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LAWS OF MOTION (ADDITIONAL)

371631 Two masses of $1 g$ and $4 g$ are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is

1

4 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : 16

Explanation:

B Given that,
Mass of bodies, $m_{1} = 1 kg$
$m_{2} = 4 kg$
Both kinetic energy are equal
Kinetic energy $(K) = \frac{P^{2}}{2 m}$
$P = \sqrt{2 m K}$
According to question kinetic energy is same.
So,
$P \propto \sqrt{m}$
$∴ \frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$= \sqrt{\frac{1}{4}}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2}$
$P_{1} : P_{2} = 1 : 2$

TS EAMCET 28.09.2020

LAWS OF MOTION (ADDITIONAL)

371632 Two bodies of masses $m$ and $4 m$ have kinetic energies in the ratio $1 : 2$ . Their momentum $p_{1}$ and $p_{2}$ are in the ratio

1

1 : 2 \sqrt{2}

2

1 : 2 \sqrt{3}

3

2 \sqrt{2} : 1

4

3 \sqrt{2} : 1

5

1 : 3 \sqrt{2}

Explanation:

A Given,
$m_{1} = m \frac{(K.E.)_{1}}{m_{2}} = 4 m, \frac{1}{2}$
$\frac{P_{1}}{P_{2}} = ?$
We know that,
$K . E = \frac{P^{2}}{2 m}$
$K . E \propto \frac{P^{2}}{m}$
$\frac{(K . E)_{1}}{(K . E)_{2}} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{m_{2}}{m_{1}})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{4 m}{m})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{1}{8} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2 \sqrt{2}}$

Kerala CEE 04.07.2022

LAWS OF MOTION (ADDITIONAL)

371633 A body of mass 2 kg moves with an acceleration $3 {ms}^{- 2}$ . The change in momentum in one second is:

1

\frac{2}{3} kg m s^{- 1}

2

\frac{3}{2} kg m s^{- 1}

3

6 kg m s^{- 1}

4

5 kg m s^{- 1}

Explanation:

C Given that,
Mass of the body $(m) = 2 kg$
Acceleration of the body (a) $= 3 m / \sec^{2}$
By Newton's second law
Rate of change of momentum is equal to applied force
$F = \frac{dp}{dt}$
$\int_{0}^{1} F d t = \int_{P_{1}}^{P_{2}} d p$
$F \int_{0}^{1} dt = \int_{P_{1}}^{P_{2}} dp$
$F [t]_{0}^{1} = [p]_{p_{1}}^{p_{2}}$
$F (1 - 0) = p_{2} - p_{1}$
$F = Δ p$
$∵ F = ma$
$F = 2 \times 3$
$F = 6 N$
$∴ Δ p = 6 kg m s$
$So, change of momentum in one second is 6 kg {ms}^{- 1}$

AP EAMCET-06.09.2021

LAWS OF MOTION (ADDITIONAL)

371634 Area under the force-time graph gives the change in

1 velocity

2 acceleration

3 linear momentum

4 angular momentum

5 impulsive force

Explanation:

E
If we calculate area of triangle $OAB$ ,
Area of shaped portion,
$= \frac{1}{2} \times Base \times Altitude$
$= \frac{1}{2} \times time \times F_{peak}$
The above term gives impulse or change in momentum of body.
Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.

Kerala CEE 2021

LAWS OF MOTION (ADDITIONAL)

371635 A ball of mass $0.15 kg$ is dropped from a height $10 m$ , strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly $(g = 10 m / s^{2})$

1 0

2

4.2 kg - m / s

3

2.1 kg - m / s

4

1.4 kg - m / s

Explanation:

B Given,
Mass of the ball $= 0.15 kg$
Height from which ball is dropped $= 10 m$

We know that,
Impulse, $I =$ change in linear momentum $(Δ p)$
Velocity of ball at ground, $v = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 10}$
$v = 10 \sqrt{2} m / s$
$I = Δ p$
$= {mv}_{f} - {mv}_{i}$
$= mv - [m (- v)]$
$= 2 mv$
$= 2 \times 0.15 \times 10 \sqrt{2}$
$= 4.2 kg - m / s$

NEET- 2021

LAWS OF MOTION (ADDITIONAL)

