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PHXI05:LAWS OF MOTION

363472 A $0.5 k g$ ball moving with speed of $1.2 m / s$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is:
supporting img

1

12 N

2

24 N

3

48 N

4

96 N

Explanation:

Change in velocity
$= 2 v \sin 30^{\circ} = 2 \times 12 \times \frac{1}{2} = 12 m / s$
$F o r c e = \frac{m Δ v}{Δ t} = \frac{0.5 \times 12}{0.25} = 24 N$
supporting img

PHXI05:LAWS OF MOTION

363473 Assertion :
There is no appreciable 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 assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363474 Figures (a), (b), (c) and (d) show variation of force with time.
supporting img

The impulse in highest in figure.

1 (b)

2 (c)

3 (a)

4 (d)

Explanation:

Impulse is given by the area under force-time graph. $i . e .$ $I = \int_{0}^{t} F d t .$
(a) : $I = (1 / 2) \times 1 \times 0.5 = 0.25 N s$
(b) : $I = 0.5 \times 2 = 1 N s$
(c) : $I = (1 / 2) \times 1 \times 0.75 = 0.375 N s$
(d) : $I = (1 / 2) \times 2 \times 0.5 = 0.5 N s$
Impulse is highest in (b).

JEE - 2023

PHXI05:LAWS OF MOTION

363475 A ball of mass 50 $g m$ is dropped from a height $h = 10 m$ . It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for $Δ t = 0.01 \sec .$ The impulse of the impact force is: (take $g = 10 m / s^{2}$ )

1

105 N - s

2

1.3 N - s

3

1300 N - s

4

0.35 N - s

Explanation:

The velocity before collision is
$v_{1} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 10} = 10 \sqrt{2}$
$K . E_{1} = \frac{1}{2} m {(10 \sqrt{2})}^{2}$
Given that
$K . E_{2} = \frac{1}{4} (\frac{1}{2} m {(10 \sqrt{2})}^{2}) = \frac{1}{2} m v_{2}^{2} \Rightarrow v_{2} = 5 \sqrt{2}$
$\Rightarrow Δ P = 0.05 \times 5 \sqrt{2}$
$= 0.35 N s$

PHXI05:LAWS OF MOTION

363472 A $0.5 k g$ ball moving with speed of $1.2 m / s$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is:
supporting img

1

12 N

2

24 N

3

48 N

4

96 N

Explanation:

Change in velocity
$= 2 v \sin 30^{\circ} = 2 \times 12 \times \frac{1}{2} = 12 m / s$
$F o r c e = \frac{m Δ v}{Δ t} = \frac{0.5 \times 12}{0.25} = 24 N$
supporting img

PHXI05:LAWS OF MOTION

363473 Assertion :
There is no appreciable 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 assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363474 Figures (a), (b), (c) and (d) show variation of force with time.
supporting img

The impulse in highest in figure.

1 (b)

2 (c)

3 (a)

4 (d)

Explanation:

Impulse is given by the area under force-time graph. $i . e .$ $I = \int_{0}^{t} F d t .$
(a) : $I = (1 / 2) \times 1 \times 0.5 = 0.25 N s$
(b) : $I = 0.5 \times 2 = 1 N s$
(c) : $I = (1 / 2) \times 1 \times 0.75 = 0.375 N s$
(d) : $I = (1 / 2) \times 2 \times 0.5 = 0.5 N s$
Impulse is highest in (b).

JEE - 2023

PHXI05:LAWS OF MOTION

363475 A ball of mass 50 $g m$ is dropped from a height $h = 10 m$ . It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for $Δ t = 0.01 \sec .$ The impulse of the impact force is: (take $g = 10 m / s^{2}$ )

1

105 N - s

2

1.3 N - s

3

1300 N - s

4

0.35 N - s

Explanation:

The velocity before collision is
$v_{1} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 10} = 10 \sqrt{2}$
$K . E_{1} = \frac{1}{2} m {(10 \sqrt{2})}^{2}$
Given that
$K . E_{2} = \frac{1}{4} (\frac{1}{2} m {(10 \sqrt{2})}^{2}) = \frac{1}{2} m v_{2}^{2} \Rightarrow v_{2} = 5 \sqrt{2}$
$\Rightarrow Δ P = 0.05 \times 5 \sqrt{2}$
$= 0.35 N s$

PHXI05:LAWS OF MOTION

363472 A $0.5 k g$ ball moving with speed of $1.2 m / s$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is:
supporting img

1

12 N

2

24 N

3

48 N

4

96 N

Explanation:

Change in velocity
$= 2 v \sin 30^{\circ} = 2 \times 12 \times \frac{1}{2} = 12 m / s$
$F o r c e = \frac{m Δ v}{Δ t} = \frac{0.5 \times 12}{0.25} = 24 N$
supporting img

PHXI05:LAWS OF MOTION

363473 Assertion :
There is no appreciable 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 assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363474 Figures (a), (b), (c) and (d) show variation of force with time.
supporting img

The impulse in highest in figure.

1 (b)

2 (c)

3 (a)

4 (d)

Explanation:

Impulse is given by the area under force-time graph. $i . e .$ $I = \int_{0}^{t} F d t .$
(a) : $I = (1 / 2) \times 1 \times 0.5 = 0.25 N s$
(b) : $I = 0.5 \times 2 = 1 N s$
(c) : $I = (1 / 2) \times 1 \times 0.75 = 0.375 N s$
(d) : $I = (1 / 2) \times 2 \times 0.5 = 0.5 N s$
Impulse is highest in (b).

JEE - 2023

PHXI05:LAWS OF MOTION

363475 A ball of mass 50 $g m$ is dropped from a height $h = 10 m$ . It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for $Δ t = 0.01 \sec .$ The impulse of the impact force is: (take $g = 10 m / s^{2}$ )

1

105 N - s

2

1.3 N - s

3

1300 N - s

4

0.35 N - s

Explanation:

The velocity before collision is
$v_{1} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 10} = 10 \sqrt{2}$
$K . E_{1} = \frac{1}{2} m {(10 \sqrt{2})}^{2}$
Given that
$K . E_{2} = \frac{1}{4} (\frac{1}{2} m {(10 \sqrt{2})}^{2}) = \frac{1}{2} m v_{2}^{2} \Rightarrow v_{2} = 5 \sqrt{2}$
$\Rightarrow Δ P = 0.05 \times 5 \sqrt{2}$
$= 0.35 N s$

PHXI05:LAWS OF MOTION

363472 A $0.5 k g$ ball moving with speed of $1.2 m / s$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is:
supporting img

1

12 N

2

24 N

3

48 N

4

96 N

Explanation:

Change in velocity
$= 2 v \sin 30^{\circ} = 2 \times 12 \times \frac{1}{2} = 12 m / s$
$F o r c e = \frac{m Δ v}{Δ t} = \frac{0.5 \times 12}{0.25} = 24 N$
supporting img

PHXI05:LAWS OF MOTION

363473 Assertion :
There is no appreciable 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 assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI05:LAWS OF MOTION

363474 Figures (a), (b), (c) and (d) show variation of force with time.
supporting img

The impulse in highest in figure.

1 (b)

2 (c)

3 (a)

4 (d)

Explanation:

Impulse is given by the area under force-time graph. $i . e .$ $I = \int_{0}^{t} F d t .$
(a) : $I = (1 / 2) \times 1 \times 0.5 = 0.25 N s$
(b) : $I = 0.5 \times 2 = 1 N s$
(c) : $I = (1 / 2) \times 1 \times 0.75 = 0.375 N s$
(d) : $I = (1 / 2) \times 2 \times 0.5 = 0.5 N s$
Impulse is highest in (b).

JEE - 2023

PHXI05:LAWS OF MOTION

363475 A ball of mass 50 $g m$ is dropped from a height $h = 10 m$ . It rebounds losing 75 percent of its kinetic energy. If it remains in contact with the ground for $Δ t = 0.01 \sec .$ The impulse of the impact force is: (take $g = 10 m / s^{2}$ )

1

105 N - s

2

1.3 N - s

3

1300 N - s

4

0.35 N - s

Explanation:

The velocity before collision is
$v_{1} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 10} = 10 \sqrt{2}$
$K . E_{1} = \frac{1}{2} m {(10 \sqrt{2})}^{2}$
Given that
$K . E_{2} = \frac{1}{4} (\frac{1}{2} m {(10 \sqrt{2})}^{2}) = \frac{1}{2} m v_{2}^{2} \Rightarrow v_{2} = 5 \sqrt{2}$
$\Rightarrow Δ P = 0.05 \times 5 \sqrt{2}$
$= 0.35 N s$