Q: Given \$k_1 = 1500 \ N/m,\$ \$k_2 =500 \ N/m,\$ \$m_1 = 2 \ kg\$ and \$m_2 = 1 \ kg\$. Find potential energy stored in equilibrium: (Take \$g = 10 \ m/s^2\$)

Soppose the elongation in first string be \$x_1\$ and in the second string be \$x_2\$Now for second string, in equilibrium\$K_2x_2=m_2g\$ or \$x_2= m_2g/k_2\$ (1)For first string in equilibrium\$K_1x_1=(m_1+m_2)g\$ or \$x_1= (m_2+m_1)g/k_1\$ (2)Now potential energy stored will be\$U=\frac{K_1x_1^2}{2}\$ + \$\frac{K_2x_2^2}{2}\$ Putting the values of \$k_1\$,\$k_2\$,\$x_1\$ and \$x_2\$We get U= 0.4J

Q: A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is

Using momentum conservation,\$\displaystyle m_{car} \displaystyle u _{car} + \displaystyle m_{ truck} \displaystyle u_{ truck}= \displaystyle m_{ car} \displaystyle \displaystyle v _{car} + \displaystyle m_{truck} \displaystyle v_{truck}\$\$400 \times 72 + 4000 \times 9 = -18 \times 400 + 4000 \times v\$\$v = 18 \ kmph\$

Q: A perfectly elastic ball \$p_{1}\$of mass 'm' moving with velocity v collides elastically with three exactly similar balls \$p_{2}\$ , \$p_{3}\$ , \$p_{4}\$ lying on a smooth table. Velocities of the four balls after collision are

\$\textbf{Hint}\$: Use law of conservation of momentum\$\textbf{Step}\$:Here when \$P_{1}\$ and \$P_{2}\$ will collide, their momentum will be exchanged because the collision is elastic and the two bodies are similar. When \$P_{2}\$ and \$P_{3}\$ collide same happens. So, momentum of \$P_{1}\$ is transferred to \$P_{4}\$So, \$P_{1}\$, \$P_{2}\$, \$P_{3}\$ will have zero velocity and \$P_{4}\$ has a velocity of \$V\$\$\textbf{Step 2}: \textbf{Calculation}\$:Let us take two balls. Let the first ball has a mass of 'm' and velocity 'v'. Let the second ball has a mass of 'm' and velocity is \$0\$ that is it is at rest. For perfectly elastic collision coefficient of restitution is \$e=1\$Total momentum before collision (Initial momentum)=Total momentum after collision(Final momentum)\$P_i=P_f\$\$mV=mV_1+mV_2\$\$\implies V_1+V_2=V\$ \$(1)\$We know coefficient of restitution \$e=1\$So, \$e= \dfrac{V_{separation}}{V_{approach}}\$\$1=\dfrac{V_1-V_2}{V}\$\$V_1-V_2=V\$ \$(2)\$Solving equation \$(1)\$ and \$(2)\$ we get\$2V_1=2V\$\$\implies V_1=V\$\$V_2=0\$So \$P_{1}\$, \$P_{2}\$, \$P_{3}\$ will have zero velocity and \$P_{4}\$ has a velocity of \$V\$Thus option D is correct.

Q: A particle moves under the effect of a force \$F = C x\$ from \$x = 0\$ to \$ x = x_{1}\$. The work done in the process is (treat \$C\$ as a constant):

Work Done \$\displaystyle =\int_{0}^{x_1} \vec{F}.d\vec{x}\$ \$\displaystyle=\int_{0}^{x_1}Cx.dx\$ \$= \dfrac {Cx^{2}_{1}}{2}\$

Q: A body of mass 5 kg moving with a speed of \$ 3 ms^{-1}\$ collides head on with a body of mass 3 kg moving in the opposite direction at a speed of \$2 ms^{-1}\$. The first body stops after the collision. Find the final velocity of the second body.

