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Laws of Motion

145791 A bullet of mass $20 g$ moving with a velocity of $200 m / s$ strikes a target and is brought to rest in ${(\frac{1}{50})}^{th}$ of a second. The impulse and average
force of impact are respectively.

1

2 Ns, 100 N

2

4 Ns, 200 N

3

2 Ns, 200 N

4

4 Ns, 100 N

Explanation:

B Given, $m = 20 g, v = 200 m / s$
momentum $(p) = mv$
$p = 20 \times 10^{- 3} kg \times 200 m / s$
$p = 4 kg m / s$
Impulse $=$ change in momentum $(p) = 4 kg m / s$
$∴ Force = \frac{Impulse}{Time}$
$= \frac{4}{(\frac{1}{50})} = 200 N$
The impulse and average force of impact are respectively $4 Ns, 200 N$ .

MHT-CET 2020

Laws of Motion

145792 Let a force $\vec{F} = - F k$ acts on the origin of Cartesian frame of reference. The moment of force about a point $(1, - 1)$ will be

1

F (\hat{i} - \hat{j})

2

F (\hat{i} + \hat{j})

3

- F (\hat{i} - \hat{j})

4

- F (\hat{i} + \hat{j})

Explanation:

B
The torque about the given position, $τ = \vec{r} \times F$
Here, $\vec{r} = \hat{i} - \hat{j}$ and $F = - F \hat{k}$
$∴ τ = (\hat{i} - \hat{j}) \times (- F \hat{k})$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 0 \\ 0 & 0 & - F \end{array} |$
$= F \hat{i} - \hat{j} (- F)$
$τ = F (\hat{i} + \hat{j})$

MHT-CET 2020

Laws of Motion

145793 A body of mass $2 kg$ is acted upon by two perpendicular forces $4 N$ and $3 N$ . The magnitude and direction of the acceleration of the body respectively are.

1

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

4 N

force

2

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

3

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{4}{3})

with respect to the direction of

4 N

force

4

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

Explanation:

A Given that
mass of body $(m) = 2 kg$
two perpendicular forces $= 4 N, 3 N$

$F_{net} = \sqrt{4^{2} + 3^{2}}$
$= 5 N$
$∴ F_{net} = ma$
$5 = 2 . a \Rightarrow a = 2.5 m / \sec^{2}$
direction of resultant
$\tan θ = \frac{F_{2}}{F_{1}}$
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} (\frac{3}{4})$

TS EAMCET 28.09.2020

Laws of Motion

145794 The resultant force of $5 N$ and $10 N$ cannot be

1

12 N

2

4 N

3

8 N

4

5 N

Explanation:

B The resultant force of $5 N$ and $10 N$ ,
$F_{max} = 10 + 5$
$F_{max} = 15 N$
$F_{min} = 10 - 5$
$F_{min} = 5 N$
Range of force $= 5 \leq F \leq 15$
Hence, option (b) is correct.

AP EAMCET (Medical)-07.10.2020

Laws of Motion

145796 Find the apparent weight of a body of mass, 1.0 $kg$ falling with an acceleration of $10 {ms}^{- 2}$ . $(g \approx 10 {ms}^{- 2})$

1

1 kg

-wt

2

2 kg - wt

3 0

4

0.5 kg - wt

Explanation:

C Given, mass $(m) = 1 kg$ , acceleration $(a) = 10$ $m / s^{2}, g = 10 m / s^{2}$
As the body is falling so using second law of motion
$mg - N = ma$
$N = m (g - a)$
$N = 1 (10 - 10)$
$N = 0$

AP EAMCET (18.09.2020) Shift-I

Laws of Motion

145791 A bullet of mass $20 g$ moving with a velocity of $200 m / s$ strikes a target and is brought to rest in ${(\frac{1}{50})}^{th}$ of a second. The impulse and average
force of impact are respectively.

1

2 Ns, 100 N

2

4 Ns, 200 N

3

2 Ns, 200 N

4

4 Ns, 100 N

Explanation:

B Given, $m = 20 g, v = 200 m / s$
momentum $(p) = mv$
$p = 20 \times 10^{- 3} kg \times 200 m / s$
$p = 4 kg m / s$
Impulse $=$ change in momentum $(p) = 4 kg m / s$
$∴ Force = \frac{Impulse}{Time}$
$= \frac{4}{(\frac{1}{50})} = 200 N$
The impulse and average force of impact are respectively $4 Ns, 200 N$ .

