datavorous/entrance-exam-dataset huggingface dataset:
A ball is projected from the ground with a speed 15 m s-1 at an angle θ with horizontal so that its range and maximum height are equal, then tanθ will be equal to

Options:
A 14
B 12
C 2
D 4

The correct answer is:
4

Solution:
The range and the maximum height of the projectile are given as below,

R=u2sin2θg=2u2sinθcosθgH=u2sin2θ2g

Therefore,

As H=R

⇒tanθ=4

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A ball is projected upwards. Its acceleration at the highest point is

Options:
A Zero
B directed upwards
C directed downwards
D such as cannot be predicted

The correct answer is:
directed downwards

Solution:
A force is only required to maintain an acceleration. And in the case of a projectile that is moving upward, there is a downward force and a downward acceleration. That is, the object is moving upward and slowing down.

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
A ball is projected vertically up from ground. Boy 'A' standing at the window of first floor of a nearby building observes that the time interval between the ball crossing him while going up and the ball crossing him while going down is 2 s. Another boy B standing on the second floor notices the time interval between the ball passing him twice, during up and down motion is 1 s. Calculate the difference between the vertical positions of boy B and boy A. (Assume g=10 m s-2)

Options:
A 8.45 m
B 3.75 m
C 4.25
D 2.50 m

The correct answer is:
3.75 m

Solution:
Let's assume v1 and v2 are the speeds of the ball while crossing the Boy A and B respectively during its motion.

Applying the equation of motion, S=ut+12at2 for the motion of ball

while passing the Boy A is

⇒ 0=v12-121022

⇒ v1=10 m s-1.

while passing the Boy B is

⇒ 0=v21-121012

⇒ v1=5 m s-1.

The difference between the vertical positions of the boys A and B is

v22=v12-2gh

⇒ 52=102-210 h

⇒ h=100-2520=3.75 m.

Topic: Physics, Work Power Energy, JEE Main
A ball is projected vertically upward with an initial velocity of 50 m s-1 at t=0 s. At t=2 s, another ball is projected vertically upward with same velocity. At t=____s, second ball will meet the first ball g=10 m s-2.

Options:

The correct answer is:
6

Solution:
Distance covered by ball A is h=502-12×10×22=100-20=80 m

Now, at time t=2 s, velocity of A is vA=50-10×2=30 m s-1

Relative velocity of both balls after 2 s is vrel=50-30=20 m s-1

Here, relative displacement xrel=80 m s-1 and relative accelerationarel=0.

Time taken t=8020=4 s

Time at which they will meet is t=4+2=6 s

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A ball is projected vertically upwards from ground. It reaches a height ' $h$ ' in time $t_1$, continues its motion and then takes a time $t_2$ to reach ground. The height $h$ in terms of $g, t_1$ and $t_2$ is $(g=$ acceleration due to gravity)

Options:
A $\frac{1}{2} \frac{\mathrm{gt}_1}{\mathrm{t}_2}$
B $\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2$
C $\mathrm{g} \mathrm{t}_1 \mathrm{t}_2$
D $2 \mathrm{gt}_1 \mathrm{t}_2$

The correct answer is:
$\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2$

Solution:
We know,
$$
\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2
$$
The total time required for the ball to go up and reach the ground is $t=t_1+t_2$, and the total displacement is zero.
$$
\begin{aligned}
& \therefore \quad 0=\mathrm{u}\left(\mathrm{t}_1+\mathrm{t}_2\right)+\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right)^2 \\
& \therefore \quad \mathrm{u}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right)
\end{aligned}
$$
The displacement in time $t_1$ is
$$
\begin{aligned}
\mathrm{h} & =\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_1+\mathrm{t}_2\right) \mathrm{t}_1-\frac{1}{2} \mathrm{gt}_1^2 \\
\mathrm{~h} & =\frac{1}{2} \mathrm{gt}_1\left(\mathrm{t}_1+\mathrm{t}_2-\mathrm{t}_1\right) \\
\therefore \quad \mathrm{h} & =\frac{1}{2} \mathrm{gt}_1 \mathrm{t}_2
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A ball is projected with a speed of 2 m sâ1 with respect to a trolley which is moving with 4 m sâ1, as shown in the figure. The angle at which ball is projected with respect to the horizontal in ground frame is

