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Motion In A Straight Line

Q: A person sitting in the rear end of the compartment throws a ball towards the front end.The ball follows a parabolic path.The train is moving with velocity of \$20\ m/s\$.A person standing outside on the ground also observes the ball.How will the maximum heights \$y_m\$ attained and the ranges \$R\$ seen by the thrower and the outside observer compare with each other?

The motion of train will affect only the horizontal component of the velocity of the ball.Since , vertical component is same for both observers, the \$y_m\$ will be same ,but range R will be different.

Q: A particle moves with a non zero initial velocity \${ v }_{ 0 }\$ and retardation \$kv\$, where \$v\$ is the velocity at any time \$t\$.

\$u=v_0, \$ retardation \$a=kv\$ and \$v\$ is the velocity at time \$t\$.using \$v=u-at,\$ we will get \$v=v_0-kvt \Rightarrow t=\dfrac{v_0}{kv}-\frac{1}{k}\$if \$v=0, t =\infty\$. So the particle will be moved for a long time.

Q: Mass \$A\$ is released from rest at the top of a frictionless inclined plane \$18 m\$ long and reaches the bottom \$3 s\$ later. At the instant when \$A\$ is released, a second mass \$B\$ is projected upwards along the plate from the bottom with a certain initial velocity. Mass \$B\$ travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with \$A\$. The two masses do not collide. Initial velocity of \$B\$ is

Here, for \$A, 18=0 \times 3+\displaystyle \frac{1}{2}a \times 3^{2}\$ or \$a=4 ms^{-2}\$ For \$B\$, time taken to move up is given by , \$t_1=u/a\$ (\$\therefore\$ the relation \$v=u+at\$ here becomes \$0=u-at_{1}\$). Distance moved up is given by the relation \$0=u^{2}-2a S\$i.e \$S=u^{2}/2a\$For coming down the inclined plane for \$S=\displaystyle \frac{1}{2}at_{2}^{2}\$and \$t_{2}=\sqrt{\displaystyle \frac{2S}{a}}\$or \$t_{2}=\sqrt{\displaystyle \frac{2u^{2}}{a.2a}} =\displaystyle \frac{u}{a}\$But \$\displaystyle \frac{u}{a}=t_{1}\$, then\$t_1+t_2=3\$ or \$\displaystyle \frac{2u}{a}=3\$ or\$u=\displaystyle \frac{3a}{2}\$ or\$u=\displaystyle \frac{3}{2} \times 4=6 \: ms^{-1}\$

Q: 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.STATEMENT-2: If the observer and the object are moving at velocities \$\vec{\mathrm{V}}_{1}\$ and \$\vec{\mathrm{V}}_{2}\$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is \$\vec{\mathrm{V}}_{2}-\vec{\mathrm{V}}_{1}\$.

Both the statements are true,but correct explanation of statement one is we,know that \${\theta}=\frac{arc}{radius}\$so value of \${\theta}\$ will be more for nearby object and radius will be less wile value of \${\theta}\$ will be less for distant object and radius will be more.that is the reason behind this phenomenon and near one will move faster and distant object will move relatively less(i.e both are moving w.r.t any frame of refrence)

Q: The displacement of a particle moving in a straight line is given by \$x=16t-2t^{2}\$ (where, \$x\$ is in meters and \$t\$ is in second). The distance traveled by the particle in \$8\$ seconds [starting from \$t\$ \$=\$ 0] is

Given: \$x=16 t - 2t^{2}\$Comparing with second equation of motion: \$s=ut+\cfrac{1}{2}at^2\$It can be concluded that: \$u=16\ m/s,\quad a=4\ m/s^2\$Now, using first equation of motion:\$v = u+at\$\$v=16 - 4t\$For \$v = 0 \$ we get, \$t= 4 s\$Here, direction of velocity will reverse. So total distance travelled will be the sum of displacement in first four seconds and displacement in next four seconds. Now,Displacement at\$t=0,\quad x=0\$\$t=4,\quad x=32\$\$t=8,\quad x=0\$\$\therefore\$ Distance \$=\$ (distance from \$x=0\$ to \$x=32\$) + (distance from \$x=32\$ to \$x=0\$) \$\therefore\$ Distance \$= 32+32 =64\ m\$

