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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> $Medium Questions$

Work Energy And Power

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

When\quad { P }_{ 1 }\quad and\quad { P }_{ 2 }\quad collide,\quad their\quad momentum\quad is\quad exchanged\quad (\quad since\quad the\quad \\ collsion\quad is\quad elastic\quad and\quad the\quad two\quad bodies\quad are\quad similar).\quad When\quad { P }_{ 2 }\quad and\quad { P }_{ 3 }\quad \\ same\quad thing\quad happens.\quad Thus\quad the\quad momentum\quad of\quad { P }_{ 1 }\quad is\quad finally\quad transferred\\ to\quad { P }_{ 4 }.\quad { P }_{ 1 },\quad { P }_{ 2 }\quad and\quad { P }_{ 3 }\quad will\quad have\quad zero\quad velocity\quad and\quad { P }_{ 4 }\quad will\quad have\quad velocity\quad =v.

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 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 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

Q: A sphere of mass m moving with constant velocity u, collides with another stationary sphere of same mass. If e is the coefficient of restitution, the ratio of the final velocities of the first and second sphere is

using momentum conservation v_{1}+v_{2}=u e=\dfrac{v_{2}-v_{1}}{u-0} \Rightarrow v_{2}-v_{1}=eu \Rightarrow v_{2}=\dfrac{(1+e)}{2}u \Rightarrow v_{1}=\dfrac{(1-e)}{2}u v_{1} : v_{2}=\dfrac{1-e}{1+e}

Q: The change in gravitational potential energy per unit space (distance) of a body represents:

PE=mgh Change in PE w.r.t. height is: \dfrac{d \left ( PE \right )}{d h}=mg=weight

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 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

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

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

Let the velocity of bigger ball be v .Mass be M >> mass of small ball( m )After collision, let the velocities be v_1 and v_2 respectively.Hence, conserving momentum: Mv = Mv_1+ mv_2 And for elastic collision: v= v_2-v_1 => v_1= v_2-v Replacing this in momentum equation: Mv = Mv_2 -Mv + mv_2 => v_2 = 2v(M/(M+m)) which approximates to 2v , the limit M>> m

Physics
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JEE Mains JEE Advanced AIIMS BITSAT Easy Qs Med Qs Hard Qs

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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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.

3 m s^{- 1}

5 m s^{- 1}

- 9 m s^{- 1}

30 m s^{- 1}

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 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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A sphere of mass m moving with constant velocity u, collides with another stationary sphere of same mass. If e is the coefficient of restitution, the ratio of the final velocities of the first and second sphere is

\frac{1 + e}{1 - e}

\frac{1 - e}{1 + e}

\frac{e}{1 - e}

\frac{1 + e}{e}

Medium

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The change in gravitational potential energy per unit space (distance) of a body represents:

Mass

Kinetic energy

Linear momentum

Weight

Medium

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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 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

9 kmph

18 kmph

27 kmph

36 kmph

Medium

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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

2 m / s

4 m / s

2 \times 1 0^{4} m / s

2 \times 1 0^{3} m / s

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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Work Energy And Power

Physics 205 Views

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 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 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 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

A sphere of mass m moving with constant velocity u, collides with another stationary sphere of same mass. If e is the coefficient of restitution, the ratio of the final velocities of the first and second sphere is

The change in gravitational potential energy per unit space (distance) of a body represents:

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

Physics
205 Views

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 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 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 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: