ans.

Given:

• Mass of particle P, mP = 20 g = 0.02 kg
• Mass of particle Q, mQ = 40 g = 0.04 kg
• Initial speed of both particles: 49 m/s
• Angle of projection of P: 45°
• Angle of projection of Q: 135°
• Horizontal separation between points A and B: 245 m
• Acceleration due to gravity, g = 9.8 m/s²

Step 1: Initial velocity components

Particle P:

• uPx = 49 × cos 45° = 34.65 m/s
• uPy = 49 × sin 45° = 34.65 m/s

Particle Q:

• uQx = 49 × cos 135° = -34.65 m/s
• uQy = 49 × sin 135° = 34.65 m/s

Step 2: Time of collision (tc)

At collision, horizontal positions of P and Q are equal:

34.65 × tc = 245 – 34.65 × tc

69.3 × tc = 245 ⇒ tc = 3.535 seconds

Step 3: Vertical position at collision

For both particles, vertical position at tc:

y = 34.65 × 3.535 – 0.5 × 9.8 × (3.535)² = 61.2 meters

Step 4: Velocities just before collision

• Velocity of P: vPx = 34.65 m/s, vPy = 0 m/s
• Velocity of Q: vQx = -34.65 m/s, vQy = 0 m/s

Step 5: Velocity of P after collision

P retraces its path, so:

vPx’ = -34.65 m/s, vPy’ = 0 m/s

Step 6: Velocity of Q after collision (using conservation of momentum)

0.02 × 34.65 + 0.04 × (-34.65) = 0.02 × (-34.65) + 0.04 × vQx’

-0.693 = -0.693 + 0.04 × vQx’ ⇒ vQx’ = 0 m/s

Vertical velocity of Q remains vQy’ = 0 m/s

Step 7: Position of Q when it hits the ground

Q falls vertically from height 61.2 m with zero horizontal velocity.

Time to fall:

tf = √(2 × 61.2 / 9.8) = 3.535 seconds

Step 8: Horizontal position of Q when it lands

Position remains constant during fall:

xQ = 245 – 34.65 × 3.535 = 122.6 meters

Final Answers:

• Position of Q on ground: 122.6 meters from point A
• Time taken by Q to reach ground after collision: 3.535 seconds

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.

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

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]

Ans. True

Explanation:

Since air resistance is neglected, the only force acting on both balls is gravity, which accelerates them equally regardless of their masses. Therefore, both balls will return to the point of projection with the same speed they were thrown upwards, irrespective of their different masses.

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]

Ans. True

Explanation:
Since air resistance is neglected, the only force acting on both balls is gravity, which accelerates them equally regardless of their masses. Therefore, both balls will return to the point of projection with the same speed they were thrown upwards, irrespective of their different masses.

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]

Ans. False

Explanation:

Because of the Earth’s rotation, trains moving in opposite directions along the equator experience different effective gravitational forces due to the Coriolis effect and centrifugal force. The train moving eastward (in the direction of Earth’s rotation) experiences a slightly reduced effective weight, while the train moving westward experiences a slightly increased effective weight. As pressure depends on the normal force (weight) exerted on the rails, the pressures will not be exactly the same.

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]

Ans. Displacement: The displacement is the straight-line distance between the starting point and the point after half a revolution, which is the diameter of the circle.

Displacement=2R

Distance covered: The distance covered is half the circumference of the circle.

Distance=πR

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]

Ans. The four persons will meet at the center of the square after a time

t= v/d

Explanation:

Each person moves towards the next with speed v, and due to symmetry, their paths curve inward, forming a logarithmic spiral. The distance between them decreases uniformly, and they meet exactly after time t= v/d