Q: A water tap leaks such that water drops fall at regular intervals. Tap is fixed \$5\ m\$ above the ground. First drop reaches the ground and at that very instant third drop leaves the tap. At this instant the second drop is at a height of

Velocity of drop 5m below the tap can be obtained by energy conservation.Assume that potential energy is zero 5m below the tap.Now, \$v=\sqrt { 2gh } =10\quad m/s\$Now, time taken to achieve this velocity\$=\dfrac { 10 }{ 10 } =1s\$it is given that drops leave the tap at a regular time interval. Thus, second drop will leave at 0.5s as the third drop is leaving at 1s.Now, total time of motion of second drop is 0.5sThus, distance moved by second drop in 0.5s is\$s=\dfrac { 1 }{ 2 } \times 10\times { (0.5) }^{ 2 }=1.25\$Thus height from ground level is \$5-1.25=3.75m\$

Q: A train of length \$l=350\:m\$ starts moving rectilinearly with constant acceleration \$\omega=3.0\cdot10^{-2}\:m/s^2\$; \$t=30\:s\$ after the start the locomotive headlight is switched on (event 1), and \$\tau=60\:s\$ after that event the tail signal light is switched on (event 2) . At what constant velocity \$V\$ (in m/s) relative to the Earth must a certain reference frame \$K\$ move for the two events to occur in it at the same point? (round off your answer to the nearest integer)

In the reference frame fixed to the train, the distance between the two events is obviously equal to \$l\$. Suppose the train starts moving at time \$t=0\$ in the positive x direction and take the origin (\$x=0\$) at the head-light of the train at \$t=0\$. Then the coordinate of first event in the earth's frame is\$\displaystyle x_1=\frac{1}{2}wt^2\$and similarly the coordinate of the second event is\$\displaystyle x_2=\frac{1}{2}w{(t+\tau)}^2-l\$The distance between the two events is obviously,\$\displaystyle x_1-x_2=l-w\tau\left(t+\frac{\tau}{2}\right)=0.242\:km\$in the reference frame fixed on the earth.For the two events to occur at the same point in the reference frame \$K\$, moving with constant velocity \$V\$ relative to the earth, the distance travelled by the frame in the time interval \$T\$ must be equal to the above distance.Thus \$\displaystyle V\tau=l-w\tau\left(t+\frac{\tau}{2}\right)\$So, \$\displaystyle V=\frac{1}{\tau}-w\tau\left(t+\frac{\tau}{2}\right)=4.03\:m/s\$The frame \$K\$ must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point \$x_1\$ on the earth at time \$t\$, it coincides with the point \$x_2\$ at time \$t+\tau\$.

Q: A ball is thrown vertically upwards (relative to the train) in a compartment of a moving train. The face of the person sitting inside the compartment is towards engine of the train.

At the time of throwing the ball, it inherits velocity of platform at that moment . And, the horizontal velocity of the ball is constant in the air, because no acceleration in horizontal direction for ball. If train is accelerating in forward direction then from frame of train, the ball has horizontal component of acceleration in backward direction and vice versa.

Physics
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Easy Qs Med Qs Hard Qs

A person sitting in the rear end of the compartment throws a ball towards the front end.The ball follows a parabolic path.The train is moving with velocity of $20 m / s$ .A person standing outside on the ground also observes the ball.How will the maximum heights $y_{m}$ attained and the ranges $R$ seen by the thrower and the outside observer compare with each other?

same

y_{m}

different

R

same

y_{m}

and

R

differently

y_{m}

same

R

differently

y_{m}

and

R

Hard

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A particle moves with a non zero initial velocity $v_{0}$ and retardation $k v$ , where $v$ is the velocity at any time $t$ .

The particle will cover a total distance

\frac{v _{0}}{k}

The particle comes to rest at

t = \frac{1}{k}

Particle continues to move for long time

at time

\frac{1}{α}, v = \frac{v _{0}}{2}

Hard

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Mass $A$ is released from rest at the top of a frictionless inclined plane $18 m$ long and reaches the bottom $3 s$ later. At the instant when $A$ is released, a second mass $B$ is projected upwards along the plate from the bottom with a certain initial velocity. Mass $B$ travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with $A$ . The two masses do not collide. Initial velocity of $B$ is

4 m s^{- 1}

5 m s^{- 1}

6 m s^{- 1}

7 m s^{- 1}

Hard

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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.