\$\underline {Given:}\$\$m_1=5kg\$\$m_2=3kg\$\$u_1=3ms^{-1}\$\$u_2=-2ms^{-1}\$ Before collision, the 2nd body is moving backwards. Hence its velocity is negative.\$v_1=0ms^{-1}\$After collision, the 1st body stops moving. So its velocity is zero.\$To\$ \$find:\$ \$v_2=\$ \$?\$\$\underline {Solution:}\$\$m_1 u_1+m_2 u_2\$ \$=\$ \$m_1 v_1+m_2 v_2\$Substituting the values, we get, \$(5\times3)\$ \$ +\$ \$(3\times(-2))\$ \$=\$ \$5\times 0\$ \$+\$ \$3\times v_2\$\$\therefore15\$ \$-\$ \$6\$ \$=\$ \$3\times v_2\$\$\therefore\$ \$9\$ \$=\$ \$3 v_2\$\$\therefore\$ \$v_2\$ \$=\$ \$9\$ \$\div\$ \$3\$ \$=\$ \$3\$\$\therefore\$ \$v_2\$ \$=\$ \$3ms^{-1}\$Since we got a positive value for \$v_2,\$ it means that the 2nd body moves forwards after collision.Velocity being a vector quantity, the direction is also important.After collision, the 2nd body moves with a velocity \$3ms^{-1}\$ in the forward direction.

Q: A heavier body moving with certain velocity collides head on elastically with a lighter body at rest. Then

\$\textbf{Hint}\$: Apply Law of conservation of momentum\$\textbf{Step 1:Assumptions}\$Let us assume that there are two bodies. Let the first body has a mass of 'M' and initial velocity=\$V\$ and final velocity=\$V_1\$ (after collision). Let the second body has a mass of 'm' and initial velocity=0 ; final velocity=\$V_2\$ (after collision)\$\textbf{Step 2:Apply Law of conservation of Linear Momentum}\$We know according to the law of conservation of momentum Initial momentum=Final Momentum\$MV+m(0)=MV_1+mV_2\$ \$(1)\$\$\textbf{Step3:Coefficient of Restitution}\$We know for elastic collision e=1\$e=1=\dfrac{V_2-V_1}{V-0}\$\$V=V_2-V_1\$\$V_1=V_2-V\$ \$(2)\$\$\textbf{Step 4:Solve above equations to obtain answer}\$Substitute \$V_1=V_2-V\$ in \$(1)\$we get \$MV=M(V_1)+mV_2\$\$MV=M(V_2-V)+mV_2\$\$MV=MV_2-MV+mV_2\$\$2MV=MV_2+mV_2\$\$V_2=\dfrac{2VM}{M+m}\$\$V_2=2V\$ if \$M>>m\$Thus option C is correct

Q: A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. After collision,

Using conservation of linear momentum we have:\$M_1u_1-M_2u_2=-M_1v_1+M_2v_2\$\$MV-Mu_2=-Mv+Mv_2\$\$V-u_2=-v+v_2\$.............(i)and as for a perfect elastic collision we have \$e=1\$, thus we get\$v_2+v=u_2+V\$.................(ii)Using these two equations we get:\$u_2=v\$ and \$v_2=V\$Thus, the velocities are exchanged between the two balls.

Question 1

Given $k_1 = 1500 \ N/m,$ $k_2 =500 \ N/m,$ $m_1 = 2 \ kg$ and $m_2 = 1  \ kg$. Find potential energy stored in equilibrium: (Take $g = 10 \ m/s^2$)

Accepted Answer

Soppose the elongation in first string be $x_1$ and in the second string be $x_2$Now for second string, in equilibrium$K_2x_2=m_2g$ or $x_2= m_2g/k_2$                  (1)For first string in equilibrium$K_1x_1=(m_1+m_2)g$ or $x_1= (m_2+m_1)g/k_1$               (2)Now potential energy stored will be$U=\frac{K_1x_1^2}{2}$ + $\frac{K_2x_2^2}{2}$ Putting the values of $k_1$,$k_2$,$x_1$ and $x_2$We get U= 0.4J

Question 2

A ball of mass $'m'$ moves towards a wall with a velocity $'u'$, the direction of motion making an angle $\theta$ with the surface of the wall and rebounds with the same period. The change in momentum of the ball during the collision is:

Accepted Answer

As the ball hits the wall at an angle and rebounds, there is a change in the momentum only along the direction perpendicular to the wall. The change along the wall is zero. In diagram $\vec p_1$ and $\vec p_2$ is the momentum before and after the collision with wall.Since , the angle with the wall is $\theta $, the change is only in $sin\theta $ component, which is across $\vec F_{wall}$.Change in momentum$=m[usin\theta -(-usin\theta )]=2mu\ sin\theta $ away from the wall, as the final velocity is away from the wall.

Question 3

A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is

Accepted Answer

Using momentum conservation,$\displaystyle  m_{car} \displaystyle u _{car} + \displaystyle m_{ truck} \displaystyle u_{ truck}= \displaystyle m_{ car} \displaystyle \displaystyle v _{car} + \displaystyle m_{truck} \displaystyle v_{truck}$$400 \times 72 + 4000 \times 9 = -18 \times 400 + 4000 \times v$$v = 18 \ kmph$

Question 4

A perfectly elastic ball $p_{1}$of mass 'm' moving with velocity v collides elastically with three exactly similar balls $p_{2}$ , $p_{3}$ , $p_{4}$  lying on a smooth table. Velocities of the four balls after collision are

Accepted Answer

$\textbf{Hint}$: Use law of conservation of momentum$\textbf{Step}$:Here when $P_{1}$ and $P_{2}$ will collide, their momentum will be exchanged because the collision is elastic and the two bodies are similar. When $P_{2}$ and $P_{3}$ collide same happens. So, momentum of $P_{1}$ is transferred to $P_{4}$So, $P_{1}$, $P_{2}$, $P_{3}$ will have zero velocity and $P_{4}$ has a velocity of $V$$\textbf{Step 2}: \textbf{Calculation}$:Let us take two balls. Let the first ball has a mass of 'm' and velocity 'v'. Let the second ball has a mass of 'm' and velocity is $0$ that is it is at rest. For perfectly elastic collision coefficient of restitution is $e=1$Total momentum before collision (Initial momentum)=Total momentum after collision(Final momentum)$P_i=P_f$$mV=mV_1+mV_2$$\implies V_1+V_2=V$   $(1)$We know coefficient of restitution $e=1$So, $e= \dfrac{V_{separation}}{V_{approach}}$$1=\dfrac{V_1-V_2}{V}$$V_1-V_2=V$        $(2)$Solving equation $(1)$ and $(2)$ we get$2V_1=2V$$\implies V_1=V$$V_2=0$So $P_{1}$, $P_{2}$, $P_{3}$ will have zero velocity and $P_{4}$ has a velocity of $V$Thus option D is correct.

Question 5

A particle moves under the effect of a force $F = C x$ from $x = 0$ to $ x = x_{1}$. The work done in the process is (treat $C$ as a constant):

Accepted Answer

Work Done $\displaystyle =\int_{0}^{x_1} \vec{F}.d\vec{x}$ $\displaystyle=\int_{0}^{x_1}Cx.dx$ $= \dfrac {Cx^{2}_{1}}{2}$

Question 6

A body of mass 5 kg moving with a speed of $3 m s^{- 1}$ collides head on with a body of mass 3 kg moving in the opposite direction at a speed of $2 m s^{- 1}$ . The first body stops after the collision. Find the final velocity of the second body.

Accepted Answer

$\underline {Given:}$$m_1=5kg$$m_2=3kg$$u_1=3ms^{-1}$$u_2=-2ms^{-1}$  Before collision, the 2nd body is moving backwards. Hence its velocity is negative.$v_1=0ms^{-1}$After collision, the 1st body stops moving. So its velocity is zero.$To$ $find:$  $v_2=$ $?$$\underline {Solution:}$$m_1 u_1+m_2 u_2$  $=$  $m_1 v_1+m_2 v_2$Substituting the values, we get,   $(5\times3)$  $ +$ $(3\times(-2))$   $=$   $5\times 0$   $+$  $3\times v_2$$\therefore15$  $-$ $6$   $=$   $3\times v_2$$\therefore$  $9$   $=$   $3 v_2$$\therefore$  $v_2$  $=$  $9$  $\div$  $3$  $=$  $3$$\therefore$  $v_2$  $=$ $3ms^{-1}$Since we got a positive value for $v_2,$ it means that the 2nd body moves forwards after collision.Velocity being a vector quantity, the direction is also important.After collision, the 2nd body moves with a velocity $3ms^{-1}$ in the forward direction.

Question 7

A heavier body moving with certain velocity collides head on elastically with a lighter body at rest. Then

Accepted Answer

$\textbf{Hint}$: Apply Law of conservation of momentum$\textbf{Step 1:Assumptions}$Let us assume that there are two bodies. Let the first body has a mass of 'M' and initial velocity=$V$ and final velocity=$V_1$ (after collision). Let the second body has a mass of 'm' and initial velocity=0 ; final velocity=$V_2$ (after collision)$\textbf{Step 2:Apply Law of conservation of Linear Momentum}$We know according to the law of conservation of momentum Initial momentum=Final Momentum$MV+m(0)=MV_1+mV_2$       $(1)$$\textbf{Step3:Coefficient of Restitution}$We know for elastic collision e=1$e=1=\dfrac{V_2-V_1}{V-0}$$V=V_2-V_1$$V_1=V_2-V$                         $(2)$$\textbf{Step 4:Solve above equations to obtain answer}$Substitute $V_1=V_2-V$ in $(1)$we get $MV=M(V_1)+mV_2$$MV=M(V_2-V)+mV_2$$MV=MV_2-MV+mV_2$$2MV=MV_2+mV_2$$V_2=\dfrac{2VM}{M+m}$$V_2=2V$ if $M>>m$Thus option C is correct

Question 8

A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. After collision,

Accepted Answer

Using conservation of linear momentum we have:$M_1u_1-M_2u_2=-M_1v_1+M_2v_2$$MV-Mu_2=-Mv+Mv_2$$V-u_2=-v+v_2$.............(i)and as for a perfect elastic collision we have $e=1$, thus we get$v_2+v=u_2+V$.................(ii)Using these two equations we get:$u_2=v$ and $v_2=V$Thus, the velocities are exchanged between the two balls.

Question 9

A heavy steel ball of mass greater than 1 kg moving with a speed of 2m/ s collides head on with a stationary ping pong ball of mass less than 0.1 g. The collision is elastic. After the collision the ping pong ball moves approximately with a speed

Accepted Answer

Since the body is much heavy these won't be much change in velocity & $e = 1$i.e., $\dfrac{v-2}{0-2} = -1$$\Rightarrow v = 4 m/s$

Question 10

A rain drop of mass (1/10) gram falls vertically at constant speed under the influence of the forces of gravity and viscous drag. In falling through 100 m, the work done by gravity is

Accepted Answer

work done = mghm = mass of drop = 0.1 g = 0.00001kgg =9.8m/sh = 100mwork done = $0.00001 \times 9.8 \times 100$                  = 0.098 J

Work Energy And Power

Physics
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Given $k_{1} = 1500 N / m,$ $k_{2} = 500 N / m,$ $m_{1} = 2 k g$ and $m_{2} = 1 k g$ . Find potential energy stored in equilibrium: (Take $g = 10 m / s^{2}$ )

A ball of mass $^{'} m^{'}$ moves towards a wall with a velocity $^{'} u^{'}$ , the direction of motion making an angle $θ$ with the surface of the wall and rebounds with the same period. The change in momentum of the ball during the collision is:

A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is

A perfectly elastic ball $p_{1}$ of mass 'm' moving with velocity v collides elastically with three exactly similar balls $p_{2}$ , $p_{3}$ , $p_{4}$ lying on a smooth table. Velocities of the four balls after collision are

A particle moves under the effect of a force $F = C x$ from $x = 0$ to $x = x_{1}$ . The work done in the process is (treat $C$ as a constant):

A body of mass 5 kg moving with a speed of $3 m s^{- 1}$ collides head on with a body of mass 3 kg moving in the opposite direction at a speed of $2 m s^{- 1}$ . The first body stops after the collision. Find the final velocity of the second body.

A heavier body moving with certain velocity collides head on elastically with a lighter body at rest. Then

A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. After collision,

A heavy steel ball of mass greater than 1 kg moving with a speed of 2m/ s collides head on with a stationary ping pong ball of mass less than 0.1 g. The collision is elastic. After the collision the ping pong ball moves approximately with a speed

A rain drop of mass (1/10) gram falls vertically at constant speed under the influence of the forces of gravity and viscous drag. In falling through 100 m, the work done by gravity is

Work Energy And Power

Physics 9441 Views

Given k1​=1500 N/m, k2​=500 N/m, m1​=2 kg and m2​=1 kg. Find potential energy stored in equilibrium: (Take g=10 m/s2)

A ball of mass ′m′ moves towards a wall with a velocity ′u′, the direction of motion making an angle θ with the surface of the wall and rebounds with the same period. The change in momentum of the ball during the collision is:

A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is

A perfectly elastic ball p1​of mass 'm' moving with velocity v collides elastically with three exactly similar balls p2​ , p3​ , p4​ lying on a smooth table. Velocities of the four balls after collision are

A particle moves under the effect of a force F=Cx from x=0 to x=x1​. The work done in the process is (treat C as a constant):

A body of mass 5 kg moving with a speed of 3ms−1 collides head on with a body of mass 3 kg moving in the opposite direction at a speed of 2ms−1. The first body stops after the collision. Find the final velocity of the second body.

A heavier body moving with certain velocity collides head on elastically with a lighter body at rest. Then

A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. After collision,

A heavy steel ball of mass greater than 1 kg moving with a speed of 2m/ s collides head on with a stationary ping pong ball of mass less than 0.1 g. The collision is elastic. After the collision the ping pong ball moves approximately with a speed

A rain drop of mass (1/10) gram falls vertically at constant speed under the influence of the forces of gravity and viscous drag. In falling through 100 m, the work done by gravity is

Physics
9441 Views

Given $k_{1} = 1500 N / m,$ $k_{2} = 500 N / m,$ $m_{1} = 2 k g$ and $m_{2} = 1 k g$ . Find potential energy stored in equilibrium: (Take $g = 10 m / s^{2}$ )

A ball of mass $^{'} m^{'}$ moves towards a wall with a velocity $^{'} u^{'}$ , the direction of motion making an angle $θ$ with the surface of the wall and rebounds with the same period. The change in momentum of the ball during the collision is:

A perfectly elastic ball $p_{1}$ of mass 'm' moving with velocity v collides elastically with three exactly similar balls $p_{2}$ , $p_{3}$ , $p_{4}$ lying on a smooth table. Velocities of the four balls after collision are

A particle moves under the effect of a force $F = C x$ from $x = 0$ to $x = x_{1}$ . The work done in the process is (treat $C$ as a constant):

A body of mass 5 kg moving with a speed of $3 m s^{- 1}$ collides head on with a body of mass 3 kg moving in the opposite direction at a speed of $2 m s^{- 1}$ . The first body stops after the collision. Find the final velocity of the second body.