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PHXI06:WORK ENERGY AND POWER

355204 Assertion :
Quick collision between two bodies is more violent than a slow collision; even when the initial and final velocities are identical
Reason :
The momentum is greater in first case.

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:

As momentum, $p = m v$ or $p \propto v$ , i.e., momentum is directly proportional to its velocity, so the momentum is greater in a quicker collision between two bodies than in slower one.
Hence, due to greater momentum quicker collision between two bodies will be more violent even inital and final velocities are identical.

AIIMS - 2008

PHXI06:WORK ENERGY AND POWER

355205 Two particles masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time $t = 0$ . They collide at time $t_{0}$ . Their velocities become ${\vec{v}}_{1}^{'}$ and ${\vec{v}}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of $∣ (m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) ∣$ is

1

(m_{1} + m_{2}) g t_{0}

2 Zero

3

\frac{1}{2} (m_{1} + m_{2}) g t_{0}

4

2 (m_{1} + m_{2}) g t_{0}

Explanation:

A constant external force $(m_{1} + m_{2}) g$ acts on the system. The impluse given by this force, in time $t = 0$ to $t = 2 t_{0}$ is $(m_{1} + m_{2}) g \times 2 t_{0}$
$∴$ Change in montemum in this interval
$| m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} \vec{v_{2}}) | = 2 (m_{1} + m_{2}) g t_{0} .$

PHXI06:WORK ENERGY AND POWER

355206 Two balls at the same temperature collide, the conservable quantity is

1 Velocity

2 Momentum

3 Kinetic energy

4 Temperature

Explanation:

When two balls at the same temperature collide, some dfraction of the $K E$ appears in other forms of energy, like heat energy, sound energy. Hence, neither temperature, nor velocity or $K E$ will remain conserved. The only quantity which will remain conserved is their momentum.

PHXI06:WORK ENERGY AND POWER

355207 Two astronauts, $A$ and $B$ , both with mass of $60 k g$ , are moving along a straight line in the same direction in a "weightless" spaceship. Relative to the spaceship the speed of $A$ is $2 m / s$ and that of $B$ is $1 m / s$ . A is carrying a bag of mass $5 k g$ with him. To avoid collision with $B, A$ throws the bag with a speed $v$ relative to the spaceship towards $B$ and $B$ catches it. Find the minimum value of $v$ .
supporting img

1

7.8 m / s

2

26.0 m / s

3

14.0 m / s

4

9.2 m / s

Explanation:

$65 \times 2 = 5 v + 60 v^{1}$
$5 v + 60 = 65 v^{11}$
for no collision to take place
$v^{1} < v^{11} \Rightarrow \frac{130 - 5 v}{60} < \frac{5 v + 60}{65}$
$\Rightarrow 1690 - 65 v < 60 v + 720 < 970 < 125 v$
$v > \frac{970}{125} \Rightarrow v > 7.8$

PHXI06:WORK ENERGY AND POWER

355208 For a system to follow the law of conservation of linear momentum during a collision, the condition is
(i) total external force acting on the system is zero.
(ii) total external force acting on the system is finite and time of collision is negligible.
(iii) total internal force acting on the system is zero.

1 Only (i)

2 Only (ii)

3 Only(iii)

4 Only (i) or (ii)

Explanation:

From Newton's second law,
$F = \frac{d p}{d t}$
If $F = 0$ , then $\frac{d p}{d t} = 0$
$\Rightarrow p =$ constant
Thus, if total external force acting on the system is zero during collision,then the linear momentum of the system remains conserved.

PHXI06:WORK ENERGY AND POWER

355204 Assertion :
Quick collision between two bodies is more violent than a slow collision; even when the initial and final velocities are identical
Reason :
The momentum is greater in first case.

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:

As momentum, $p = m v$ or $p \propto v$ , i.e., momentum is directly proportional to its velocity, so the momentum is greater in a quicker collision between two bodies than in slower one.
Hence, due to greater momentum quicker collision between two bodies will be more violent even inital and final velocities are identical.

