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PHXI06:WORK ENERGY AND POWER

355286 A particle is moving with velocity $u = 2 m / s$ towards a heavy wall moving towards the particle with constant speed $v_{0} = 1 m / s$ as shown in the figure. Assuming collision to be elastic, find the velocity of the particle immediately after the collision.
supporting img

1

1 m / s

2

4 m / s

3

7 m / s

4

10 m / s

Explanation:

Let the velocity of the particle immediately after the collision be $v_{1}$ .
supporting img
For calculating velocity of the ball just after collision, we use coefficient of restitution equation
$e = (\frac{v_{2} - v_{1}}{u_{1} - u_{2}})$
$= \frac{relative velocity of separation}{relative velocity of approach}$
This equation is applicable along the direction of impulsive force.
Here $u_{1} = u, u_{2} = - v_{0}$
$v_{1} = - v$ and
$v_{2} = - v_{0}$
Here $v_{2} = - v_{0}$ .
since the wall being very heavy (infinite mass).
moves with constant velocity.
Hence $e = \frac{- v_{0} - (- v)}{u - (- v_{0})} = \frac{v - v_{0}}{u + v_{0}} (1)$
As collision is elastic.Putting the value of $e = 1$ ,we obtain
$1 = \frac{v - v_{0}}{u + v_{0}} \Rightarrow v = u + 2 v_{0}$
Putting $u = 2 m / s, v_{0} = 1 m / s$ ,we get
$v = 2 + 2 (1) = 4 m / s e c$

PHXI06:WORK ENERGY AND POWER

355287 A particle of mass ' $m$ ' is moving horizontally $w i$ th a speed ' $v$ ' towards a rigid wall which is moving in the opposite direction with a constant velocity $u$ . Assuming elastic impact between the wall and the particle, the work done by the wall in reflecting the particle is :

1

\frac{1}{2} m (v + u)^{2}

2

\frac{1}{2} m (v + u)

3 2 muv

4

2 m u (u + v)

Explanation:

Since the collision between the particle and the wall is elastic,
Velocity of separation $=$ velocity of approach $v^{'} - u = v + u$ $[v^{'} = velocity of particle after collision]$
Work done $(W)$ by the wall in reflecting the particle will be given by change in the $K . E$ . of the particle
$- W = \frac{1}{2} m v^{2} - \frac{1}{2} m v^{' 2} = \frac{1}{2} m v^{2} - \frac{1}{2} m (v + 2 u)^{2}$
$= \frac{1}{2} m v^{2} - \frac{1}{2} m (v^{2} + 4 u^{2} + 4 u v)$ $= \frac{1}{2} m (- 4 u^{2} - 4 u v)$
$= \frac{- 4}{2} m u (u + v) = - 2 m u (u + v)$
$∴ W = 2 m u (u + v)$ .
So correct option is (4)

PHXI06:WORK ENERGY AND POWER

355288 A ball of mass $m$ is moving with velocity $u$ towards another stationary ball of mass 2 $m$ . If collision is elastic then the percentage $K E$ transferred by ball 1 to ball 2 is

1

68.8 %

2

78.8 %

3

88.8 %

4

100 %

Explanation:

supporting img
From conservation $C M$
$m u = m v_{1} + 2 m v_{2}$
$u = v_{1} + 2 v_{2} (1)$
From coefficient of restitution
$v_{2} - v_{1} = u (2)$
From eq's (1) & (2)
$v_{2} = \frac{2}{3} v, v_{1} = \frac{- 1}{3} v$
$% Δ K = \frac{K_{2}}{K_{i}} 100 = \frac{\frac{1}{2} m \frac{8}{9} u^{2}}{\frac{1}{2} m u^{2}} \times 100 = 88.8 %$

PHXI06:WORK ENERGY AND POWER

355289 A particle of mass $m$ moves with $a$ velocity of $v_{0} = 20 m / s$ towards a wall, which is moving with velocity $v = 5 m / s$ . If the particle collides with the wall elastically, the speed of the particle just after the collision is
supporting img

1

30 m / s

2

20 m / s

3

25 m / s

4

22 m / s

Explanation:

supporting img
Velocity of seperation $=$ Velocity of approach
( $∵$ collision is elastic)
$\begin{array}{ll} ∴ & v - 5 = 20 + 5 \\ \Rightarrow & v = 30 m / s \end{array}$

