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PHXI05:LAWS OF MOTION

363480 A ball of mass 100 $g m$ is released from a height $h_{1} = 5 m$ from the ground level and then rebounds to a height $h_{2} = 1.25 m$ . The time of contact of the ball and the ground is $Δ t = 0.01 \sec$ .The impulsive force offered by the ball on the ground is :

1

105 N

2

150 N

3

20.8 N

4

208 N

Explanation:

$F Δ t = Δ P \Rightarrow F Δ t = m (v_{1} + v_{2})$
$v_{1} = \sqrt{2 g h_{1}}$
$v_{2} = \sqrt{2 g h_{2}}$
$F = \frac{m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})}{Δ t}$
$= \frac{0.1 (\sqrt{2 \times 10 \times 5} + \sqrt{2 \times 10 \times 1.25})}{0.01}$
$= 150 N$

PHXI05:LAWS OF MOTION

363481 In the figure given, the position-time graph of a particle of mass $0.1 k g$ is shown. The impulse at $t = 2 s$ is
supporting img

1

0.2 k g - m s^{- 1}

2

- 0.2 k g - m s^{- 1}

3

0.1 k g - m s^{- 1}

4

- 0.4 k g - m s^{- 1}

Explanation:

From the graph, the body comes to rest after $t = 4 s e c$ .
$\Rightarrow$ Impulse $= P_{f} - P_{i}$
$= 0 - m v$
$= - m v$
but $v$ here is slope of $x - t$ graph
$\Rightarrow$ impulse $= - 0.1 \times 2 = - 0.2$

AIIMS - 2005

PHXI05:LAWS OF MOTION

363482 Assertion :
There will be large change in the position of the body during the action of the impulsive force.
Reason :
In case of impulsive force the time of action of the force is very short.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$Δ p = \int F d t$ is small during the action of an impulsive force, there is no appreciable change in the position of the body because the time of action of the force is indeed very short.Large $F$ causes much acceleration. But, during small $Δ t$ , the value of displacement $Δ s$ is not large. So correct option is (4)

PHXI05:LAWS OF MOTION

363483 A body of mass $1 k g$ moving with velocity $1 m / s$ makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time $1 s e c$ . Their force of interaction increases from zero to $F_{0}$ linearly in time $0.5 s$ and decreases linearly to zero in further time $0.5 s e c$ as shown in figure. Find the magnitude of force $F_{0}$
supporting img

1

1 N

2

5 N

3

2 N

4

4 N

Explanation:

In the one dimensional elastic collision with one body at rest, the body moving initially comes to rest and the one which was at rest earlier starts moving with the velocity that first body had before collision.
if $m$ and $V_{0}$ be the mass & velocity of body, the change in momentum
$= m V_{0} \Rightarrow \int F d t = m V_{0}$
$\Rightarrow \int F d t = m V_{0} \Rightarrow F = \frac{2 m V_{0}}{Δ t} = 2 N$

PHXI05:LAWS OF MOTION

363484 A force of $10 N$ acts on a body of mass $20 k g$ for $10 s .$ The magnitude of impulse is

1

50 k g - m s^{- 1}

2

100 k g - m s^{- 1}

3

300 k g - m s^{- 1}

4

1000 k g - m s^{- 1}

Explanation:

The impulse of the body is given by
$J = F \times Δ t$
$= 10 \times 10$
$= 100 k g - m s^{- 1}$

PHXI05:LAWS OF MOTION

363480 A ball of mass 100 $g m$ is released from a height $h_{1} = 5 m$ from the ground level and then rebounds to a height $h_{2} = 1.25 m$ . The time of contact of the ball and the ground is $Δ t = 0.01 \sec$ .The impulsive force offered by the ball on the ground is :

1

105 N

2

150 N

3

20.8 N

4

208 N

Explanation:

$F Δ t = Δ P \Rightarrow F Δ t = m (v_{1} + v_{2})$
$v_{1} = \sqrt{2 g h_{1}}$
$v_{2} = \sqrt{2 g h_{2}}$
$F = \frac{m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})}{Δ t}$
$= \frac{0.1 (\sqrt{2 \times 10 \times 5} + \sqrt{2 \times 10 \times 1.25})}{0.01}$
$= 150 N$