Options:
A tan-1⁡328
B tan-1⁡53
C sin-1⁡328
D sin-1⁡528

The correct answer is:
sin-1⁡328

Solution:
vâb=vâbT+vâT
tanâ¡Ï=2sinâ¡60o4+2cosâ¡60o=35
sinâ¡Ï=328
Ï=sin-1â¡328

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball is projected with kinetic energy E at an angle of $45^{\circ}$ to the horizontal. At the highest point during its flight, its kinetic energy will be

Options:
A Zero
B $\frac{E}{2}$
C $\frac{E}{\sqrt{2}}$
D $E$

The correct answer is:
$\frac{E}{2}$

Solution:
$E^{\prime \prime}=E \cos ^2 \theta=E \cos ^2\left(45^{\circ}\right)=\frac{E}{2}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball is projected with kinetic energy E, at an angle of 60° to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :

Options:
A Zero
B E2
C E4
D E

The correct answer is:
E4

Solution:
E=12mu2

At the Highest point, velocity is equal to the horizontal component of the velocity of projection. Therefore,

v=ucos60°=u2

∴ Kinetic energy at the topmost point =12 mu22=E4

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (29 Jul Shift 1)
A ball is released from a height h. If t1 and t2 be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t1 and t2.

Options:
A t1=2t2
B t1=2-1t2
C t2=2+1t1
D t2=2-1t1

The correct answer is:
t2=2-1t1

Solution:
As the body is dropped from a certain height h, its initial velocity will be zero.

For first half, h1=h2.

Using second equation of motion, s=ut+12at2

We have, h2=12gt12.

Time taken in travel first half of the height, t1=hg ...1.

For total height h, using again second equation of motion,

h=12gt1+t22⇒t1+t2=2hg ...2

From equation 1 and 2,

⇒12=t1t1+t2

⇒1+t2t1=2

⇒t1t2=12-1

Thus, the relation is t2=2-1t1.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A ball is released from the top of a tower of height $h$ metres. It takes $T$ seconds to reach the ground. What is the position of the ball in $\mathrm{T} / 3$ seconds?

Options:
A
$\mathrm{h} / 9$ metres from the ground

B
$7 \mathrm{~h} / 9$ metres from the ground

C
$8 \mathrm{~h} / 9$ metres from the ground

D
$17 \mathrm{~h} / 18$ metres from the ground.

The correct answer is:
$8 \mathrm{~h} / 9$ metres from the ground

Solution:
No Solution Available

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2004
A ball is spun with angular acceleration α=6t2-2t where t is in second and α is in rad s-2. At t=0, the ball has angular velocity of 10 rad s-1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is

Options:
A 32t4-t2+10t
B t42-t33+10t+4
C 2t43-t36+10t+12
D 2t4-t32+5t+4

The correct answer is:
t42-t33+10t+4

Solution:
Given: α=6t2-2t

Using relation α=dωdt=6t2-2t

Integrating the above,

∫10ωdω=∫0t6t2-2tdt

⇒ω-10=2t3-t2

Now, ω=dθdt=10+2t3-t2

∫4θdθ=∫0t10+2t3-t2dt

Integrating the above relation,

θ-4=10t+t42-t33

Thus, angular position of the ball is θ=t42-t33+10t+4.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A ball is thrown from a point with a speed $v_0$ at an angle of projection $\theta$. From the same point and at the same instant person starts running with a constant speed $v_0 / 2$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

Options:
A
yes, $60^{\circ}$

B
yes, $30^{\circ}$

C
no

D
yes, $45^{\circ}$

The correct answer is:
yes, $60^{\circ}$

Solution:
$\mathrm{v}_0 \cos \theta=\frac{\mathrm{v}_0}{2}$
$\Rightarrow \theta=60^{\circ}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2004
A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall, the angle of projection of ball is