PHXI06:WORK ENERGY AND POWER

355286 A particle is moving with velocity $u = 2 m / s$ towards a heavy wall moving towards the particle with constant speed $v_{0} = 1 m / s$ as shown in the figure. Assuming collision to be elastic, find the velocity of the particle immediately after the collision.
supporting img

1

1 m / s

2

4 m / s

3

7 m / s

4

10 m / s

Explanation:

Let the velocity of the particle immediately after the collision be $v_{1}$ .
supporting img
For calculating velocity of the ball just after collision, we use coefficient of restitution equation
$e = (\frac{v_{2} - v_{1}}{u_{1} - u_{2}})$
$= \frac{relative velocity of separation}{relative velocity of approach}$
This equation is applicable along the direction of impulsive force.
Here $u_{1} = u, u_{2} = - v_{0}$
$v_{1} = - v$ and
$v_{2} = - v_{0}$
Here $v_{2} = - v_{0}$ .
since the wall being very heavy (infinite mass).
moves with constant velocity.
Hence $e = \frac{- v_{0} - (- v)}{u - (- v_{0})} = \frac{v - v_{0}}{u + v_{0}} (1)$
As collision is elastic.Putting the value of $e = 1$ ,we obtain
$1 = \frac{v - v_{0}}{u + v_{0}} \Rightarrow v = u + 2 v_{0}$
Putting $u = 2 m / s, v_{0} = 1 m / s$ ,we get
$v = 2 + 2 (1) = 4 m / s e c$

PHXI06:WORK ENERGY AND POWER

355287 A particle of mass ' $m$ ' is moving horizontally $w i$ th a speed ' $v$ ' towards a rigid wall which is moving in the opposite direction with a constant velocity $u$ . Assuming elastic impact between the wall and the particle, the work done by the wall in reflecting the particle is :

1

\frac{1}{2} m (v + u)^{2}

2

\frac{1}{2} m (v + u)

3 2 muv

4

2 m u (u + v)

Explanation:

Since the collision between the particle and the wall is elastic,
Velocity of separation $=$ velocity of approach $v^{'} - u = v + u$ $[v^{'} = velocity of particle after collision]$
Work done $(W)$ by the wall in reflecting the particle will be given by change in the $K . E$ . of the particle
$- W = \frac{1}{2} m v^{2} - \frac{1}{2} m v^{' 2} = \frac{1}{2} m v^{2} - \frac{1}{2} m (v + 2 u)^{2}$
$= \frac{1}{2} m v^{2} - \frac{1}{2} m (v^{2} + 4 u^{2} + 4 u v)$ $= \frac{1}{2} m (- 4 u^{2} - 4 u v)$
$= \frac{- 4}{2} m u (u + v) = - 2 m u (u + v)$
$∴ W = 2 m u (u + v)$ .
So correct option is (4)

PHXI06:WORK ENERGY AND POWER

355288 A ball of mass $m$ is moving with velocity $u$ towards another stationary ball of mass 2 $m$ . If collision is elastic then the percentage $K E$ transferred by ball 1 to ball 2 is

1

68.8 %

2

78.8 %

3

88.8 %

4

100 %

Explanation:

supporting img
From conservation $C M$
$m u = m v_{1} + 2 m v_{2}$
$u = v_{1} + 2 v_{2} (1)$
From coefficient of restitution
$v_{2} - v_{1} = u (2)$
From eq's (1) & (2)
$v_{2} = \frac{2}{3} v, v_{1} = \frac{- 1}{3} v$
$% Δ K = \frac{K_{2}}{K_{i}} 100 = \frac{\frac{1}{2} m \frac{8}{9} u^{2}}{\frac{1}{2} m u^{2}} \times 100 = 88.8 %$

PHXI06:WORK ENERGY AND POWER

355289 A particle of mass $m$ moves with $a$ velocity of $v_{0} = 20 m / s$ towards a wall, which is moving with velocity $v = 5 m / s$ . If the particle collides with the wall elastically, the speed of the particle just after the collision is
supporting img

1

30 m / s

2

20 m / s

3

25 m / s

4

22 m / s

Explanation:

supporting img
Velocity of seperation $=$ Velocity of approach
( $∵$ collision is elastic)
$\begin{array}{ll} ∴ & v - 5 = 20 + 5 \\ \Rightarrow & v = 30 m / s \end{array}$

PHXI06:WORK ENERGY AND POWER

355286 A particle is moving with velocity $u = 2 m / s$ towards a heavy wall moving towards the particle with constant speed $v_{0} = 1 m / s$ as shown in the figure. Assuming collision to be elastic, find the velocity of the particle immediately after the collision.
supporting img

1

1 m / s

2

4 m / s

3

7 m / s

4

10 m / s

Explanation:

Let the velocity of the particle immediately after the collision be $v_{1}$ .
supporting img
For calculating velocity of the ball just after collision, we use coefficient of restitution equation
$e = (\frac{v_{2} - v_{1}}{u_{1} - u_{2}})$
$= \frac{relative velocity of separation}{relative velocity of approach}$
This equation is applicable along the direction of impulsive force.
Here $u_{1} = u, u_{2} = - v_{0}$
$v_{1} = - v$ and
$v_{2} = - v_{0}$
Here $v_{2} = - v_{0}$ .
since the wall being very heavy (infinite mass).
moves with constant velocity.
Hence $e = \frac{- v_{0} - (- v)}{u - (- v_{0})} = \frac{v - v_{0}}{u + v_{0}} (1)$
As collision is elastic.Putting the value of $e = 1$ ,we obtain
$1 = \frac{v - v_{0}}{u + v_{0}} \Rightarrow v = u + 2 v_{0}$
Putting $u = 2 m / s, v_{0} = 1 m / s$ ,we get
$v = 2 + 2 (1) = 4 m / s e c$

PHXI06:WORK ENERGY AND POWER

355287 A particle of mass ' $m$ ' is moving horizontally $w i$ th a speed ' $v$ ' towards a rigid wall which is moving in the opposite direction with a constant velocity $u$ . Assuming elastic impact between the wall and the particle, the work done by the wall in reflecting the particle is :

1

\frac{1}{2} m (v + u)^{2}

2

\frac{1}{2} m (v + u)

3 2 muv

4

2 m u (u + v)

Explanation:

Since the collision between the particle and the wall is elastic,
Velocity of separation $=$ velocity of approach $v^{'} - u = v + u$ $[v^{'} = velocity of particle after collision]$
Work done $(W)$ by the wall in reflecting the particle will be given by change in the $K . E$ . of the particle
$- W = \frac{1}{2} m v^{2} - \frac{1}{2} m v^{' 2} = \frac{1}{2} m v^{2} - \frac{1}{2} m (v + 2 u)^{2}$
$= \frac{1}{2} m v^{2} - \frac{1}{2} m (v^{2} + 4 u^{2} + 4 u v)$ $= \frac{1}{2} m (- 4 u^{2} - 4 u v)$
$= \frac{- 4}{2} m u (u + v) = - 2 m u (u + v)$
$∴ W = 2 m u (u + v)$ .
So correct option is (4)

PHXI06:WORK ENERGY AND POWER

355288 A ball of mass $m$ is moving with velocity $u$ towards another stationary ball of mass 2 $m$ . If collision is elastic then the percentage $K E$ transferred by ball 1 to ball 2 is

1

68.8 %

2

78.8 %

3

88.8 %

4

100 %

Explanation:

supporting img
From conservation $C M$
$m u = m v_{1} + 2 m v_{2}$
$u = v_{1} + 2 v_{2} (1)$
From coefficient of restitution
$v_{2} - v_{1} = u (2)$
From eq's (1) & (2)
$v_{2} = \frac{2}{3} v, v_{1} = \frac{- 1}{3} v$
$% Δ K = \frac{K_{2}}{K_{i}} 100 = \frac{\frac{1}{2} m \frac{8}{9} u^{2}}{\frac{1}{2} m u^{2}} \times 100 = 88.8 %$

PHXI06:WORK ENERGY AND POWER

355289 A particle of mass $m$ moves with $a$ velocity of $v_{0} = 20 m / s$ towards a wall, which is moving with velocity $v = 5 m / s$ . If the particle collides with the wall elastically, the speed of the particle just after the collision is
supporting img

1

30 m / s

2

20 m / s

3

25 m / s

4

22 m / s

Explanation:

supporting img
Velocity of seperation $=$ Velocity of approach
( $∵$ collision is elastic)
$\begin{array}{ll} ∴ & v - 5 = 20 + 5 \\ \Rightarrow & v = 30 m / s \end{array}$

PHXI06:WORK ENERGY AND POWER

355286 A particle is moving with velocity $u = 2 m / s$ towards a heavy wall moving towards the particle with constant speed $v_{0} = 1 m / s$ as shown in the figure. Assuming collision to be elastic, find the velocity of the particle immediately after the collision.
supporting img

1

1 m / s

2

4 m / s

3

7 m / s

4

10 m / s

Explanation:

Let the velocity of the particle immediately after the collision be $v_{1}$ .
supporting img
For calculating velocity of the ball just after collision, we use coefficient of restitution equation
$e = (\frac{v_{2} - v_{1}}{u_{1} - u_{2}})$
$= \frac{relative velocity of separation}{relative velocity of approach}$
This equation is applicable along the direction of impulsive force.
Here $u_{1} = u, u_{2} = - v_{0}$
$v_{1} = - v$ and
$v_{2} = - v_{0}$
Here $v_{2} = - v_{0}$ .
since the wall being very heavy (infinite mass).
moves with constant velocity.
Hence $e = \frac{- v_{0} - (- v)}{u - (- v_{0})} = \frac{v - v_{0}}{u + v_{0}} (1)$
As collision is elastic.Putting the value of $e = 1$ ,we obtain
$1 = \frac{v - v_{0}}{u + v_{0}} \Rightarrow v = u + 2 v_{0}$
Putting $u = 2 m / s, v_{0} = 1 m / s$ ,we get
$v = 2 + 2 (1) = 4 m / s e c$

PHXI06:WORK ENERGY AND POWER

355287 A particle of mass ' $m$ ' is moving horizontally $w i$ th a speed ' $v$ ' towards a rigid wall which is moving in the opposite direction with a constant velocity $u$ . Assuming elastic impact between the wall and the particle, the work done by the wall in reflecting the particle is :

1

\frac{1}{2} m (v + u)^{2}

2

\frac{1}{2} m (v + u)

3 2 muv

4

2 m u (u + v)

Explanation:

Since the collision between the particle and the wall is elastic,
Velocity of separation $=$ velocity of approach $v^{'} - u = v + u$ $[v^{'} = velocity of particle after collision]$
Work done $(W)$ by the wall in reflecting the particle will be given by change in the $K . E$ . of the particle
$- W = \frac{1}{2} m v^{2} - \frac{1}{2} m v^{' 2} = \frac{1}{2} m v^{2} - \frac{1}{2} m (v + 2 u)^{2}$
$= \frac{1}{2} m v^{2} - \frac{1}{2} m (v^{2} + 4 u^{2} + 4 u v)$ $= \frac{1}{2} m (- 4 u^{2} - 4 u v)$
$= \frac{- 4}{2} m u (u + v) = - 2 m u (u + v)$
$∴ W = 2 m u (u + v)$ .
So correct option is (4)

PHXI06:WORK ENERGY AND POWER

355288 A ball of mass $m$ is moving with velocity $u$ towards another stationary ball of mass 2 $m$ . If collision is elastic then the percentage $K E$ transferred by ball 1 to ball 2 is

1

68.8 %

2

78.8 %

3

88.8 %

4

100 %

Explanation:

supporting img
From conservation $C M$
$m u = m v_{1} + 2 m v_{2}$
$u = v_{1} + 2 v_{2} (1)$
From coefficient of restitution
$v_{2} - v_{1} = u (2)$
From eq's (1) & (2)
$v_{2} = \frac{2}{3} v, v_{1} = \frac{- 1}{3} v$
$% Δ K = \frac{K_{2}}{K_{i}} 100 = \frac{\frac{1}{2} m \frac{8}{9} u^{2}}{\frac{1}{2} m u^{2}} \times 100 = 88.8 %$

PHXI06:WORK ENERGY AND POWER

355289 A particle of mass $m$ moves with $a$ velocity of $v_{0} = 20 m / s$ towards a wall, which is moving with velocity $v = 5 m / s$ . If the particle collides with the wall elastically, the speed of the particle just after the collision is
supporting img

1

30 m / s

2

20 m / s

3

25 m / s

4

22 m / s

Explanation:

supporting img
Velocity of seperation $=$ Velocity of approach
( $∵$ collision is elastic)
$\begin{array}{ll} ∴ & v - 5 = 20 + 5 \\ \Rightarrow & v = 30 m / s \end{array}$