A force \$\vec{F}=(3t\hat{i}+5\hat{j})\$ N acts on a particle whose position vector varies as \$\vec{S}=(2t^{2}\hat{i}+5\hat{j})\$ m, where \$t\$ is time in seconds. The work done by this force from \$t=0\$ to \$t=2s\$ is:

The work done by the force \$\vec{F}=(3t\hat{i}+5\hat{j})\$ is \$W=\int \vec{F}.d\vec{s}=\int (3t \hat{i}+5\hat{j}).(4t dt \hat{i})\$ \$=\int_{0}^{2}12t^{2}dt=\dfrac{12[t^{3}]_{0}^{2}}{3}=\dfrac{12(8-0)}{3}=32 \ J\$

The balls, having linear momenta \$\vec{\mathrm{p}}_{1}=\mathrm{p}\hat{i}\$ and \$\vec{\mathrm{p}}_{2}=-\mathrm{p}\hat{\mathrm{i}}\$, undergo a collision in free space. There is no extemal force acting on the balls. Let \$\vec{\mathrm{p}}_{1}\$ and \$\vec{\mathrm{p}}_{2}\$ be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-zero value of \$\mathrm{p},\ \mathrm{a}_{1},\ \mathrm{a}_{2},\ \mathrm{b}_{1},\ \mathrm{b}_{2},\ \mathrm{c}_{1}\$ and \$\mathrm{c}_{2}\$.

Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in \$j\$ and \$k\$ directions as both momenta are in \$i\$ direction. Hence options \$A\$ and \$D\$ are not possible.

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>>Class 11
>>Physics
>>Work, Energy and Power
>> $Hard Questions$

Work Energy And Power

Q: If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is

After collision, vertical velocity will be \$vsin({ 30 }^{ 0 })cos({ 30 }^{ 0 })-vcos({ 30 }^{ 0 })cos({ 60 }^{ 0 })=0\$.

Q: The figure shows a block having a small basket in it and an small inclined plane (with fixed angle) is rigidly attached to the block. The combined mass (including incline plane) of all these is M = 5 kg. Initially they are at rest. Now identical balls each of mass m=1 kg are thrown horizontally with velocity v = 5m/s (with respect to ground) which strike the inclined plane and then are collected in the small basket. Assume that in the basket balls come to rest with respect to basket. Choose the correct statement(s) from the following :

Total momentum of \$N\$ balls\$=N m v=N (1\times5)=5N\$\$\therefore \$ Momentum of the block along with \$N\$ balls\$=5N\$ or\$(M+Nm)\$ \${v}'=5N\$ where \${v}'\$ is the velocity of the blockAfter \$N\$ balls are collected in the basket.\$\therefore {v}' =\displaystyle \dfrac{5N}{5+N}=\dfrac{5}{\dfrac{5}{N}+1}\$For \$N=5\$; \${v}'\$ \$=2.5\ m/s\$For \$N=10\$; \${v}'\$ \$=10/3\ m/s\$Also \${v}'\$ is never exactly equal to \$5 m/s.\$From above we can say that increase in speed of the ball is not same every time for collection of each ball in the basket, it varies.

Q: The work done by the force \$\vec{F}=A(y^{2}\hat{i}+2x^{2}\hat{j})\$, where \$A\$ is a constant and \$x\$ & \$y\$ are in meters around the path shown is :