STATEMENT-2: If the observer and the object are moving at velocities $V_{1}$ and $V_{2}$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is $V_{2} - V_{1}$ .

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT -1 is True, STATEMENT-2 is False

STATEMENT -1 is False, STATEMENT-2 is True

Hard

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The displacement of a particle moving in a straight line is given by $x = 16 t - 2 t^{2}$ (where, $x$ is in meters and $t$ is in second). The distance traveled by the particle in $8$ seconds [starting from $t$ $=$ 0] is

24 m

40 m

64 m

80 m

Hard

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A water tap leaks such that water drops fall at regular intervals. Tap is fixed $5 m$ above the ground. First drop reaches the ground and at that very instant third drop leaves the tap. At this instant the second drop is at a height of

3 m

4.5 m

3.75 m

2.5 m

Hard

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A train of length $l = 350 m$ starts moving rectilinearly with constant acceleration $ω = 3.0 \cdot 1 0^{- 2} m / s^{2}$ ; $t = 30 s$ after the start the locomotive headlight is switched on (event 1), and $τ = 60 s$ after that event the tail signal light is switched on (event 2) . At what constant velocity $V$ (in m/s) relative to the Earth must a certain reference frame $K$ move for the two events to occur in it at the same point? (round off your answer to the nearest integer)

Hard

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A ball is thrown vertically upwards (relative to the train) in a compartment of a moving train. The face of the person sitting inside the compartment is towards engine of the train.

The ball will maintain the same horizontal velocity as that of the person (or the compartment) at the time of throwing.

If the train is accelerating then the horizontal velocity of the ball will be different from that of the train velocity, at the time of throwing.

If the ball appears to be moving backward to the person sitting in the compartment it means that speed of the train is increasing.

If the ball appears to be moving ahead of the person sitting in the compartment it means the train's motion is retarding.

Hard

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Using the table given below where the values of velocity at the end of t seconds for a body under linear motion are given

$V (m s^{- 1})$
0

6
12
24
30
36
42

$t (s)$
0

2
4
8
10
12
14

What can be concluded about the motion of the body?

It moves with uniform speed.

It moves with uniform motion

It moves with uniform velocity

It moves with uniform acceleration

Hard

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The acceleration (a) of a particle is plotted on the Y-axis while the time (t) elapsed is plotted along the X - axis. What does the a - t graph give?

The distance covered

The difference in velocities

The difference in acceleration

The difference in force

Hard

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Motion In A Straight Line

Physics 3828 Views

A particle moves with a non zero initial velocity v0​ and retardation kv, where v is the velocity at any time t.

The displacement of a particle moving in a straight line is given by x=16t−2t2 (where, x is in meters and t is in second). The distance traveled by the particle in 8 seconds [starting from t = 0] is

A water tap leaks such that water drops fall at regular intervals. Tap is fixed 5 m above the ground. First drop reaches the ground and at that very instant third drop leaves the tap. At this instant the second drop is at a height of

A ball is thrown vertically upwards (relative to the train) in a compartment of a moving train. The face of the person sitting inside the compartment is towards engine of the train.

Using the table given below where the values of velocity at the end of t seconds for a body under linear motion are given V(ms−1) 0 6 12 24 30 36 42 t(s) 0 2 4 8 10 12 14 What can be concluded about the motion of the body?

The acceleration (a) of a particle is plotted on the Y-axis while the time (t) elapsed is plotted along the X - axis. What does the a - t graph give?

Physics
3828 Views

A particle moves with a non zero initial velocity $v_{0}$ and retardation $k v$ , where $v$ is the velocity at any time $t$ .

The displacement of a particle moving in a straight line is given by $x = 16 t - 2 t^{2}$ (where, $x$ is in meters and $t$ is in second). The distance traveled by the particle in $8$ seconds [starting from $t$ $=$ 0] is

A water tap leaks such that water drops fall at regular intervals. Tap is fixed $5 m$ above the ground. First drop reaches the ground and at that very instant third drop leaves the tap. At this instant the second drop is at a height of

Using the table given below where the values of velocity at the end of t seconds for a body under linear motion are given

$V (m s^{- 1})$
0

6
12
24
30
36
42

$t (s)$
0

2
4
8
10
12
14

What can be concluded about the motion of the body?