PHXI05:LAWS OF MOTION

363481 In the figure given, the position-time graph of a particle of mass $0.1 k g$ is shown. The impulse at $t = 2 s$ is
supporting img

1

0.2 k g - m s^{- 1}

2

- 0.2 k g - m s^{- 1}

3

0.1 k g - m s^{- 1}

4

- 0.4 k g - m s^{- 1}

Explanation:

From the graph, the body comes to rest after $t = 4 s e c$ .
$\Rightarrow$ Impulse $= P_{f} - P_{i}$
$= 0 - m v$
$= - m v$
but $v$ here is slope of $x - t$ graph
$\Rightarrow$ impulse $= - 0.1 \times 2 = - 0.2$

AIIMS - 2005

PHXI05:LAWS OF MOTION

363482 Assertion :
There will be large change in the position of the body during the action of the impulsive force.
Reason :
In case of impulsive force the time of action of the force is very short.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$Δ p = \int F d t$ is small during the action of an impulsive force, there is no appreciable change in the position of the body because the time of action of the force is indeed very short.Large $F$ causes much acceleration. But, during small $Δ t$ , the value of displacement $Δ s$ is not large. So correct option is (4)

PHXI05:LAWS OF MOTION

363483 A body of mass $1 k g$ moving with velocity $1 m / s$ makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time $1 s e c$ . Their force of interaction increases from zero to $F_{0}$ linearly in time $0.5 s$ and decreases linearly to zero in further time $0.5 s e c$ as shown in figure. Find the magnitude of force $F_{0}$
supporting img

1

1 N

2

5 N

3

2 N

4

4 N

Explanation:

In the one dimensional elastic collision with one body at rest, the body moving initially comes to rest and the one which was at rest earlier starts moving with the velocity that first body had before collision.
if $m$ and $V_{0}$ be the mass & velocity of body, the change in momentum
$= m V_{0} \Rightarrow \int F d t = m V_{0}$
$\Rightarrow \int F d t = m V_{0} \Rightarrow F = \frac{2 m V_{0}}{Δ t} = 2 N$

PHXI05:LAWS OF MOTION

363484 A force of $10 N$ acts on a body of mass $20 k g$ for $10 s .$ The magnitude of impulse is

1

50 k g - m s^{- 1}

2

100 k g - m s^{- 1}

3

300 k g - m s^{- 1}

4

1000 k g - m s^{- 1}

Explanation:

The impulse of the body is given by
$J = F \times Δ t$
$= 10 \times 10$
$= 100 k g - m s^{- 1}$

PHXI05:LAWS OF MOTION

363480 A ball of mass 100 $g m$ is released from a height $h_{1} = 5 m$ from the ground level and then rebounds to a height $h_{2} = 1.25 m$ . The time of contact of the ball and the ground is $Δ t = 0.01 \sec$ .The impulsive force offered by the ball on the ground is :

1

105 N

2

150 N

3

20.8 N

4

208 N

Explanation:

$F Δ t = Δ P \Rightarrow F Δ t = m (v_{1} + v_{2})$
$v_{1} = \sqrt{2 g h_{1}}$
$v_{2} = \sqrt{2 g h_{2}}$
$F = \frac{m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})}{Δ t}$
$= \frac{0.1 (\sqrt{2 \times 10 \times 5} + \sqrt{2 \times 10 \times 1.25})}{0.01}$
$= 150 N$

PHXI05:LAWS OF MOTION

363481 In the figure given, the position-time graph of a particle of mass $0.1 k g$ is shown. The impulse at $t = 2 s$ is
supporting img

1

0.2 k g - m s^{- 1}

2

- 0.2 k g - m s^{- 1}

3

0.1 k g - m s^{- 1}

4

- 0.4 k g - m s^{- 1}

Explanation:

From the graph, the body comes to rest after $t = 4 s e c$ .
$\Rightarrow$ Impulse $= P_{f} - P_{i}$
$= 0 - m v$
$= - m v$
but $v$ here is slope of $x - t$ graph
$\Rightarrow$ impulse $= - 0.1 \times 2 = - 0.2$

AIIMS - 2005

PHXI05:LAWS OF MOTION

363482 Assertion :
There will be large change in the position of the body during the action of the impulsive force.
Reason :
In case of impulsive force the time of action of the force is very short.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$Δ p = \int F d t$ is small during the action of an impulsive force, there is no appreciable change in the position of the body because the time of action of the force is indeed very short.Large $F$ causes much acceleration. But, during small $Δ t$ , the value of displacement $Δ s$ is not large. So correct option is (4)

PHXI05:LAWS OF MOTION

363483 A body of mass $1 k g$ moving with velocity $1 m / s$ makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time $1 s e c$ . Their force of interaction increases from zero to $F_{0}$ linearly in time $0.5 s$ and decreases linearly to zero in further time $0.5 s e c$ as shown in figure. Find the magnitude of force $F_{0}$
supporting img

1

1 N

2

5 N

3

2 N

4

4 N

Explanation:

In the one dimensional elastic collision with one body at rest, the body moving initially comes to rest and the one which was at rest earlier starts moving with the velocity that first body had before collision.
if $m$ and $V_{0}$ be the mass & velocity of body, the change in momentum
$= m V_{0} \Rightarrow \int F d t = m V_{0}$
$\Rightarrow \int F d t = m V_{0} \Rightarrow F = \frac{2 m V_{0}}{Δ t} = 2 N$

PHXI05:LAWS OF MOTION

363484 A force of $10 N$ acts on a body of mass $20 k g$ for $10 s .$ The magnitude of impulse is

1

50 k g - m s^{- 1}

2

100 k g - m s^{- 1}

3

300 k g - m s^{- 1}

4

1000 k g - m s^{- 1}

Explanation:

The impulse of the body is given by
$J = F \times Δ t$
$= 10 \times 10$
$= 100 k g - m s^{- 1}$

PHXI05:LAWS OF MOTION

363480 A ball of mass 100 $g m$ is released from a height $h_{1} = 5 m$ from the ground level and then rebounds to a height $h_{2} = 1.25 m$ . The time of contact of the ball and the ground is $Δ t = 0.01 \sec$ .The impulsive force offered by the ball on the ground is :

1

105 N

2

150 N

3

20.8 N

4

208 N

Explanation:

$F Δ t = Δ P \Rightarrow F Δ t = m (v_{1} + v_{2})$
$v_{1} = \sqrt{2 g h_{1}}$
$v_{2} = \sqrt{2 g h_{2}}$
$F = \frac{m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})}{Δ t}$
$= \frac{0.1 (\sqrt{2 \times 10 \times 5} + \sqrt{2 \times 10 \times 1.25})}{0.01}$
$= 150 N$

PHXI05:LAWS OF MOTION

363481 In the figure given, the position-time graph of a particle of mass $0.1 k g$ is shown. The impulse at $t = 2 s$ is
supporting img

1

0.2 k g - m s^{- 1}

2

- 0.2 k g - m s^{- 1}

3

0.1 k g - m s^{- 1}

4

- 0.4 k g - m s^{- 1}

Explanation:

From the graph, the body comes to rest after $t = 4 s e c$ .
$\Rightarrow$ Impulse $= P_{f} - P_{i}$
$= 0 - m v$
$= - m v$
but $v$ here is slope of $x - t$ graph
$\Rightarrow$ impulse $= - 0.1 \times 2 = - 0.2$

AIIMS - 2005

PHXI05:LAWS OF MOTION

363482 Assertion :
There will be large change in the position of the body during the action of the impulsive force.
Reason :
In case of impulsive force the time of action of the force is very short.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$Δ p = \int F d t$ is small during the action of an impulsive force, there is no appreciable change in the position of the body because the time of action of the force is indeed very short.Large $F$ causes much acceleration. But, during small $Δ t$ , the value of displacement $Δ s$ is not large. So correct option is (4)

PHXI05:LAWS OF MOTION

363483 A body of mass $1 k g$ moving with velocity $1 m / s$ makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time $1 s e c$ . Their force of interaction increases from zero to $F_{0}$ linearly in time $0.5 s$ and decreases linearly to zero in further time $0.5 s e c$ as shown in figure. Find the magnitude of force $F_{0}$
supporting img

1

1 N

2

5 N

3

2 N

4

4 N

Explanation:

In the one dimensional elastic collision with one body at rest, the body moving initially comes to rest and the one which was at rest earlier starts moving with the velocity that first body had before collision.
if $m$ and $V_{0}$ be the mass & velocity of body, the change in momentum
$= m V_{0} \Rightarrow \int F d t = m V_{0}$
$\Rightarrow \int F d t = m V_{0} \Rightarrow F = \frac{2 m V_{0}}{Δ t} = 2 N$

PHXI05:LAWS OF MOTION

363484 A force of $10 N$ acts on a body of mass $20 k g$ for $10 s .$ The magnitude of impulse is

1

50 k g - m s^{- 1}

2

100 k g - m s^{- 1}

3

300 k g - m s^{- 1}

4

1000 k g - m s^{- 1}

Explanation:

The impulse of the body is given by
$J = F \times Δ t$
$= 10 \times 10$
$= 100 k g - m s^{- 1}$

PHXI05:LAWS OF MOTION

363480 A ball of mass 100 $g m$ is released from a height $h_{1} = 5 m$ from the ground level and then rebounds to a height $h_{2} = 1.25 m$ . The time of contact of the ball and the ground is $Δ t = 0.01 \sec$ .The impulsive force offered by the ball on the ground is :

1

105 N

2

150 N

3

20.8 N

4

208 N

Explanation:

$F Δ t = Δ P \Rightarrow F Δ t = m (v_{1} + v_{2})$
$v_{1} = \sqrt{2 g h_{1}}$
$v_{2} = \sqrt{2 g h_{2}}$
$F = \frac{m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})}{Δ t}$
$= \frac{0.1 (\sqrt{2 \times 10 \times 5} + \sqrt{2 \times 10 \times 1.25})}{0.01}$
$= 150 N$

PHXI05:LAWS OF MOTION

363481 In the figure given, the position-time graph of a particle of mass $0.1 k g$ is shown. The impulse at $t = 2 s$ is
supporting img

1

0.2 k g - m s^{- 1}

2

- 0.2 k g - m s^{- 1}

3

0.1 k g - m s^{- 1}

4

- 0.4 k g - m s^{- 1}

Explanation:

From the graph, the body comes to rest after $t = 4 s e c$ .
$\Rightarrow$ Impulse $= P_{f} - P_{i}$
$= 0 - m v$
$= - m v$
but $v$ here is slope of $x - t$ graph
$\Rightarrow$ impulse $= - 0.1 \times 2 = - 0.2$

AIIMS - 2005

PHXI05:LAWS OF MOTION

363482 Assertion :
There will be large change in the position of the body during the action of the impulsive force.
Reason :
In case of impulsive force the time of action of the force is very short.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

$Δ p = \int F d t$ is small during the action of an impulsive force, there is no appreciable change in the position of the body because the time of action of the force is indeed very short.Large $F$ causes much acceleration. But, during small $Δ t$ , the value of displacement $Δ s$ is not large. So correct option is (4)

PHXI05:LAWS OF MOTION

363483 A body of mass $1 k g$ moving with velocity $1 m / s$ makes an elastic one dimensional collision with an identical stationary body. They are in contact for brief time $1 s e c$ . Their force of interaction increases from zero to $F_{0}$ linearly in time $0.5 s$ and decreases linearly to zero in further time $0.5 s e c$ as shown in figure. Find the magnitude of force $F_{0}$
supporting img

1

1 N

2

5 N

3

2 N

4

4 N

Explanation:

In the one dimensional elastic collision with one body at rest, the body moving initially comes to rest and the one which was at rest earlier starts moving with the velocity that first body had before collision.
if $m$ and $V_{0}$ be the mass & velocity of body, the change in momentum
$= m V_{0} \Rightarrow \int F d t = m V_{0}$
$\Rightarrow \int F d t = m V_{0} \Rightarrow F = \frac{2 m V_{0}}{Δ t} = 2 N$

PHXI05:LAWS OF MOTION

363484 A force of $10 N$ acts on a body of mass $20 k g$ for $10 s .$ The magnitude of impulse is

1

50 k g - m s^{- 1}

2

100 k g - m s^{- 1}

3

300 k g - m s^{- 1}

4

1000 k g - m s^{- 1}

Explanation:

The impulse of the body is given by
$J = F \times Δ t$
$= 10 \times 10$
$= 100 k g - m s^{- 1}$