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:

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.

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

Work Energy And Power

Q: A \$500gm\$ ball moving at \$15 m/s\$ slows down uniformly until it stops. If the ball travels \$15 m\$, what was the average net force applied while it was coming to a stop?

Using IIIrd kinematic equation:\$v^{2}=u ^{2}+2as\$\$-255=2\times a\times 15\$\$a=-\dfrac{15}{2}m/s\$Force is given as :\$F=ma\$\$=\dfrac{-15}{2}\times 0.500\$\$=3.75N\$

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: Same force acts on two bodies of masses \$m_{1}\$ and \$m_{2}\$ (\$m_{2}\$> \$m_{1}\$). After travelling the same distance,

\$s=\dfrac{1}{2}at^{2}\$Now, \$K.E_{1}=\dfrac{1}{2}m_{1}(at)^{2}\$\$=\dfrac{1}{2}\dfrac{m_{1}}{a_{1}}sa_{1}=F.s\$\$K.E_{2}=\dfrac{1}{2}mg(a_{2}t)^{2}\$\$=F.s\$i.e \$K.E_{1}=KE_{2}\$

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 is moving with a velocity \$1 \ ms^{-1}\$ a force \$F\$ is needed to stop it within a distance \$x\$. If the speed of the body is \$3 \ ms^{-1}\$, the force needed to stop it with in the same distance (\$x\$) will be:

Work done by force \$Fx=\frac { 1 }{ 2 } m{ v }^{ 2 }\$ Given \$v=1 \ m/s\$\$\Rightarrow Fx=\dfrac { 1 }{ 2 } m\$Now, suppose force needed to stop the same body moving at \$3m/s\$, is \$f\$.Using force energy theorem:\$\Rightarrow fx=\dfrac { 1 }{ 2 } 9m\$\$\Rightarrow f=9F\$

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.

Q: Two identical balls collide head on. The initial velocity of one is \$0.75\$cm\$^{-1}\$. While that of the other is \$-0.43\$ms\$^{-1}\$ . If the collision is perfectly elastic. Their respective final velocities are

In an elastic collision, using momentum conservation and equation for coefficient of restitution, we get,\$v_1 = \dfrac{m_1-m_2}{m_1+m_2}u_1 + \dfrac{2m_2}{m_1+m_2}u_2 \$\$ v_2 = \dfrac {2m_1}{m_1+m_2}u_1 + \dfrac{m_1-m_2}{m_1+m_2}u_2 \$But \$m_1=m_2=m\$\$\Rightarrow v_1 = \dfrac{m-m}{m+m}u_1 + \dfrac{2m}{m+m}u_2, u_2= -0.43\$ms\$^{-1}\$\$\Rightarrow v_2 = \dfrac {2m}{m+m}u_1 + \dfrac{m-m}{m+m}u_2, u_1=0.75\$ms\$^{-1}\$When two bodies of equal masses undergo elastic collision their velocities get interchanged.

Q: 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 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

Physics
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JEE Mains NEET Easy Qs Med Qs Hard Qs

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:

2 m u s i n

θ

towards the wall

2 m u s i n

θ

away from the wall

2 m u c o s

θ

towards the wall

2 m u c o s

θ

away from the wall

Medium

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A $500 g m$ ball moving at $15 m / s$ slows down uniformly until it stops. If the ball travels $15 m$ , what was the average net force applied while it was coming to a stop?

0.375 N

3.75 N

37.5 N

6.25 N

Medium

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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

0,0,0,0

v,v,v,v

v,v,v,0

0,0,0,v

Medium

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Same force acts on two bodies of masses $m_{1}$ and $m_{2}$ ( $m_{2}$ > $m_{1}$ ). After travelling the same distance,

Kinetic energy of

m_{2}

is greater than kinetic energy of

m_{1}

Kinetic energy of

m_{1}

is greater than kinetic energy of

m_{2}

Ratio of kinetic energies of those two is equal to ratio of their momentum

Kinetic energy of

m_{1}

is equal to kinetic energy of

m_{2}

Medium

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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):

\frac{C ^{2}}{x _{1}^{2}}

C x_{1}^{2}

\frac{C x _{1}^{2}}{2}

\frac{C}{x _{1}^{2}}

Medium

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A body is moving with a velocity $1 m s^{- 1}$ a force $F$ is needed to stop it within a distance $x$ . If the speed of the body is $3 m s^{- 1}$ , the force needed to stop it with in the same distance ( $x$ ) will be:

9 F

6 F

3 F

1.5 F

Medium

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

smaller body continues to be in the same state of rest

smaller body starts to move in the same direction with same velocity as that of bigger body

the smaller body starts to move with twice the velocity of the bigger body in the same direction

the bigger body comes to rest

Medium

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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,

the velocities are exchanged between the two balls

both the balls come to rest

both of them move at right angles to the original line of motion

one ball comes to rest and another ball travels back with velocity 2V

Medium

View solution>

Two identical balls collide head on. The initial velocity of one is $0.75$ cm $^{- 1}$ . While that of the other is $- 0.43$ ms $^{- 1}$ . If the collision is perfectly elastic. Their respective final velocities are

0.75

^{- 1}, - 0.43

^{- 1}

- 0.43

^{- 1}, 0.75

^{- 1}

- 0.75

^{- 1}, 0.43

^{- 1}

0.43

^{- 1}, 0.75

^{- 1}

Medium

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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

0.98 J

0.098 J

9.8 J

98 J

Medium

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Work Energy And Power

Physics 4859 Views

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 500gm ball moving at 15m/s slows down uniformly until it stops. If the ball travels 15m, what was the average net force applied while it was coming to a stop?

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

Same force acts on two bodies of masses m1​ and m2​ (m2​> m1​). After travelling the same distance,

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 is moving with a velocity 1 ms−1 a force F is needed to stop it within a distance x. If the speed of the body is 3 ms−1, the force needed to stop it with in the same distance (x) will be:

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,

Two identical balls collide head on. The initial velocity of one is 0.75cm−1. While that of the other is −0.43ms−1 . If the collision is perfectly elastic. Their respective final velocities are

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
4859 Views

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 $500 g m$ ball moving at $15 m / s$ slows down uniformly until it stops. If the ball travels $15 m$ , what was the average net force applied while it was coming to a stop?

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

Same force acts on two bodies of masses $m_{1}$ and $m_{2}$ ( $m_{2}$ > $m_{1}$ ). After travelling the same distance,

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 is moving with a velocity $1 m s^{- 1}$ a force $F$ is needed to stop it within a distance $x$ . If the speed of the body is $3 m s^{- 1}$ , the force needed to stop it with in the same distance ( $x$ ) will be:

Two identical balls collide head on. The initial velocity of one is $0.75$ cm $^{- 1}$ . While that of the other is $- 0.43$ ms $^{- 1}$ . If the collision is perfectly elastic. Their respective final velocities are