Options:
A tan-1⁡32
B tan-1⁡23
C tan-1⁡12
D tan-1⁡34

The correct answer is:
tan-1⁡23

Solution:
We know that range of a projectile
R=u2sinâ¡2Î¸g
Here R=6+18=24
â´ u2sinâ¡2Î¸g=24 ...... (i)
The equation trajectory of a projectile
y=xtanâ¡Î¸-gx22u2cos2â¡Î¸
3=6tanâ¡Î¸-36g2u2cos2â¡Î¸ ...... (ii)
From equation (i), gu2=sinâ¡2Î¸24
= sinâ¡Î¸cosâ¡Î¸12
Substituting in equation (ii), we get
3=6tanâ¡Î¸-32tanâ¡Î¸
3=92tanâ¡Î¸
â Î¸=tan-1â¡23

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball is thrown horizontally from a height h above a staircase as shown in the figure. If the coefficient of restitution for any collision between the ball and the staircase is e, then the value of h for which the ball will bounce the same height above each step, is

Options:
A d1-e2
B d1+e2
C de21+e2
D de21-e2

The correct answer is:
d1-e2

Solution:
Vertical velocity of the ball just before each collision is

vapp=2gh

The height up to which the ball bounces after each collision is

h'=h-d

Vertical velocity of the ball just after the collision is

vsep=evapp=e2gh

Using the principle of conservation of mechanical energy we get

12me2gh2=mgh-d

12e22gh=gh-d

e2h=h-d

d=h1-e2

h=d1-e2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball is thrown horizontally from a height with a certain initial velocity at time $t=0$. The ball bounces repeatedly from the ground with the coefficient of restitution less than $1$ as shown.
Neglect air resistance and taking the upward direction as positive, which figure qualitatively depicts the vertical component of the ball's velocity $\left(V_{y}\right)$ as a function of time $(t)$?

Options:
A
B
C
D

The correct answer is:

Solution:
as $\mathrm{V_{y}=U_{y}-g t}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height h3 while going up and coming down respectively.

Options:
A 2-12+1
B 3-23+2
C 3-13+1
D 13

The correct answer is:
3-23+2

Solution:
The maximum height of a projectile thrown vertically upward is given by, h=u22g.

⇒u=2gh

Using equation of motion with constant acceleration,

s=ut+12at2

⇒h3=2ght-12gt2

⇒gt22-2ght+h3=0

Solving the above quadratic equation we will get the times t1 and t2. ⇒t2t1=2gh+2gh-4×g2×h32gh-2gh-4×g2×h3=2gh+4gh32gh-4gh3=3+23-2

Thus, the ratio t1t2=3-23+2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (29 Jul Shift 1)
A ball is thrown upward from the top of a building at angle of $30^{\circ}$ to the horizontal with an initial speed of 15 $\mathrm{ms}^{-1}$. If the ball hits the ground after $3 \mathrm{~s}$, then the height of building is (acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Options:
A $30 \mathrm{~m}$
B $12.5 \mathrm{~m}$
C $25.5 \mathrm{~m}$
D $22.5 \mathrm{~m}$

The correct answer is:
$22.5 \mathrm{~m}$

Solution:
$\frac{15}{2} \mathrm{~m} / \mathrm{s}$
Time taken by ball to reach the same height of building,
$\mathrm{t}=\frac{2 \mu \sin \theta}{\mathrm{g}}=\frac{2 \times 15 \times \sin 30^{\circ}}{10}=1.5 \mathrm{sec}$
Now, remaining $1.5 \mathrm{sec}$ will be taken by ball to reach the ground a Considering only vertical motion of ball
$\begin{aligned} & \mathrm{h}=\frac{15}{2} \times 1.5+\frac{1}{2} \times 10 \times 1.5^2 \\ & =22.5 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In One Dimension, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 1)
A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to mγv2 (where m is mass of the ball, v is its instantaneous velocity and γ is a constant). Time taken by the ball to rise to its zenith is:

Options:
A 1γgln⁡1+γgV0
B 1γgtan-1⁡γgV0
C 1γgsin-1⁡γgV0
D 12γgtan-1⁡2γgV0

The correct answer is:
1γgtan-1⁡γgV0

Solution:
Net force on ball,
F=-mg-mγv2
a=-g-γv2
dvdt=-g-γv2
∫v=V0v=0dvg+γv2=∫t=0t-dt
1gγtan-1γgvv=V0v=0=-t
∴t=1γgtan-1 γgV0

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A ball is thrown vertically upward from the ground at time, $t=0 \mathrm{~s}$. It passes the top of a tower at $t=3 \mathrm{~s}$ and $2 \mathrm{~s}$ later it reaches and its maximum height. The height of the tower is (Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $105 \mathrm{~m}$
B $125 \mathrm{~m}$
C $85 \mathrm{~m}$
D $65 \mathrm{~m}$

The correct answer is:
$105 \mathrm{~m}$

Solution:
According to question, ball reaches at top of a
tower at the time, $t_1=3 \mathrm{~s}$ and further reaches at the maximum height at time, $t_2=2 \mathrm{~s}$.
$\therefore$ Total time taken by ball to reached the maximum height,
$\begin{array}{rlrl} & & t & =t_1+t_2=3+2 \\ \therefore & & t=5 \mathrm{~s}\end{array}$
If $u$ be velocity of ball at time, $t=0$, then from the first equation of the motion,
$\begin{aligned} & v=u-g t \quad\left[\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right] \\ & 0=u-10 \times 5 \Rightarrow u=50 \mathrm{~m} / \mathrm{s}\end{aligned}$
If $h$ be the height of the tower, then from the second equation of the motion,
$\begin{aligned} h=u t-\frac{1}{2} \times g t^2 & =50 \times 3-\frac{1}{2} \times 10 \times 3^2 \\ & =150-45=105 \mathrm{~m}\end{aligned}$
Hence, the height of the tower is $105 \mathrm{~m}$.

Topic: Physics, Motion In One Dimension, JEE Main
A ball is thrown vertically upward with an initial velocity of 150 m s-1. The ratio of velocity after 3 s and 5 s is x+1x. The value of x is _____. {take, g=10 m s-2}

Options:
A 10
B -5
C 6
D 5

The correct answer is:
5

Solution:
The formula to calculate the velocity of the ball after 3 s is given by

v3=u-3g ...1

The formula to calculate the velocity of the ball after 5 s is given by

v5=u-5g ...2

Divide equation (1) by equation (2) to obtain the required ratio.

v3v5=u-3gu-5g ...3

Substitute the values of the known parameters into equation (3) to calculate the required ratio.

v3v5=150-30150-50=120100=65= 5+15

Comparing the above equation with the given expression, it can be written that x=5.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A ball is thrown vertically upwards with a velocity of 19.6 m s-1 from the top of a tower. The ball strikes the ground after 6 s. The height from the ground up to which the ball can rise will be k5 m. The value of k is _____ (use g=9.8 m s-2)

Options:

The correct answer is:
392

Solution:
Initial velocity of ball is u=19.6 m s-1, final velocity v=0.

Using v=u+at, here, acceleration a=-g.

Time taken in upward motion above tower is ta=ug=19.69.8=2 s

Now, time taken from top most point to ground is td=6-2=4 s.

Or, td=2hmaxg (Using s=ut+12gt2)

⇒hmax=16×9.82=3925 m.

Hence, the value of k=392.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A ball is thrown with a speed u, at an angle θ with the horizontal. At the highest point of its motion, the strength of gravity is somehow doubled. Taking this change into account, the total time of flight of the projectile is

Options:
A 2usinθg
B 32usinθg
C 34usinθg
D 2+12usinθg

The correct answer is:
2+12usinθg

Solution:
Time of ascent
t1=usinθg
Time of descent
t2=2H2g
t2=u2sin2θ2g2=usinθ2g
t=t1+t2=usinθg+usinθ2g
⇒usinθg2+12

Topic: Physics, Motion In Two Dimensions, JEE Main
A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The centre of mass

Options:
A Of the box remains constant
B Of the box plus the ball system remains constant
C Of the ball remains constant
D Of the ball relative to the box remains constant