AIIMS - 2008

PHXI06:WORK ENERGY AND POWER

355205 Two particles masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time $t = 0$ . They collide at time $t_{0}$ . Their velocities become ${\vec{v}}_{1}^{'}$ and ${\vec{v}}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of $∣ (m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) ∣$ is

1

(m_{1} + m_{2}) g t_{0}

2 Zero

3

\frac{1}{2} (m_{1} + m_{2}) g t_{0}

4

2 (m_{1} + m_{2}) g t_{0}

Explanation:

A constant external force $(m_{1} + m_{2}) g$ acts on the system. The impluse given by this force, in time $t = 0$ to $t = 2 t_{0}$ is $(m_{1} + m_{2}) g \times 2 t_{0}$
$∴$ Change in montemum in this interval
$| m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} \vec{v_{2}}) | = 2 (m_{1} + m_{2}) g t_{0} .$

PHXI06:WORK ENERGY AND POWER

355206 Two balls at the same temperature collide, the conservable quantity is

1 Velocity

2 Momentum

3 Kinetic energy

4 Temperature

Explanation:

When two balls at the same temperature collide, some dfraction of the $K E$ appears in other forms of energy, like heat energy, sound energy. Hence, neither temperature, nor velocity or $K E$ will remain conserved. The only quantity which will remain conserved is their momentum.

PHXI06:WORK ENERGY AND POWER

355207 Two astronauts, $A$ and $B$ , both with mass of $60 k g$ , are moving along a straight line in the same direction in a "weightless" spaceship. Relative to the spaceship the speed of $A$ is $2 m / s$ and that of $B$ is $1 m / s$ . A is carrying a bag of mass $5 k g$ with him. To avoid collision with $B, A$ throws the bag with a speed $v$ relative to the spaceship towards $B$ and $B$ catches it. Find the minimum value of $v$ .
supporting img

1

7.8 m / s

2

26.0 m / s

3

14.0 m / s

4

9.2 m / s

Explanation:

$65 \times 2 = 5 v + 60 v^{1}$
$5 v + 60 = 65 v^{11}$
for no collision to take place
$v^{1} < v^{11} \Rightarrow \frac{130 - 5 v}{60} < \frac{5 v + 60}{65}$
$\Rightarrow 1690 - 65 v < 60 v + 720 < 970 < 125 v$
$v > \frac{970}{125} \Rightarrow v > 7.8$

PHXI06:WORK ENERGY AND POWER

355208 For a system to follow the law of conservation of linear momentum during a collision, the condition is
(i) total external force acting on the system is zero.
(ii) total external force acting on the system is finite and time of collision is negligible.
(iii) total internal force acting on the system is zero.

1 Only (i)

2 Only (ii)

3 Only(iii)

4 Only (i) or (ii)

Explanation:

From Newton's second law,
$F = \frac{d p}{d t}$
If $F = 0$ , then $\frac{d p}{d t} = 0$
$\Rightarrow p =$ constant
Thus, if total external force acting on the system is zero during collision,then the linear momentum of the system remains conserved.

PHXI06:WORK ENERGY AND POWER

355204 Assertion :
Quick collision between two bodies is more violent than a slow collision; even when the initial and final velocities are identical
Reason :
The momentum is greater in first case.

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:

As momentum, $p = m v$ or $p \propto v$ , i.e., momentum is directly proportional to its velocity, so the momentum is greater in a quicker collision between two bodies than in slower one.
Hence, due to greater momentum quicker collision between two bodies will be more violent even inital and final velocities are identical.

AIIMS - 2008

PHXI06:WORK ENERGY AND POWER

355205 Two particles masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time $t = 0$ . They collide at time $t_{0}$ . Their velocities become ${\vec{v}}_{1}^{'}$ and ${\vec{v}}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of $∣ (m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) ∣$ is

1

(m_{1} + m_{2}) g t_{0}

2 Zero

3

\frac{1}{2} (m_{1} + m_{2}) g t_{0}

4

2 (m_{1} + m_{2}) g t_{0}

Explanation:

A constant external force $(m_{1} + m_{2}) g$ acts on the system. The impluse given by this force, in time $t = 0$ to $t = 2 t_{0}$ is $(m_{1} + m_{2}) g \times 2 t_{0}$
$∴$ Change in montemum in this interval
$| m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} \vec{v_{2}}) | = 2 (m_{1} + m_{2}) g t_{0} .$

PHXI06:WORK ENERGY AND POWER

355206 Two balls at the same temperature collide, the conservable quantity is

1 Velocity

2 Momentum

3 Kinetic energy

4 Temperature

Explanation:

When two balls at the same temperature collide, some dfraction of the $K E$ appears in other forms of energy, like heat energy, sound energy. Hence, neither temperature, nor velocity or $K E$ will remain conserved. The only quantity which will remain conserved is their momentum.