Variable force is apllied across the path \$OA\$ to \$AB\$ to \$BC\$ to \$CO\$\$W=\int \bar { F } .d\bar { r }\$ \$ \\ =\int A(y^{ 2 }\bar { i } +2x^{ 2 }\bar { j } ).(dx\bar { i } +dy.\bar { j } )\$\$\\ =\int A(y^{ 2 }dx+2x^{ 2 }dy)\$ Now following the path of displacement along \$OA\$ the \$y=0\$\$\\ W_{ OA }=\int _{ x=0 }^{ x=d }{ A(0\quad +\quad 2 } x^{ 2 }.0)\quad =\quad 0+0\quad =0\$Similerly, for \$W_{ AB },\quad x=d\$\$\\ W_{ AB }=A[0+2d^{ 2 }d]\quad =\quad 2Ad^{ 3 }\$Similarly, for \$W_{ BC },\ x=d,\quad y=d\$\$\\ W_{ BC }=\int _{ x=d }^{ x=0 }{ A(d^{ 2 }dx\quad +\quad } 0)\$\$\\ W_{ BC }=A[d^{ 2 }(-d)+0]=-Ad^{ 3 }\$\$\\ W_{ CO }=A[0+0]\$\$\\ W=0+2Ad^{ 3 }-Ad^{ 3 }+0=Ad^{ 3 }\$

Q: A ball of mass m moves towards a moving wall of infinite mass with a speed 'v' along the normal to the wall. The speed of the wall is 'u' toward the ball. The speed of the ball after elastic collision with wall is

In the frame of wall, the ball moves towards it with speed \$u + v\$ and due to momentum conservation velocity after collision returns with velocity \$u + v\$ in the direction in which wall is moving. So from the ground frame, the velocity of ball after collision with wall is \$2u + v.\$

Q: The force exerted by a compression device is given by \$ F(x) = kx (x-l)\$ for \$ 0 \leq x \leq l\$, where \$l\$ is the maximum possible compression, \$x\$ is the compression and \$k\$ is the constant. Work done to compress the device by a distance \$d\$ will be maximum when

For W to be maximum ; \$\dfrac{dW}{dx} = 0\$;i.e. \$ F(x) = 0 \Rightarrow x= l, x = 0\$clearly for \$d = l \$, the work done is maximum.Alternate Solution :External force and displacement are in the same direction\$\therefore\$ Work will be positive continuously so it will be maximum when displacement is maximum.

Q: A particle of mass \$m_0,\$ travelling at speed \$v_0,\$ strikes a stationary particle of mass \$2m_0.\$ As a result, the particle of mass \$m_0\$ is deflected through \$45^o\$ and has a final speed of \$\dfrac{v_0}{\sqrt2}\$. Then the speed of the particle of mass \$2m_0\$ after this collision is

A particle of mass \$m_0,\$Before collision\$mv_{0}=2mv cos\theta+\displaystyle \dfrac{mv_{0}}{2}..........(i)\$\$\displaystyle \theta=\dfrac{mv_{0}}{2}-2mv sin \theta............(ii)\$By (i) & (ii), \$\theta=45^{0}\$Now again momentum conservation in y direction\$\displaystyle \dfrac{mv_{0}}{2}=2mv.\displaystyle \dfrac{1}{\sqrt2}\Rightarrow v=\displaystyle \dfrac{v_0}{2\sqrt2}\$

Q: A particle of mass \$m\$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is \$a\$. A force \$\vec{F}=y\hat{i}-x\hat{j}\$ newton acts on the particle, where \$x, y\$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A (\$a, 0\$) to point B (\$0, a\$) along the circular path is

\$\displaystyle W=\mathbf{\int F.dr}=\int (y\hat{i}-x\hat{j}).(dx\hat{i}+dy\hat{j})\$\$=\displaystyle \int (ydx-xdy)\$ .........(1)\$\because x^{2}+y^{2}=a^{2} \therefore xdx+y dy=0\$\$\displaystyle \Rightarrow W=\int \left ( y\left ( \frac{-ydy}{x} \right )-xdy \right )=-\int \dfrac{x^{2}+y^{2}}{x}dy\$\$\displaystyle =-\int_{0}^{a}\dfrac{a^{2}}{\sqrt{a^{2}-y^{2}}}dy=-\dfrac{\pi a^{2}}{2}\$Alternate Method :It can be observed that the force is tangent to the curve ateach point and the magnitude is constant. The direction offorce is opposite to the direction of motion of the particle.\$\therefore\$ work done = (force) (distance)\$=-\sqrt{x^{2}+y^{2}}=\dfrac{\pi a}{2}=-a\times \dfrac{\pi a}{2}=-\dfrac{\pi a^{2}}{2}J\$\$W=-\dfrac{\pi a^{2}}{2}J\$

