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: A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at height \$\dfrac {h}{2}\$. If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be-

m mass falls a distance \$\dfrac {h}{2}\$ in time, say t. Then,\$\dfrac {h}{2}=\frac {1}{2}gt^2\therefore t=\sqrt {\dfrac {h}{g}}\$For mass 2m,\$\dfrac {h}{2}=vt-\dfrac {1}{2}gt^2=v\sqrt {\dfrac {h}{g}}-\dfrac {h}{2}\$\$\therefore v=\sqrt {gh}\$Now, let u is the velocity with which combined mass collides with ground, then\$u=\sqrt {(\dfrac {2}{3}v)^2+2g(\dfrac {h}{3})}\$\$=\sqrt {\dfrac {4}{9}gh+\dfrac {2}{3} gh}\$\$=\dfrac {\sqrt {10}}{3}\sqrt {gh}\$Since the collision takes place in vertical direction and external force is not zero in this direction, momentum conservation cannot be applied.If the gravitational force( weight) is much less than the impulse force, it may be ignored and conservation of momentum can be applied.

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 circus girl throws three rings upwards one after the other at equal intervals of half a second. She catches the first ring half second after the third was thrown. Then,(\$\mathrm{g}=\$ acceleration due to gravity)

The circus girl catches the first ring after \$\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\$ second\$\Rightarrow \$ the ring was in air for\$\dfrac{3}{2} \$ second in which for first \$\dfrac{3}{4} \$ second it was going up and for the next \$\dfrac{3}{4} \$ second it was coming down.Let the height attained by first ring be h in \$\dfrac{3}{4} \$ second, obviously, its velocity at the instant will be zero.if u be the velocity of projection of rings, then\$0=u-g\left ( \dfrac{3}{4} \right )\Rightarrow u=\dfrac{3g}{4}\$\$\Rightarrow (A)\$ is correct.The maximum height h attained by the ring is given by,\$0=\left ( \dfrac{3g}{4} \right )^{2}-2gh\$\$\Rightarrow h=\dfrac{9g}{32}\Rightarrow (B)\$ is not correct.When the first ring returns to her hand, the second ring has traveled for 1 second. Of that 1 second, \$\dfrac{3}{4}\$ seconds will be spent in going up and for the rest it is coming down.\$\Rightarrow h'=\dfrac{1}{2}g\left ( \dfrac{1}{4} \right )^{2}=\dfrac{g}{32}\$\$\Rightarrow \$ Distance travelled by second ring by that instant is\$=\dfrac{9g}{32}-\dfrac{g}{32}=\dfrac{g}{4}\$\$\Rightarrow \$ (C) is correct.The third ring has travelled for \$t=\dfrac{1}{2}\$ second by that instant. it will be going up.Corresponding distance from ground\$t=\left ( \dfrac{3g}{4} \right )\left ( \dfrac{1}{2} \right )-\dfrac{1}{2}g\left ( \dfrac{1}{2} \right )^{2}\$\$=\dfrac{3g}{8}-\dfrac{g}{8}=\dfrac{g}{4}\$\$\Rightarrow (D)\$ is correct.

Question 1

Relative to another coordinate system $\mathrm{S}'$ (denoted by single prime) moving with a vertical velocity $\vec{\mathrm{v}}_{0}$, the equation of motion of the object becomes

Accepted Answer

The acceleration due to gravity will be the same in both systems$\vec{v}+\vec{v_{0}}=\vec{v'}$$\Rightarrow \vec{v}=\vec{v'}-\vec{v_{0}}$$\frac{d\vec{v}}{dt}=\frac{d\vec{v}}{dt}-\frac{d\vec{v_{0}}}{dt}$$\frac{d\vec{v}}{dt}=\frac{d\vec{v'}}{dt}$$\Rightarrow m\frac{d\vec{v}}{dt}=m(\frac{d\vec{v'}}{dt})$$\Rightarrow m\left ( \frac{d\vec{v'}}{dt} \right )=-mg-b(\vec{v'-\vec{v_{0}}})$

Question 2

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

Accepted Answer

$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.

Question 3

A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at height $\dfrac {h}{2}$. If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be-

Accepted Answer

m mass falls a distance $\dfrac {h}{2}$ in time, say t. Then,$\dfrac {h}{2}=\frac {1}{2}gt^2\therefore t=\sqrt {\dfrac {h}{g}}$For mass 2m,$\dfrac {h}{2}=vt-\dfrac {1}{2}gt^2=v\sqrt {\dfrac {h}{g}}-\dfrac {h}{2}$$\therefore v=\sqrt {gh}$Now, let u is the velocity with which combined mass collides with ground, then$u=\sqrt {(\dfrac {2}{3}v)^2+2g(\dfrac {h}{3})}$$=\sqrt {\dfrac {4}{9}gh+\dfrac {2}{3} gh}$$=\dfrac {\sqrt {10}}{3}\sqrt {gh}$Since the collision takes place in vertical direction and external force is not zero in this direction, momentum conservation cannot be applied.If the gravitational force( weight) is much less than the impulse force, it may be ignored and conservation of momentum can be applied.

Question 4

A particle is projected with a velocity $u$ in horizontal direction as shown in the figure. Find $u$(approx.) so that the particle collides orthogonally with the inclined  plane of the fixed wedge.

Accepted Answer

Considering plane along Wedge axis,when particle collides with the plane, its component of velocity parallel to inclined plane should be zero.$\Rightarrow u \cos\theta - g \sin\theta \times t =0$ {Equation along the plane}$t=\displaystyle \dfrac { u }{ g \tan \theta } $$u=\displaystyle gt\times \tan\theta  = \dfrac{gt}{\sqrt{3}}$Now, Consider the plane along original x-y axis,Thus,Horizontal displacement before it hits the plane $=ut$vertical displacement during the time t $=\dfrac{1}{2} g t^2 $Now , $ \tan 30^\circ{} =\displaystyle  \dfrac{ h- \frac{1}{2} g t^2  } { ut }$$ \dfrac{1}{\sqrt{3}} = \displaystyle \dfrac{ h- \dfrac{1}{2} g t^2  } { ut }$Hence , $ \dfrac{1}{3} = \displaystyle ( \dfrac{h} {gt^2} - \dfrac{1}{2} ) $$t=\displaystyle \sqrt{ \dfrac{6h}{5g}} $ Hence ,  $t = 2.5938 $ sSo,$u = gt \, \tan 30^\circ{} = \displaystyle \dfrac{9.8 \times 2.5938}{\sqrt 3} = 14.69 \simeq 10 \sqrt{2}$ m/s

Question 5

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 $V_{2} - V_{1}$ .

Accepted Answer

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)

Question 6

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

Accepted Answer

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 $= 32+32 =64\ m$

Question 7

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)

Accepted Answer

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 $t+\tau$.

Question 8

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.

Accepted Answer

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.

Question 9

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?

Accepted Answer

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Question 10

A circus girl throws three rings upwards one after the other at equal intervals of half a second. She catches the first ring half second after the third was thrown. Then,
( $g =$ acceleration due to gravity)

Accepted Answer

The circus girl catches the first ring after $\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}$ second$\Rightarrow $ the ring was in air for$\dfrac{3}{2} $ second in which for first $\dfrac{3}{4} $ second it was going up and for the next $\dfrac{3}{4} $ second it was coming down.Let the height attained by first ring be h in $\dfrac{3}{4} $ second, obviously, its velocity at the instant will be zero.if u be the velocity of projection of rings, then$0=u-g\left ( \dfrac{3}{4} \right )\Rightarrow u=\dfrac{3g}{4}$$\Rightarrow (A)$ is correct.The maximum height h attained by the ring is given by,$0=\left ( \dfrac{3g}{4} \right )^{2}-2gh$$\Rightarrow h=\dfrac{9g}{32}\Rightarrow (B)$ is not correct.When the first ring returns to her hand, the second ring has traveled for 1 second. Of that 1 second, $\dfrac{3}{4}$ seconds will be spent in going up and for the rest it is coming down.$\Rightarrow h'=\dfrac{1}{2}g\left ( \dfrac{1}{4} \right )^{2}=\dfrac{g}{32}$$\Rightarrow $ Distance travelled by second ring by that instant is$=\dfrac{9g}{32}-\dfrac{g}{32}=\dfrac{g}{4}$$\Rightarrow $ (C) is correct.The third ring has travelled for $t=\dfrac{1}{2}$ second by that instant. it will be going up.Corresponding distance from ground$t=\left ( \dfrac{3g}{4} \right )\left ( \dfrac{1}{2} \right )-\dfrac{1}{2}g\left ( \dfrac{1}{2} \right )^{2}$$=\dfrac{3g}{8}-\dfrac{g}{8}=\dfrac{g}{4}$$\Rightarrow (D)$ is correct.

Motion In A Straight Line

Physics
20119 Views

Relative to another coordinate system $S^{'}$ (denoted by single prime) moving with a vertical velocity $v_{0}$ , the equation of motion of the object becomes

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

A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at height $\frac{h}{2}$ . If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be-

A particle is projected with a velocity $u$ in horizontal direction as shown in the figure. Find $u$ (approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.

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 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

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

A circus girl throws three rings upwards one after the other at equal intervals of half a second. She catches the first ring half second after the third was thrown. Then,
( $g =$ acceleration due to gravity)

Motion In A Straight Line

Physics 20119 Views

Relative to another coordinate system S′ (denoted by single prime) moving with a vertical velocity v0​, the equation of motion of the object becomes

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

A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at height 2h​. If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be-

A particle is projected with a velocity u in horizontal direction as shown in the figure. Find u(approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.

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 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?

A circus girl throws three rings upwards one after the other at equal intervals of half a second. She catches the first ring half second after the third was thrown. Then, (g= acceleration due to gravity)

Physics
20119 Views

Relative to another coordinate system $S^{'}$ (denoted by single prime) moving with a vertical velocity $v_{0}$ , the equation of motion of the object becomes

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

A body of mass m is dropped from a height of h. Simultaneously another body of mass 2m is thrown up vertically with such a velocity v that they collide at height $\frac{h}{2}$ . If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be-

A particle is projected with a velocity $u$ in horizontal direction as shown in the figure. Find $u$ (approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.

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

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?

A circus girl throws three rings upwards one after the other at equal intervals of half a second. She catches the first ring half second after the third was thrown. Then,
( $g =$ acceleration due to gravity)