PHXI06:WORK ENERGY AND POWER

355207 Two astronauts, $A$ and $B$ , both with mass of $60 k g$ , are moving along a straight line in the same direction in a "weightless" spaceship. Relative to the spaceship the speed of $A$ is $2 m / s$ and that of $B$ is $1 m / s$ . A is carrying a bag of mass $5 k g$ with him. To avoid collision with $B, A$ throws the bag with a speed $v$ relative to the spaceship towards $B$ and $B$ catches it. Find the minimum value of $v$ .
supporting img

1

7.8 m / s

2

26.0 m / s

3

14.0 m / s

4

9.2 m / s

Explanation:

$65 \times 2 = 5 v + 60 v^{1}$
$5 v + 60 = 65 v^{11}$
for no collision to take place
$v^{1} < v^{11} \Rightarrow \frac{130 - 5 v}{60} < \frac{5 v + 60}{65}$
$\Rightarrow 1690 - 65 v < 60 v + 720 < 970 < 125 v$
$v > \frac{970}{125} \Rightarrow v > 7.8$

PHXI06:WORK ENERGY AND POWER

355208 For a system to follow the law of conservation of linear momentum during a collision, the condition is
(i) total external force acting on the system is zero.
(ii) total external force acting on the system is finite and time of collision is negligible.
(iii) total internal force acting on the system is zero.

1 Only (i)

2 Only (ii)

3 Only(iii)

4 Only (i) or (ii)

Explanation:

From Newton's second law,
$F = \frac{d p}{d t}$
If $F = 0$ , then $\frac{d p}{d t} = 0$
$\Rightarrow p =$ constant
Thus, if total external force acting on the system is zero during collision,then the linear momentum of the system remains conserved.

PHXI06:WORK ENERGY AND POWER

355204 Assertion :
Quick collision between two bodies is more violent than a slow collision; even when the initial and final velocities are identical
Reason :
The momentum is greater in first case.

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:

As momentum, $p = m v$ or $p \propto v$ , i.e., momentum is directly proportional to its velocity, so the momentum is greater in a quicker collision between two bodies than in slower one.
Hence, due to greater momentum quicker collision between two bodies will be more violent even inital and final velocities are identical.

AIIMS - 2008

PHXI06:WORK ENERGY AND POWER

355205 Two particles masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time $t = 0$ . They collide at time $t_{0}$ . Their velocities become ${\vec{v}}_{1}^{'}$ and ${\vec{v}}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of $∣ (m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) ∣$ is

1

(m_{1} + m_{2}) g t_{0}

2 Zero

3

\frac{1}{2} (m_{1} + m_{2}) g t_{0}

4

2 (m_{1} + m_{2}) g t_{0}

Explanation:

A constant external force $(m_{1} + m_{2}) g$ acts on the system. The impluse given by this force, in time $t = 0$ to $t = 2 t_{0}$ is $(m_{1} + m_{2}) g \times 2 t_{0}$
$∴$ Change in montemum in this interval
$| m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} \vec{v_{2}}) | = 2 (m_{1} + m_{2}) g t_{0} .$

PHXI06:WORK ENERGY AND POWER

355206 Two balls at the same temperature collide, the conservable quantity is

1 Velocity

2 Momentum

3 Kinetic energy

4 Temperature

Explanation:

When two balls at the same temperature collide, some dfraction of the $K E$ appears in other forms of energy, like heat energy, sound energy. Hence, neither temperature, nor velocity or $K E$ will remain conserved. The only quantity which will remain conserved is their momentum.

PHXI06:WORK ENERGY AND POWER

355207 Two astronauts, $A$ and $B$ , both with mass of $60 k g$ , are moving along a straight line in the same direction in a "weightless" spaceship. Relative to the spaceship the speed of $A$ is $2 m / s$ and that of $B$ is $1 m / s$ . A is carrying a bag of mass $5 k g$ with him. To avoid collision with $B, A$ throws the bag with a speed $v$ relative to the spaceship towards $B$ and $B$ catches it. Find the minimum value of $v$ .
supporting img

1

7.8 m / s

2

26.0 m / s

3

14.0 m / s

4

9.2 m / s

Explanation:

$65 \times 2 = 5 v + 60 v^{1}$
$5 v + 60 = 65 v^{11}$
for no collision to take place
$v^{1} < v^{11} \Rightarrow \frac{130 - 5 v}{60} < \frac{5 v + 60}{65}$
$\Rightarrow 1690 - 65 v < 60 v + 720 < 970 < 125 v$
$v > \frac{970}{125} \Rightarrow v > 7.8$

PHXI06:WORK ENERGY AND POWER

355208 For a system to follow the law of conservation of linear momentum during a collision, the condition is
(i) total external force acting on the system is zero.
(ii) total external force acting on the system is finite and time of collision is negligible.
(iii) total internal force acting on the system is zero.

1 Only (i)

2 Only (ii)

3 Only(iii)

4 Only (i) or (ii)

Explanation:

From Newton's second law,
$F = \frac{d p}{d t}$
If $F = 0$ , then $\frac{d p}{d t} = 0$
$\Rightarrow p =$ constant
Thus, if total external force acting on the system is zero during collision,then the linear momentum of the system remains conserved.

PHXI06:WORK ENERGY AND POWER

355204 Assertion :
Quick collision between two bodies is more violent than a slow collision; even when the initial and final velocities are identical
Reason :
The momentum is greater in first case.

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:

As momentum, $p = m v$ or $p \propto v$ , i.e., momentum is directly proportional to its velocity, so the momentum is greater in a quicker collision between two bodies than in slower one.
Hence, due to greater momentum quicker collision between two bodies will be more violent even inital and final velocities are identical.

AIIMS - 2008

PHXI06:WORK ENERGY AND POWER

355205 Two particles masses $m_{1}$ and $m_{2}$ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time $t = 0$ . They collide at time $t_{0}$ . Their velocities become ${\vec{v}}_{1}^{'}$ and ${\vec{v}}_{2}^{'}$ at time $2 t_{0}$ while still moving in air. The value of $∣ (m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}) ∣$ is

1

(m_{1} + m_{2}) g t_{0}

2 Zero

3

\frac{1}{2} (m_{1} + m_{2}) g t_{0}

4

2 (m_{1} + m_{2}) g t_{0}

Explanation:

A constant external force $(m_{1} + m_{2}) g$ acts on the system. The impluse given by this force, in time $t = 0$ to $t = 2 t_{0}$ is $(m_{1} + m_{2}) g \times 2 t_{0}$
$∴$ Change in montemum in this interval
$| m_{1} {\vec{v}}_{1}^{'} + m_{2} {\vec{v}}_{2}^{'} - (m_{1} {\vec{v}}_{1} + m_{2} \vec{v_{2}}) | = 2 (m_{1} + m_{2}) g t_{0} .$

PHXI06:WORK ENERGY AND POWER

355206 Two balls at the same temperature collide, the conservable quantity is

1 Velocity

2 Momentum

3 Kinetic energy

4 Temperature

Explanation:

When two balls at the same temperature collide, some dfraction of the $K E$ appears in other forms of energy, like heat energy, sound energy. Hence, neither temperature, nor velocity or $K E$ will remain conserved. The only quantity which will remain conserved is their momentum.

PHXI06:WORK ENERGY AND POWER

355207 Two astronauts, $A$ and $B$ , both with mass of $60 k g$ , are moving along a straight line in the same direction in a "weightless" spaceship. Relative to the spaceship the speed of $A$ is $2 m / s$ and that of $B$ is $1 m / s$ . A is carrying a bag of mass $5 k g$ with him. To avoid collision with $B, A$ throws the bag with a speed $v$ relative to the spaceship towards $B$ and $B$ catches it. Find the minimum value of $v$ .
supporting img

1

7.8 m / s

2

26.0 m / s

3

14.0 m / s

4

9.2 m / s

Explanation:

$65 \times 2 = 5 v + 60 v^{1}$
$5 v + 60 = 65 v^{11}$
for no collision to take place
$v^{1} < v^{11} \Rightarrow \frac{130 - 5 v}{60} < \frac{5 v + 60}{65}$
$\Rightarrow 1690 - 65 v < 60 v + 720 < 970 < 125 v$
$v > \frac{970}{125} \Rightarrow v > 7.8$

PHXI06:WORK ENERGY AND POWER

355208 For a system to follow the law of conservation of linear momentum during a collision, the condition is
(i) total external force acting on the system is zero.
(ii) total external force acting on the system is finite and time of collision is negligible.
(iii) total internal force acting on the system is zero.

1 Only (i)

2 Only (ii)

3 Only(iii)

4 Only (i) or (ii)

Explanation:

From Newton's second law,
$F = \frac{d p}{d t}$
If $F = 0$ , then $\frac{d p}{d t} = 0$
$\Rightarrow p =$ constant
Thus, if total external force acting on the system is zero during collision,then the linear momentum of the system remains conserved.

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