The correct answer is:
Of the box plus the ball system remains constant

Solution:
Since no external force is present on the system. So, the centre of mass of the system will not be changed. Hence, the center of mass of the box plus the ball system remains constant.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball \( \mathrm{A} \) is thrown vertically upward with speed \( u \). At the same instant, another ball \( \mathrm{B} \) is released from rest at height \( h \). At time \( t \), the speed of A relative to B is (Assume both are in the air)

Options:
A \( u \)
B \( u-2 g t \)
C \( \sqrt{u^{2}-2 g h} \)
D \( u-g t \)

The correct answer is:
\( u-2 g t \)

Solution:
From equation of motion, we have
v=u+gt
where u is initial velocity and t is time.
For ball A,â v A =uâgt
For ball B,â v B =0+gt
â´ Relative speed
Î v â = v A â â v B â = u - gt j ^ - + gt j ^ = uâ2gt j ^

Topic: Physics, Motion In One Dimension, JEE Main
A ball moves one-fourth $\left(\frac{1^{\text {th }}}{4}\right)$ of a circle of radius $R$ in time $T$. Let $v_1$ and $v_2$ be the magnitudes of mean speed and mean velocity vector. The ratio $\frac{v_1}{v_2}$ will be

Options:
A $\frac{\pi}{2}$
B $\frac{3}{\pi}$
C $\frac{2}{\sqrt{3} \pi}$
D $\frac{\pi}{2 \sqrt{2}}$

The correct answer is:
$\frac{\pi}{2 \sqrt{2}}$

Solution:
Key idea Mean speed of a moving body,
$$
v_{\mathrm{rms}}=\frac{\text { total distance }}{\text { total time taken }}
$$
Mean velocity vector for a moving body,
$$
v_{w v}=\frac{\text { total displacement }}{\text { total time taken }}
$$
A ball moving in a circular arc is shown in the figure below,
Here, $v_{\mathrm{rms}}=\frac{2 \pi R}{4} \times \frac{1}{T}=v_1$
Similarly, $v_{\mathrm{vv}}=\frac{\sqrt{2} R}{T}=v_2$
Hence, $\frac{v_1}{v_2}=\frac{\pi}{2 \sqrt{2}}$
Option (4) is correct.

Topic: Physics, Work Power Energy, JEE Main
A ball moving with a velocity v hits a massive wall moving towards the ball with a velocity u. An elastic impact lasts for a time Δt.

Options:
A the average elastic force acting on the ball is mu+vΔt
B the average elastic force acting on the ball is 2mu+vΔt
C the kinetic energy of the ball increases by muu+v
D the kinetic energy of the ball remains the same after the collision.

The correct answer is:
the average elastic force acting on the ball is 2mu+vΔt

Solution:
As the wall is massive its velocity remains unchanged.
Speed of the ball after impact v¯f=v+2u
The average elastic force acting on the ball is,
F¯=mΔv¯Δt=mv¯f–v¯iΔt
F=mv+2u––v Δt=2mu+vΔt
F=2mu+vΔt

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball moving with velocity $2 \mathrm{~ms}^{-1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in $\mathrm{ms}^{-1}$ ) after collision will be

Options:
A 0,1
B 1,1
C $1,0.5$
D 0,2

The correct answer is:
0,1

Solution:
If two bodies collide head on with coefficient of restitution
$$
\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}
$$
From the law of conservation of linear momentum
$$
\begin{array}{r}
\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2 \\
\Rightarrow \quad \mathrm{v}_1=\left[\frac{\mathrm{m}_1-\mathrm{em}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{(1+\mathrm{e}) \mathrm{m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2
\end{array}
$$
Substituting $\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}$ and
$$
\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5
$$
we get
$$
\begin{array}{lc}
\text { we get } & \mathrm{v}_1=\left[\frac{\mathrm{m}-\mathrm{m}}{\mathrm{m}+2 \mathrm{~m}}\right] \times 2 \\
\Rightarrow & \mathrm{v}_1=0
\end{array}
$$
Similarly,
$$
\begin{aligned}
\mathrm{v}_2 & =\left[\frac{(1+\mathrm{e}) \mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{\mathrm{m}_2-\mathrm{em}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2 \\
& =\left[\frac{1.5 \times \mathrm{m}}{3 \mathrm{~m}}\right] \times 2 \\
& =1 \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A ball of 0.5 kg collided with wall at $30^{\circ}$ and bounced back elastically. The speed of ball was $12 \mathrm{~m} / \mathrm{s}$. The contact remained for 1 s . What is the force applied by wall on ball?