Q: An elastic string carrying a body of mass \$m\$ at one end extends by \$1.5\ cm\$. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

When the body is at rest, the extension is \$1.5\ cm\$ while force acting on it is \$ mg \$Force when it reaches bottom is \$F_b = \dfrac{mv^2}{r} + mg \$Now using energy balance between topmost point and bottom most point ,\$\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{rg})^2+mgh\$ (h is the height difference between the top and bottom point ; \$\sqrt{gr}\$ is critical velocity at topmost point )\$ v^2 = rg + 2gh\$or, \$ v^2 = rg + 4gr = 5 gr\$So, \$F_b = 6mg \$Using this in: \$\dfrac{e_2}{e_1} = \dfrac{F_2}{F_1} \$, we get:\$e_2 = 1.5 \times 6 = 9.0\: cm \$

Physics
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Easy Qs Med Qs Hard Qs

If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is

30 m / s

15 m / s

- 15 m / s

Hard

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The figure shows a block having a small basket in it and an small inclined plane (with fixed angle) is rigidly attached to the block. The combined mass (including incline plane) of all these is M = 5 kg. Initially they are at rest. Now identical balls each of mass m=1 kg are thrown horizontally with velocity v = 5m/s (with respect to ground) which strike the inclined plane and then are collected in the small basket. Assume that in the basket balls come to rest with respect to basket. Choose the correct statement(s) from the following :

After collecting 5 balls the speed of the block will be 2.5 m/s

After collecting 10 balls the speed of the block will be (10/3) m/s

The velocity of the block will never be exactly 5 m/s no matter how many balls are collected in the basket.

Increase in speed of the block will be same every time for collection of each ball in the basket

Hard

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The work done by the force $F = A (y^{2} \hat{i} + 2 x^{2} \hat{j})$ , where $A$ is a constant and $x$ & $y$ are in meters around the path shown is :

z e r o

A d

A d^{2}

A d^{3}

Hard

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A ball of mass m moves towards a moving wall of infinite mass with a speed 'v' along the normal to the wall. The speed of the wall is 'u' toward the ball. The speed of the ball after elastic collision with wall is

u + v

away from the wall

2 u + v

away from the wall

∣ u - v ∣

away from the wall

∣ v - 2 u ∣

away from the wall

Hard

View solution>

A force $F = (3 t \hat{i} + 5 \hat{j})$ N acts on a particle whose position vector varies as $S = (2 t^{2} \hat{i} + 5 \hat{j})$ m, where $t$ is time in seconds. The work done by this force from $t = 0$ to $t = 2 s$ is:

23 J

32 J

zero

can't be obtained

Hard

View solution>

The force exerted by a compression device is given by $F (x) = k x (x - l)$ for $0 \leq x \leq l$ , where $l$ is the maximum possible compression, $x$ is the compression and $k$ is the constant. Work done to compress the device by a distance $d$ will be maximum when

d = \frac{l}{4}

d = \frac{l}{2}

d = \frac{l}{2}

d = l

Hard

View solution>

A particle of mass $m_{0},$ travelling at speed $v_{0},$ strikes a stationary particle of mass $2 m_{0} .$ As a result, the particle of mass $m_{0}$ is deflected through $4 5^{o}$ and has a final speed of $\frac{v _{0}}{2}$ . Then the speed of the particle of mass $2 m_{0}$ after this collision is

\frac{v _{0}}{2}

\frac{v _{0}}{2 2}

2 v_{0}

\frac{v _{0}}{2}

Hard

View solution>

A particle of mass $m$ moves along the quarter section of the circular path whose centre is at the origin. The radius of the circular path is $a$ . A force $F = y \hat{i} - x \hat{j}$ newton acts on the particle, where $x, y$ denote the coordinates of position of the particle. The work done by this force in taking the particle from point A ( $a, 0$ ) to point B ( $0, a$ ) along the circular path is

\frac{π a ^{2}}{4} J

\frac{π a ^{2}}{2} J

\frac{π a ^{2}}{3} J

N o n e o f t h e s e

Hard

View solution>

The balls, having linear momenta $p_{1} = p \hat{i}$ and $p_{2} = - p \hat{i}$ , undergo a collision in free space. There is no extemal force acting on the balls. Let $p_{1}$ and $p_{2}$ be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-zero value of $p, a_{1}, a_{2}, b_{1}, b_{2}, c_{1}$ and $c_{2}$ .

p_{1} = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}

;

p_{2} = a_{2} i + b_{2} \hat{j}

p_{1} = c_{1} \hat{k}

;

p_{2} = c_{2} \hat{k}

p_{1} = a_{1} i + b_{1} \hat{j} + c_{1} \hat{k}

;

p_{2} = a_{2} \hat{i} + b_{2} \hat{j} - c_{1} \hat{k}

p_{1} = a_{1} \hat{i} + b_{1} \hat{j}

;

p_{2} = a_{2} i + b_{1} \hat{j}

Hard

View solution>

An elastic string carrying a body of mass $m$ at one end extends by $1.5 c m$ . If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

3.0 c m

4.5 c m

1.5 c m

9.0 c m

Hard

View solution>

Work Energy And Power

Physics 11393 Views

If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is

The work done by the force F=A(y2i^+2x2j^​), where A is a constant and x & y are in meters around the path shown is :

A ball of mass m moves towards a moving wall of infinite mass with a speed 'v' along the normal to the wall. The speed of the wall is 'u' toward the ball. The speed of the ball after elastic collision with wall is

A force F=(3ti^+5j^​) N acts on a particle whose position vector varies as S=(2t2i^+5j^​) m, where t is time in seconds. The work done by this force from t=0 to t=2s is:

The force exerted by a compression device is given by F(x)=kx(x−l) for 0≤x≤l, where l is the maximum possible compression, x is the compression and k is the constant. Work done to compress the device by a distance d will be maximum when

A particle of mass m0​, travelling at speed v0​, strikes a stationary particle of mass 2m0​. As a result, the particle of mass m0​ is deflected through 45o and has a final speed of 2​v0​​. Then the speed of the particle of mass 2m0​ after this collision is

An elastic string carrying a body of mass m at one end extends by 1.5 cm. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

Physics
11393 Views

The work done by the force $F = A (y^{2} \hat{i} + 2 x^{2} \hat{j})$ , where $A$ is a constant and $x$ & $y$ are in meters around the path shown is :

A force $F = (3 t \hat{i} + 5 \hat{j})$ N acts on a particle whose position vector varies as $S = (2 t^{2} \hat{i} + 5 \hat{j})$ m, where $t$ is time in seconds. The work done by this force from $t = 0$ to $t = 2 s$ is:

The force exerted by a compression device is given by $F (x) = k x (x - l)$ for $0 \leq x \leq l$ , where $l$ is the maximum possible compression, $x$ is the compression and $k$ is the constant. Work done to compress the device by a distance $d$ will be maximum when

A particle of mass $m_{0},$ travelling at speed $v_{0},$ strikes a stationary particle of mass $2 m_{0} .$ As a result, the particle of mass $m_{0}$ is deflected through $4 5^{o}$ and has a final speed of $\frac{v _{0}}{2}$ . Then the speed of the particle of mass $2 m_{0}$ after this collision is

An elastic string carrying a body of mass $m$ at one end extends by $1.5 c m$ . If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is: