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Work, Energy and Power

268616 Choose the false statement

1 In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separation

2 In an inelastic collision the relative velocity of approach is less than the relative velocity of separation

3 In an inelastic collision the relative velocity of separation is less than the relative velocity of approach

4 In perfect inelastic collision relative velocity of separation is zero

Work, Energy and Power

268678 A $6 kg$ mass travelling at $2.5 {ms}^{- 1}$ collides head on with a stationary $4 kg$ mass. After the collision the $6 kg$ mass travels in its original direction with a speed of $1 {ms}^{- 1}$ . The final velocity of $4 k g$ mass is

1

1 {ms}^{- 1}

2

2.25 {ms}^{- 1}

3

2 {ms}^{- 1}

4

0 {ms}^{- 1}

Explanation:

According to law of conservation of linear momentum, $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Work, Energy and Power

268679 A body of mass $10 kg$ moving with a velocity of $5 {ms}^{- 1}$ hits a body of $1 g m$ at rest. The velocity of the second body after collision, assuming it to be perfectly elastic is

1

10 {ms}^{- 1}

2

5 {ms}^{- 1}

3

15 {ms}^{- 1}

4

0.10 {ms}^{- 1}

Explanation:

$v_{2} = ◻ \frac{2 m_{1}}{◻ m_{1} + m_{2}} ◻ ◻^{u_{1}}$

Work, Energy and Power

268680 A block of mass $1 kg$ moving with a speed of $4 {ms}^{- 1}$ , collides with another block of mass 2 $kg$ which is at rest. The lighter block comes to rest after collision. The loss in $KE$ of the system is

1

8 J

2

4 \times 10^{- 7} J

3

4 J

4

0 J

Explanation:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Δ K E = \frac{1}{2} m_{1} u_{1}^{2} - \frac{1}{2} m_{2} v_{2}^{2}$

Work, Energy and Power

268616 Choose the false statement

1 In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separation

2 In an inelastic collision the relative velocity of approach is less than the relative velocity of separation

3 In an inelastic collision the relative velocity of separation is less than the relative velocity of approach

4 In perfect inelastic collision relative velocity of separation is zero

Work, Energy and Power

268678 A $6 kg$ mass travelling at $2.5 {ms}^{- 1}$ collides head on with a stationary $4 kg$ mass. After the collision the $6 kg$ mass travels in its original direction with a speed of $1 {ms}^{- 1}$ . The final velocity of $4 k g$ mass is

1

1 {ms}^{- 1}

2

2.25 {ms}^{- 1}

3

2 {ms}^{- 1}

4

0 {ms}^{- 1}

Explanation:

According to law of conservation of linear momentum, $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Work, Energy and Power

268679 A body of mass $10 kg$ moving with a velocity of $5 {ms}^{- 1}$ hits a body of $1 g m$ at rest. The velocity of the second body after collision, assuming it to be perfectly elastic is

1

10 {ms}^{- 1}

2

5 {ms}^{- 1}

3

15 {ms}^{- 1}

4

0.10 {ms}^{- 1}

Explanation:

$v_{2} = ◻ \frac{2 m_{1}}{◻ m_{1} + m_{2}} ◻ ◻^{u_{1}}$

Work, Energy and Power

268680 A block of mass $1 kg$ moving with a speed of $4 {ms}^{- 1}$ , collides with another block of mass 2 $kg$ which is at rest. The lighter block comes to rest after collision. The loss in $KE$ of the system is

1

8 J

2

4 \times 10^{- 7} J

3

4 J

4

0 J

Explanation:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Δ K E = \frac{1}{2} m_{1} u_{1}^{2} - \frac{1}{2} m_{2} v_{2}^{2}$

Work, Energy and Power

268616 Choose the false statement

1 In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separation

2 In an inelastic collision the relative velocity of approach is less than the relative velocity of separation

3 In an inelastic collision the relative velocity of separation is less than the relative velocity of approach

4 In perfect inelastic collision relative velocity of separation is zero

Work, Energy and Power

268678 A $6 kg$ mass travelling at $2.5 {ms}^{- 1}$ collides head on with a stationary $4 kg$ mass. After the collision the $6 kg$ mass travels in its original direction with a speed of $1 {ms}^{- 1}$ . The final velocity of $4 k g$ mass is

1

1 {ms}^{- 1}

2

2.25 {ms}^{- 1}

3

2 {ms}^{- 1}

4

0 {ms}^{- 1}

Explanation:

According to law of conservation of linear momentum, $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Work, Energy and Power

268679 A body of mass $10 kg$ moving with a velocity of $5 {ms}^{- 1}$ hits a body of $1 g m$ at rest. The velocity of the second body after collision, assuming it to be perfectly elastic is

1

10 {ms}^{- 1}

2

5 {ms}^{- 1}

3

15 {ms}^{- 1}

4

0.10 {ms}^{- 1}

Explanation:

$v_{2} = ◻ \frac{2 m_{1}}{◻ m_{1} + m_{2}} ◻ ◻^{u_{1}}$

Work, Energy and Power

268680 A block of mass $1 kg$ moving with a speed of $4 {ms}^{- 1}$ , collides with another block of mass 2 $kg$ which is at rest. The lighter block comes to rest after collision. The loss in $KE$ of the system is

1

8 J

2

4 \times 10^{- 7} J

3

4 J

4

0 J

Explanation:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Δ K E = \frac{1}{2} m_{1} u_{1}^{2} - \frac{1}{2} m_{2} v_{2}^{2}$

Work, Energy and Power

268616 Choose the false statement

1 In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separation

2 In an inelastic collision the relative velocity of approach is less than the relative velocity of separation

3 In an inelastic collision the relative velocity of separation is less than the relative velocity of approach

4 In perfect inelastic collision relative velocity of separation is zero

Work, Energy and Power

268678 A $6 kg$ mass travelling at $2.5 {ms}^{- 1}$ collides head on with a stationary $4 kg$ mass. After the collision the $6 kg$ mass travels in its original direction with a speed of $1 {ms}^{- 1}$ . The final velocity of $4 k g$ mass is

1

1 {ms}^{- 1}

2

2.25 {ms}^{- 1}

3

2 {ms}^{- 1}

4

0 {ms}^{- 1}

Explanation:

According to law of conservation of linear momentum, $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

Work, Energy and Power

268679 A body of mass $10 kg$ moving with a velocity of $5 {ms}^{- 1}$ hits a body of $1 g m$ at rest. The velocity of the second body after collision, assuming it to be perfectly elastic is

1

10 {ms}^{- 1}

2

5 {ms}^{- 1}

3

15 {ms}^{- 1}

4

0.10 {ms}^{- 1}

Explanation:

$v_{2} = ◻ \frac{2 m_{1}}{◻ m_{1} + m_{2}} ◻ ◻^{u_{1}}$

Work, Energy and Power

268680 A block of mass $1 kg$ moving with a speed of $4 {ms}^{- 1}$ , collides with another block of mass 2 $kg$ which is at rest. The lighter block comes to rest after collision. The loss in $KE$ of the system is

1

8 J

2

4 \times 10^{- 7} J

3

4 J

4

0 J

Explanation:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$Δ K E = \frac{1}{2} m_{1} u_{1}^{2} - \frac{1}{2} m_{2} v_{2}^{2}$