MHT-CET 2020

Laws of Motion

145792 Let a force $\vec{F} = - F k$ acts on the origin of Cartesian frame of reference. The moment of force about a point $(1, - 1)$ will be

1

F (\hat{i} - \hat{j})

2

F (\hat{i} + \hat{j})

3

- F (\hat{i} - \hat{j})

4

- F (\hat{i} + \hat{j})

Explanation:

B
The torque about the given position, $τ = \vec{r} \times F$
Here, $\vec{r} = \hat{i} - \hat{j}$ and $F = - F \hat{k}$
$∴ τ = (\hat{i} - \hat{j}) \times (- F \hat{k})$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 0 \\ 0 & 0 & - F \end{array} |$
$= F \hat{i} - \hat{j} (- F)$
$τ = F (\hat{i} + \hat{j})$

MHT-CET 2020

Laws of Motion

145793 A body of mass $2 kg$ is acted upon by two perpendicular forces $4 N$ and $3 N$ . The magnitude and direction of the acceleration of the body respectively are.

1

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

4 N

force

2

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

3

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{4}{3})

with respect to the direction of

4 N

force

4

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

Explanation:

A Given that
mass of body $(m) = 2 kg$
two perpendicular forces $= 4 N, 3 N$

$F_{net} = \sqrt{4^{2} + 3^{2}}$
$= 5 N$
$∴ F_{net} = ma$
$5 = 2 . a \Rightarrow a = 2.5 m / \sec^{2}$
direction of resultant
$\tan θ = \frac{F_{2}}{F_{1}}$
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} (\frac{3}{4})$

TS EAMCET 28.09.2020

Laws of Motion

145794 The resultant force of $5 N$ and $10 N$ cannot be

1

12 N

2

4 N

3

8 N

4

5 N

Explanation:

B The resultant force of $5 N$ and $10 N$ ,
$F_{max} = 10 + 5$
$F_{max} = 15 N$
$F_{min} = 10 - 5$
$F_{min} = 5 N$
Range of force $= 5 \leq F \leq 15$
Hence, option (b) is correct.

AP EAMCET (Medical)-07.10.2020

Laws of Motion

145796 Find the apparent weight of a body of mass, 1.0 $kg$ falling with an acceleration of $10 {ms}^{- 2}$ . $(g \approx 10 {ms}^{- 2})$

1

1 kg

-wt

2

2 kg - wt

3 0

4

0.5 kg - wt

Explanation:

C Given, mass $(m) = 1 kg$ , acceleration $(a) = 10$ $m / s^{2}, g = 10 m / s^{2}$
As the body is falling so using second law of motion
$mg - N = ma$
$N = m (g - a)$
$N = 1 (10 - 10)$
$N = 0$

AP EAMCET (18.09.2020) Shift-I

Laws of Motion

145791 A bullet of mass $20 g$ moving with a velocity of $200 m / s$ strikes a target and is brought to rest in ${(\frac{1}{50})}^{th}$ of a second. The impulse and average
force of impact are respectively.

1

2 Ns, 100 N

2

4 Ns, 200 N

3

2 Ns, 200 N

4

4 Ns, 100 N

Explanation:

B Given, $m = 20 g, v = 200 m / s$
momentum $(p) = mv$
$p = 20 \times 10^{- 3} kg \times 200 m / s$
$p = 4 kg m / s$
Impulse $=$ change in momentum $(p) = 4 kg m / s$
$∴ Force = \frac{Impulse}{Time}$
$= \frac{4}{(\frac{1}{50})} = 200 N$
The impulse and average force of impact are respectively $4 Ns, 200 N$ .

MHT-CET 2020

Laws of Motion

145792 Let a force $\vec{F} = - F k$ acts on the origin of Cartesian frame of reference. The moment of force about a point $(1, - 1)$ will be

1

F (\hat{i} - \hat{j})

2

F (\hat{i} + \hat{j})

3

- F (\hat{i} - \hat{j})

4

- F (\hat{i} + \hat{j})

Explanation:

B
The torque about the given position, $τ = \vec{r} \times F$
Here, $\vec{r} = \hat{i} - \hat{j}$ and $F = - F \hat{k}$
$∴ τ = (\hat{i} - \hat{j}) \times (- F \hat{k})$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 0 \\ 0 & 0 & - F \end{array} |$
$= F \hat{i} - \hat{j} (- F)$
$τ = F (\hat{i} + \hat{j})$

MHT-CET 2020

Laws of Motion

145793 A body of mass $2 kg$ is acted upon by two perpendicular forces $4 N$ and $3 N$ . The magnitude and direction of the acceleration of the body respectively are.

1

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

4 N

force

2

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

3

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{4}{3})

with respect to the direction of

4 N

force

4

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

Explanation:

A Given that
mass of body $(m) = 2 kg$
two perpendicular forces $= 4 N, 3 N$

$F_{net} = \sqrt{4^{2} + 3^{2}}$
$= 5 N$
$∴ F_{net} = ma$
$5 = 2 . a \Rightarrow a = 2.5 m / \sec^{2}$
direction of resultant
$\tan θ = \frac{F_{2}}{F_{1}}$
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} (\frac{3}{4})$

TS EAMCET 28.09.2020

Laws of Motion

145794 The resultant force of $5 N$ and $10 N$ cannot be

1

12 N

2

4 N

3

8 N

4

5 N

Explanation:

B The resultant force of $5 N$ and $10 N$ ,
$F_{max} = 10 + 5$
$F_{max} = 15 N$
$F_{min} = 10 - 5$
$F_{min} = 5 N$
Range of force $= 5 \leq F \leq 15$
Hence, option (b) is correct.

AP EAMCET (Medical)-07.10.2020

Laws of Motion

145796 Find the apparent weight of a body of mass, 1.0 $kg$ falling with an acceleration of $10 {ms}^{- 2}$ . $(g \approx 10 {ms}^{- 2})$

1

1 kg

-wt

2

2 kg - wt

3 0

4

0.5 kg - wt

Explanation:

C Given, mass $(m) = 1 kg$ , acceleration $(a) = 10$ $m / s^{2}, g = 10 m / s^{2}$
As the body is falling so using second law of motion
$mg - N = ma$
$N = m (g - a)$
$N = 1 (10 - 10)$
$N = 0$

AP EAMCET (18.09.2020) Shift-I

Laws of Motion

145791 A bullet of mass $20 g$ moving with a velocity of $200 m / s$ strikes a target and is brought to rest in ${(\frac{1}{50})}^{th}$ of a second. The impulse and average
force of impact are respectively.

1

2 Ns, 100 N

2

4 Ns, 200 N

3

2 Ns, 200 N

4

4 Ns, 100 N

Explanation:

B Given, $m = 20 g, v = 200 m / s$
momentum $(p) = mv$
$p = 20 \times 10^{- 3} kg \times 200 m / s$
$p = 4 kg m / s$
Impulse $=$ change in momentum $(p) = 4 kg m / s$
$∴ Force = \frac{Impulse}{Time}$
$= \frac{4}{(\frac{1}{50})} = 200 N$
The impulse and average force of impact are respectively $4 Ns, 200 N$ .

MHT-CET 2020

Laws of Motion

145792 Let a force $\vec{F} = - F k$ acts on the origin of Cartesian frame of reference. The moment of force about a point $(1, - 1)$ will be

1

F (\hat{i} - \hat{j})

2

F (\hat{i} + \hat{j})

3

- F (\hat{i} - \hat{j})

4

- F (\hat{i} + \hat{j})

Explanation:

B
The torque about the given position, $τ = \vec{r} \times F$
Here, $\vec{r} = \hat{i} - \hat{j}$ and $F = - F \hat{k}$
$∴ τ = (\hat{i} - \hat{j}) \times (- F \hat{k})$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 0 \\ 0 & 0 & - F \end{array} |$
$= F \hat{i} - \hat{j} (- F)$
$τ = F (\hat{i} + \hat{j})$

MHT-CET 2020

Laws of Motion

145793 A body of mass $2 kg$ is acted upon by two perpendicular forces $4 N$ and $3 N$ . The magnitude and direction of the acceleration of the body respectively are.

1

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

4 N

force

2

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

3

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{4}{3})

with respect to the direction of

4 N

force

4

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

Explanation:

A Given that
mass of body $(m) = 2 kg$
two perpendicular forces $= 4 N, 3 N$

$F_{net} = \sqrt{4^{2} + 3^{2}}$
$= 5 N$
$∴ F_{net} = ma$
$5 = 2 . a \Rightarrow a = 2.5 m / \sec^{2}$
direction of resultant
$\tan θ = \frac{F_{2}}{F_{1}}$
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} (\frac{3}{4})$

TS EAMCET 28.09.2020

Laws of Motion

145794 The resultant force of $5 N$ and $10 N$ cannot be

1

12 N

2

4 N

3

8 N

4

5 N

Explanation:

B The resultant force of $5 N$ and $10 N$ ,
$F_{max} = 10 + 5$
$F_{max} = 15 N$
$F_{min} = 10 - 5$
$F_{min} = 5 N$
Range of force $= 5 \leq F \leq 15$
Hence, option (b) is correct.

AP EAMCET (Medical)-07.10.2020

Laws of Motion

145796 Find the apparent weight of a body of mass, 1.0 $kg$ falling with an acceleration of $10 {ms}^{- 2}$ . $(g \approx 10 {ms}^{- 2})$

1

1 kg

-wt

2

2 kg - wt

3 0

4

0.5 kg - wt

Explanation:

C Given, mass $(m) = 1 kg$ , acceleration $(a) = 10$ $m / s^{2}, g = 10 m / s^{2}$
As the body is falling so using second law of motion
$mg - N = ma$
$N = m (g - a)$
$N = 1 (10 - 10)$
$N = 0$

AP EAMCET (18.09.2020) Shift-I

Laws of Motion

145791 A bullet of mass $20 g$ moving with a velocity of $200 m / s$ strikes a target and is brought to rest in ${(\frac{1}{50})}^{th}$ of a second. The impulse and average
force of impact are respectively.

1

2 Ns, 100 N

2

4 Ns, 200 N

3

2 Ns, 200 N

4

4 Ns, 100 N

Explanation:

B Given, $m = 20 g, v = 200 m / s$
momentum $(p) = mv$
$p = 20 \times 10^{- 3} kg \times 200 m / s$
$p = 4 kg m / s$
Impulse $=$ change in momentum $(p) = 4 kg m / s$
$∴ Force = \frac{Impulse}{Time}$
$= \frac{4}{(\frac{1}{50})} = 200 N$
The impulse and average force of impact are respectively $4 Ns, 200 N$ .

MHT-CET 2020

Laws of Motion

145792 Let a force $\vec{F} = - F k$ acts on the origin of Cartesian frame of reference. The moment of force about a point $(1, - 1)$ will be

1

F (\hat{i} - \hat{j})

2

F (\hat{i} + \hat{j})

3

- F (\hat{i} - \hat{j})

4

- F (\hat{i} + \hat{j})

Explanation:

B
The torque about the given position, $τ = \vec{r} \times F$
Here, $\vec{r} = \hat{i} - \hat{j}$ and $F = - F \hat{k}$
$∴ τ = (\hat{i} - \hat{j}) \times (- F \hat{k})$
$= | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 0 \\ 0 & 0 & - F \end{array} |$
$= F \hat{i} - \hat{j} (- F)$
$τ = F (\hat{i} + \hat{j})$

MHT-CET 2020

Laws of Motion

145793 A body of mass $2 kg$ is acted upon by two perpendicular forces $4 N$ and $3 N$ . The magnitude and direction of the acceleration of the body respectively are.

1

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

4 N

force

2

2.5 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

3

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{4}{3})

with respect to the direction of

4 N

force

4

2.0 m / s^{2} \cdot {Tan}^{- 1} (\frac{3}{4})

with respect to the direction of

3 N

force

Explanation:

A Given that
mass of body $(m) = 2 kg$
two perpendicular forces $= 4 N, 3 N$

$F_{net} = \sqrt{4^{2} + 3^{2}}$
$= 5 N$
$∴ F_{net} = ma$
$5 = 2 . a \Rightarrow a = 2.5 m / \sec^{2}$
direction of resultant
$\tan θ = \frac{F_{2}}{F_{1}}$
$\tan θ = \frac{3}{4}$
$θ = \tan^{- 1} (\frac{3}{4})$

TS EAMCET 28.09.2020

Laws of Motion

145794 The resultant force of $5 N$ and $10 N$ cannot be

1

12 N

2

4 N

3

8 N

4

5 N

Explanation:

B The resultant force of $5 N$ and $10 N$ ,
$F_{max} = 10 + 5$
$F_{max} = 15 N$
$F_{min} = 10 - 5$
$F_{min} = 5 N$
Range of force $= 5 \leq F \leq 15$
Hence, option (b) is correct.

AP EAMCET (Medical)-07.10.2020

Laws of Motion

145796 Find the apparent weight of a body of mass, 1.0 $kg$ falling with an acceleration of $10 {ms}^{- 2}$ . $(g \approx 10 {ms}^{- 2})$

1

1 kg

-wt

2

2 kg - wt

3 0

4

0.5 kg - wt

Explanation:

C Given, mass $(m) = 1 kg$ , acceleration $(a) = 10$ $m / s^{2}, g = 10 m / s^{2}$
As the body is falling so using second law of motion
$mg - N = ma$
$N = m (g - a)$
$N = 1 (10 - 10)$
$N = 0$

AP EAMCET (18.09.2020) Shift-I

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