Kinematics JEE Main Questions 2025: Important Practice Questions with PYQs

1.

A piece of wood of mass 0.03 kg is dropped from the top of a 100m height building. At the same time, a bullet of mass 0.02 Kg is fired vertically upward, with a velocity of 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

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)

2.

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

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)

3.

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

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)

4.

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

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)

5.

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 a vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass, hung vertically from a 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:

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.

Answer: i) The cart is moving in x-y plane

The stone, thrown from the cart, travels in the y-z plane while the z-axis is the 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. The velocity of the cart is along the x-axis

ii) It is given that the tension in the string becomes zero at the horizontal position. The combined mass therefore is at rest in this position. During the subsequent motion of the combined mass, the energy is conserved.

6.

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

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)

7.

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

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)

8.

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

False. One train is moving in the direction of the earth’s rotation and the other is moving in the opposite direction. Hence, they will exert different pressure on the rails.

9.

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)

The 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)

10.

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.

11.

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.

Answer: 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

12.

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

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

13.

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

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)

14.

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

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 the opposite directions, 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)

15.

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 bail out?

(a) 293 m

(b) 111 m

(c) 91 m

(d) 182 m

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.