Options:
A $12 \sqrt{3} N$
B $\sqrt{3} N$
C $6 \sqrt{3} N$
D $3 \sqrt{3} N$

The correct answer is:
$6 \sqrt{3} N$

Solution:
Given, $\mathrm{m}=0.5 \mathrm{~kg}, \mathrm{v}=12 \mathrm{~m} / \mathrm{s}, \Delta \mathrm{t}=1 \mathrm{~s}, \theta=30^{\circ}$
Force applied by wall on ball, $\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}$
or $\quad \mathrm{F}=\frac{\left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}-\left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}}{\Delta \mathrm{t}}$
In this elastic collision, final and initial velocity will be same but direction will changed.
$\therefore$ Horizontal component, $\left(\mathrm{p}_{\mathrm{f}}\right)_{\mathrm{H}}=\mathrm{mv} \cos \theta$ and $\left(\mathrm{p}_{\mathrm{i}}\right)_{\mathrm{H}}=-\mathrm{mv} \cos \theta$
$\begin{aligned} & \therefore \mathrm{F}=\frac{\mathrm{mv} \cos \theta-(-\mathrm{mV} \cos \theta)}{\Delta \mathrm{t}}=\frac{2 \mathrm{mV} \cos \theta}{\Delta \mathrm{t}} \\ & =\frac{2 \times 0.5 \times 12 \times \cos 30^{\circ}}{1}=6 \sqrt{3} \mathrm{~N} \\ & \end{aligned}$

Topic: Physics, Laws of Motion, JIPMER, JIPMER 2018
A ball of mass 0.2 kg is thrown vertically upward by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes up to 2 m height further, find the magnitude of the force. Consider g=10 m s-2

Options:
A 4 N

B 16 N

C 20 N

D 22 N

The correct answer is:
22 N

Solution:
W=∆TE
F×0.2=mgh
F×0.2=0.2×10×2+0.2
F=22 N

Topic: Physics, Laws of Motion, JEE Main
A ball of mass 0.5 kg is attached to a string of length 50 cm. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is 400 N. The maximum possible value of angular velocity of the ball in rad s-1 is,:

Options:
A 1600
B 40
C 1000
D 20

The correct answer is:
40

Solution:
Let the string makes angle θ with the vertical. Weight will be balanced by the vertical component of tension, which is Tcosθ=mg and the horizontal component towards the centre will provide the required centripetal force for circular motion. Therefore,

Tsinθ=mω2lsinθ, where l=length of the string.

⇒400sinθ=0.5ω2×0.5sinθ

⇒ω=40 rad s-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is _____ m. [Use g=10 m s-2 ]

Options:

The correct answer is:
5

Solution:
Consider downward direction to be positive.

Applying first equation of motion, v=u+at

⇒v=0+gt ⇒g=gt ⇒t=1 s

Distance covered in 1 s, s=ut+12at2=0+12g×12=102=5 m

Hence, height of the point from the surface =10-5 m=5 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (26 Jun Shift 1)
A ball of mass 0.15 kg hits the wall with its initial speed of 12 m s-1 and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N. calculate the time duration of the contact of ball with the wall.

Options:
A 0.018 s
B 0.036 s
C 0.009 s
D 0.072 s

The correct answer is:
0.036 s

Solution:
Given: m=1.5 kg & u=12 m s-1

Change in the momentum of the ball after the collision=2mv

=2×1.5×12

=36 N s

As the force applied during collision is equal to 100 N and if t is the duration of collision, so

100×t=Δp

⇒t=36100 s

t=36×10-2 s=0.036 s

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (26 Jul Shift 2)