371631 Two masses of $1 g$ and $4 g$ are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is

1

4 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : 16

Explanation:

B Given that,
Mass of bodies, $m_{1} = 1 kg$
$m_{2} = 4 kg$
Both kinetic energy are equal
Kinetic energy $(K) = \frac{P^{2}}{2 m}$
$P = \sqrt{2 m K}$
According to question kinetic energy is same.
So,
$P \propto \sqrt{m}$
$∴ \frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$= \sqrt{\frac{1}{4}}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2}$
$P_{1} : P_{2} = 1 : 2$

TS EAMCET 28.09.2020

LAWS OF MOTION (ADDITIONAL)

371632 Two bodies of masses $m$ and $4 m$ have kinetic energies in the ratio $1 : 2$ . Their momentum $p_{1}$ and $p_{2}$ are in the ratio

1

1 : 2 \sqrt{2}

2

1 : 2 \sqrt{3}

3

2 \sqrt{2} : 1

4

3 \sqrt{2} : 1

5

1 : 3 \sqrt{2}

Explanation:

A Given,
$m_{1} = m \frac{(K.E.)_{1}}{m_{2}} = 4 m, \frac{1}{2}$
$\frac{P_{1}}{P_{2}} = ?$
We know that,
$K . E = \frac{P^{2}}{2 m}$
$K . E \propto \frac{P^{2}}{m}$
$\frac{(K . E)_{1}}{(K . E)_{2}} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{m_{2}}{m_{1}})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{4 m}{m})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{1}{8} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2 \sqrt{2}}$

Kerala CEE 04.07.2022

LAWS OF MOTION (ADDITIONAL)

371633 A body of mass 2 kg moves with an acceleration $3 {ms}^{- 2}$ . The change in momentum in one second is:

1

\frac{2}{3} kg m s^{- 1}

2

\frac{3}{2} kg m s^{- 1}

3

6 kg m s^{- 1}

4

5 kg m s^{- 1}

Explanation:

C Given that,
Mass of the body $(m) = 2 kg$
Acceleration of the body (a) $= 3 m / \sec^{2}$
By Newton's second law
Rate of change of momentum is equal to applied force
$F = \frac{dp}{dt}$
$\int_{0}^{1} F d t = \int_{P_{1}}^{P_{2}} d p$
$F \int_{0}^{1} dt = \int_{P_{1}}^{P_{2}} dp$
$F [t]_{0}^{1} = [p]_{p_{1}}^{p_{2}}$
$F (1 - 0) = p_{2} - p_{1}$
$F = Δ p$
$∵ F = ma$
$F = 2 \times 3$
$F = 6 N$
$∴ Δ p = 6 kg m s$
$So, change of momentum in one second is 6 kg {ms}^{- 1}$

AP EAMCET-06.09.2021

LAWS OF MOTION (ADDITIONAL)

371634 Area under the force-time graph gives the change in

1 velocity

2 acceleration

3 linear momentum

4 angular momentum

5 impulsive force

Explanation:

E
If we calculate area of triangle $OAB$ ,
Area of shaped portion,
$= \frac{1}{2} \times Base \times Altitude$
$= \frac{1}{2} \times time \times F_{peak}$
The above term gives impulse or change in momentum of body.
Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.

Kerala CEE 2021

LAWS OF MOTION (ADDITIONAL)

371635 A ball of mass $0.15 kg$ is dropped from a height $10 m$ , strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly $(g = 10 m / s^{2})$

1 0

2

4.2 kg - m / s

3

2.1 kg - m / s

4

1.4 kg - m / s

Explanation:

B Given,
Mass of the ball $= 0.15 kg$
Height from which ball is dropped $= 10 m$

We know that,
Impulse, $I =$ change in linear momentum $(Δ p)$
Velocity of ball at ground, $v = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 10}$
$v = 10 \sqrt{2} m / s$
$I = Δ p$
$= {mv}_{f} - {mv}_{i}$
$= mv - [m (- v)]$
$= 2 mv$
$= 2 \times 0.15 \times 10 \sqrt{2}$
$= 4.2 kg - m / s$

NEET- 2021

LAWS OF MOTION (ADDITIONAL)

371631 Two masses of $1 g$ and $4 g$ are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is

1

4 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : 16

Explanation:

B Given that,
Mass of bodies, $m_{1} = 1 kg$
$m_{2} = 4 kg$
Both kinetic energy are equal
Kinetic energy $(K) = \frac{P^{2}}{2 m}$
$P = \sqrt{2 m K}$
According to question kinetic energy is same.
So,
$P \propto \sqrt{m}$
$∴ \frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$= \sqrt{\frac{1}{4}}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2}$
$P_{1} : P_{2} = 1 : 2$

TS EAMCET 28.09.2020

LAWS OF MOTION (ADDITIONAL)

371632 Two bodies of masses $m$ and $4 m$ have kinetic energies in the ratio $1 : 2$ . Their momentum $p_{1}$ and $p_{2}$ are in the ratio

1

1 : 2 \sqrt{2}

2

1 : 2 \sqrt{3}

3

2 \sqrt{2} : 1

4

3 \sqrt{2} : 1

5

1 : 3 \sqrt{2}

Explanation:

A Given,
$m_{1} = m \frac{(K.E.)_{1}}{m_{2}} = 4 m, \frac{1}{2}$
$\frac{P_{1}}{P_{2}} = ?$
We know that,
$K . E = \frac{P^{2}}{2 m}$
$K . E \propto \frac{P^{2}}{m}$
$\frac{(K . E)_{1}}{(K . E)_{2}} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{m_{2}}{m_{1}})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{4 m}{m})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{1}{8} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2 \sqrt{2}}$

Kerala CEE 04.07.2022

LAWS OF MOTION (ADDITIONAL)

371633 A body of mass 2 kg moves with an acceleration $3 {ms}^{- 2}$ . The change in momentum in one second is:

1

\frac{2}{3} kg m s^{- 1}

2

\frac{3}{2} kg m s^{- 1}

3

6 kg m s^{- 1}

4

5 kg m s^{- 1}

Explanation:

C Given that,
Mass of the body $(m) = 2 kg$
Acceleration of the body (a) $= 3 m / \sec^{2}$
By Newton's second law
Rate of change of momentum is equal to applied force
$F = \frac{dp}{dt}$
$\int_{0}^{1} F d t = \int_{P_{1}}^{P_{2}} d p$
$F \int_{0}^{1} dt = \int_{P_{1}}^{P_{2}} dp$
$F [t]_{0}^{1} = [p]_{p_{1}}^{p_{2}}$
$F (1 - 0) = p_{2} - p_{1}$
$F = Δ p$
$∵ F = ma$
$F = 2 \times 3$
$F = 6 N$
$∴ Δ p = 6 kg m s$
$So, change of momentum in one second is 6 kg {ms}^{- 1}$

AP EAMCET-06.09.2021

LAWS OF MOTION (ADDITIONAL)

371634 Area under the force-time graph gives the change in

1 velocity

2 acceleration

3 linear momentum

4 angular momentum

5 impulsive force

Explanation:

E
If we calculate area of triangle $OAB$ ,
Area of shaped portion,
$= \frac{1}{2} \times Base \times Altitude$
$= \frac{1}{2} \times time \times F_{peak}$
The above term gives impulse or change in momentum of body.
Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.

Kerala CEE 2021

LAWS OF MOTION (ADDITIONAL)

371635 A ball of mass $0.15 kg$ is dropped from a height $10 m$ , strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly $(g = 10 m / s^{2})$

1 0

2

4.2 kg - m / s

3

2.1 kg - m / s

4

1.4 kg - m / s

Explanation:

B Given,
Mass of the ball $= 0.15 kg$
Height from which ball is dropped $= 10 m$

We know that,
Impulse, $I =$ change in linear momentum $(Δ p)$
Velocity of ball at ground, $v = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 10}$
$v = 10 \sqrt{2} m / s$
$I = Δ p$
$= {mv}_{f} - {mv}_{i}$
$= mv - [m (- v)]$
$= 2 mv$
$= 2 \times 0.15 \times 10 \sqrt{2}$
$= 4.2 kg - m / s$

NEET- 2021

LAWS OF MOTION (ADDITIONAL)

371631 Two masses of $1 g$ and $4 g$ are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is

1

4 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : 16

Explanation:

B Given that,
Mass of bodies, $m_{1} = 1 kg$
$m_{2} = 4 kg$
Both kinetic energy are equal
Kinetic energy $(K) = \frac{P^{2}}{2 m}$
$P = \sqrt{2 m K}$
According to question kinetic energy is same.
So,
$P \propto \sqrt{m}$
$∴ \frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$= \sqrt{\frac{1}{4}}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2}$
$P_{1} : P_{2} = 1 : 2$

TS EAMCET 28.09.2020

LAWS OF MOTION (ADDITIONAL)

371632 Two bodies of masses $m$ and $4 m$ have kinetic energies in the ratio $1 : 2$ . Their momentum $p_{1}$ and $p_{2}$ are in the ratio

1

1 : 2 \sqrt{2}

2

1 : 2 \sqrt{3}

3

2 \sqrt{2} : 1

4

3 \sqrt{2} : 1

5

1 : 3 \sqrt{2}

Explanation:

A Given,
$m_{1} = m \frac{(K.E.)_{1}}{m_{2}} = 4 m, \frac{1}{2}$
$\frac{P_{1}}{P_{2}} = ?$
We know that,
$K . E = \frac{P^{2}}{2 m}$
$K . E \propto \frac{P^{2}}{m}$
$\frac{(K . E)_{1}}{(K . E)_{2}} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{m_{2}}{m_{1}})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{4 m}{m})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{1}{8} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2 \sqrt{2}}$

Kerala CEE 04.07.2022

LAWS OF MOTION (ADDITIONAL)

371633 A body of mass 2 kg moves with an acceleration $3 {ms}^{- 2}$ . The change in momentum in one second is:

1

\frac{2}{3} kg m s^{- 1}

2

\frac{3}{2} kg m s^{- 1}

3

6 kg m s^{- 1}

4

5 kg m s^{- 1}

Explanation:

C Given that,
Mass of the body $(m) = 2 kg$
Acceleration of the body (a) $= 3 m / \sec^{2}$
By Newton's second law
Rate of change of momentum is equal to applied force
$F = \frac{dp}{dt}$
$\int_{0}^{1} F d t = \int_{P_{1}}^{P_{2}} d p$
$F \int_{0}^{1} dt = \int_{P_{1}}^{P_{2}} dp$
$F [t]_{0}^{1} = [p]_{p_{1}}^{p_{2}}$
$F (1 - 0) = p_{2} - p_{1}$
$F = Δ p$
$∵ F = ma$
$F = 2 \times 3$
$F = 6 N$
$∴ Δ p = 6 kg m s$
$So, change of momentum in one second is 6 kg {ms}^{- 1}$

AP EAMCET-06.09.2021

LAWS OF MOTION (ADDITIONAL)

371634 Area under the force-time graph gives the change in

1 velocity

2 acceleration

3 linear momentum

4 angular momentum

5 impulsive force

Explanation:

E
If we calculate area of triangle $OAB$ ,
Area of shaped portion,
$= \frac{1}{2} \times Base \times Altitude$
$= \frac{1}{2} \times time \times F_{peak}$
The above term gives impulse or change in momentum of body.
Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.

Kerala CEE 2021

LAWS OF MOTION (ADDITIONAL)

371635 A ball of mass $0.15 kg$ is dropped from a height $10 m$ , strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly $(g = 10 m / s^{2})$

1 0

2

4.2 kg - m / s

3

2.1 kg - m / s

4

1.4 kg - m / s

Explanation:

B Given,
Mass of the ball $= 0.15 kg$
Height from which ball is dropped $= 10 m$

We know that,
Impulse, $I =$ change in linear momentum $(Δ p)$
Velocity of ball at ground, $v = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 10}$
$v = 10 \sqrt{2} m / s$
$I = Δ p$
$= {mv}_{f} - {mv}_{i}$
$= mv - [m (- v)]$
$= 2 mv$
$= 2 \times 0.15 \times 10 \sqrt{2}$
$= 4.2 kg - m / s$

NEET- 2021

LAWS OF MOTION (ADDITIONAL)

371631 Two masses of $1 g$ and $4 g$ are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is

1

4 : 1

2

1 : 2

3

\sqrt{2} : 1

4

1 : 16

Explanation:

B Given that,
Mass of bodies, $m_{1} = 1 kg$
$m_{2} = 4 kg$
Both kinetic energy are equal
Kinetic energy $(K) = \frac{P^{2}}{2 m}$
$P = \sqrt{2 m K}$
According to question kinetic energy is same.
So,
$P \propto \sqrt{m}$
$∴ \frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$
$= \sqrt{\frac{1}{4}}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2}$
$P_{1} : P_{2} = 1 : 2$

TS EAMCET 28.09.2020

LAWS OF MOTION (ADDITIONAL)

371632 Two bodies of masses $m$ and $4 m$ have kinetic energies in the ratio $1 : 2$ . Their momentum $p_{1}$ and $p_{2}$ are in the ratio

1

1 : 2 \sqrt{2}

2

1 : 2 \sqrt{3}

3

2 \sqrt{2} : 1

4

3 \sqrt{2} : 1

5

1 : 3 \sqrt{2}

Explanation:

A Given,
$m_{1} = m \frac{(K.E.)_{1}}{m_{2}} = 4 m, \frac{1}{2}$
$\frac{P_{1}}{P_{2}} = ?$
We know that,
$K . E = \frac{P^{2}}{2 m}$
$K . E \propto \frac{P^{2}}{m}$
$\frac{(K . E)_{1}}{(K . E)_{2}} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{m_{2}}{m_{1}})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2} (\frac{4 m}{m})$
$\frac{1}{2} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{1}{8} = {(\frac{P_{1}}{P_{2}})}^{2}$
$\frac{P_{1}}{P_{2}} = \frac{1}{2 \sqrt{2}}$

Kerala CEE 04.07.2022

LAWS OF MOTION (ADDITIONAL)

371633 A body of mass 2 kg moves with an acceleration $3 {ms}^{- 2}$ . The change in momentum in one second is:

1

\frac{2}{3} kg m s^{- 1}

2

\frac{3}{2} kg m s^{- 1}

3

6 kg m s^{- 1}

4

5 kg m s^{- 1}

Explanation:

C Given that,
Mass of the body $(m) = 2 kg$
Acceleration of the body (a) $= 3 m / \sec^{2}$
By Newton's second law
Rate of change of momentum is equal to applied force
$F = \frac{dp}{dt}$
$\int_{0}^{1} F d t = \int_{P_{1}}^{P_{2}} d p$
$F \int_{0}^{1} dt = \int_{P_{1}}^{P_{2}} dp$
$F [t]_{0}^{1} = [p]_{p_{1}}^{p_{2}}$
$F (1 - 0) = p_{2} - p_{1}$
$F = Δ p$
$∵ F = ma$
$F = 2 \times 3$
$F = 6 N$
$∴ Δ p = 6 kg m s$
$So, change of momentum in one second is 6 kg {ms}^{- 1}$

AP EAMCET-06.09.2021

LAWS OF MOTION (ADDITIONAL)

371634 Area under the force-time graph gives the change in

1 velocity

2 acceleration

3 linear momentum

4 angular momentum

5 impulsive force

Explanation:

E
If we calculate area of triangle $OAB$ ,
Area of shaped portion,
$= \frac{1}{2} \times Base \times Altitude$
$= \frac{1}{2} \times time \times F_{peak}$
The above term gives impulse or change in momentum of body.
Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.

Kerala CEE 2021

LAWS OF MOTION (ADDITIONAL)

371635 A ball of mass $0.15 kg$ is dropped from a height $10 m$ , strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly $(g = 10 m / s^{2})$

1 0

2

4.2 kg - m / s

3

2.1 kg - m / s

4

1.4 kg - m / s

Explanation:

B Given,
Mass of the ball $= 0.15 kg$
Height from which ball is dropped $= 10 m$

We know that,
Impulse, $I =$ change in linear momentum $(Δ p)$
Velocity of ball at ground, $v = \sqrt{2 gh}$
$v = \sqrt{2 \times 10 \times 10}$
$v = 10 \sqrt{2} m / s$
$I = Δ p$
$= {mv}_{f} - {mv}_{i}$
$= mv - [m (- v)]$
$= 2 mv$
$= 2 \times 0.15 \times 10 \sqrt{2}$
$= 4.2 kg - m / s$

NEET- 2021