145797
A force of \(10 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\) for \(0.25 \mathrm{~s}\) starting from rest. What is its momentum now?
1 \(0.25 \mathrm{~N}-\mathrm{s}\)
2 \(2.5 \mathrm{~N}-\mathrm{s}\)
3 \(0.5 \mathrm{~N}-\mathrm{s}\)
4 \(0.75 \mathrm{~N}-\mathrm{s}\)
Explanation:
B Given, \(\mathrm{F}=10 \mathrm{~N}, \mathrm{~m}=0.5 \mathrm{~kg}\) and \(\mathrm{t}=0.25 \mathrm{sec}\) We know that, Momentum (p) \(=\mathrm{mv}\) Then, \(\text { Force }(F)=m a\) Acceleration \((\mathrm{a})=\frac{\mathrm{F}}{\mathrm{m}}=\frac{10}{0.5}=20 \mathrm{~m} / \mathrm{s}^{2}\) From the Newton's first law of motion \(\therefore \quad \mathrm{v} =\mathrm{u}+\mathrm{at}\) \(=0+20 \times 0.25\) \(=5 \mathrm{~m} / \mathrm{s}\) Then, from equation (i) \(\mathrm{p}=0.5 \times 5\) \(=2.5 \mathrm{~N}-\mathrm{s}\)
JIPMER-2018
Laws of Motion
145798
A man throws a ball of mass \(3.0 \mathrm{~kg}\) with a speed of \(5.0 \mathrm{~ms}^{-1}\). His hand is in contact with the ball for \(0.2 \mathrm{~s}\). If the throws 4 balls in 2 seconds, the average force exerted by him in 1 second is
1 \(15 \mathrm{~N}\)
2 \(30 \mathrm{~N}\)
3 \(150 \mathrm{~N}\)
4 \(75 \mathrm{~N}\)
Explanation:
B Given, Mass of the ball \((\mathrm{m})=3 \mathrm{~kg}\) Speed \((\mathrm{v})=5 \mathrm{~m} / \mathrm{s}\) Time taken by the man to throw a ball is, \(\Delta \mathrm{t}=\frac{2}{4}=0.5 \mathrm{~s}\) Change in momentum of the ball \(=\) Mass \(\times\) Speed \(=\mathrm{m} . \mathrm{v}\) \(=3 \times 5\) \(=15 \mathrm{Ns}\) Force \(=\) Rate of change in linear momentum \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
AMU-2018
Laws of Motion
145799
When a train takes a turn, the passengers are thrown outwards because of-
1 acceleration of motion
2 speed of motion
3 inertia of direction
4 Both (a) and (c)
Explanation:
C Law of inertia states that, everybody continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force. There are three types of inertia- 1.Inertia of rest 2.Inertia of motion 3.Inertia of direction In this question third type of inertia that causes the passengers to be thrown outwards when the driver makes sudden turn.
BCECE-2018
Laws of Motion
145803
A block \(P\) of mass \(M_{P}\) is in contact with another block \(Q\) of mass \(M_{Q}\) as shown in the figure and they are placed on a smooth floor. Force on block \(Q\) is
1 \(\frac{M_{P}}{M_{P}+M_{Q}}\)
2 \(\frac{M_{Q} F}{M_{P}+M_{Q}}\)
3 \(\frac{M_{P} F}{M_{Q}}\)
4 \(\frac{M_{Q} F}{M_{P}}\)
Explanation:
B Given, mass of block \(\mathrm{P}=\mathrm{M}_{\mathrm{P}}\), mass of block \(Q=M_{Q}\) Since, \(\quad \mathrm{F}_{\mathrm{ext}}=\mathrm{ma}\) \(\mathrm{F}=\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}\right) \mathrm{a}\) \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\) Now, \(\therefore \quad \mathrm{N}=\mathrm{M}_{\mathrm{Q}} \mathrm{a}\) Put the value of ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
145797
A force of \(10 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\) for \(0.25 \mathrm{~s}\) starting from rest. What is its momentum now?
1 \(0.25 \mathrm{~N}-\mathrm{s}\)
2 \(2.5 \mathrm{~N}-\mathrm{s}\)
3 \(0.5 \mathrm{~N}-\mathrm{s}\)
4 \(0.75 \mathrm{~N}-\mathrm{s}\)
Explanation:
B Given, \(\mathrm{F}=10 \mathrm{~N}, \mathrm{~m}=0.5 \mathrm{~kg}\) and \(\mathrm{t}=0.25 \mathrm{sec}\) We know that, Momentum (p) \(=\mathrm{mv}\) Then, \(\text { Force }(F)=m a\) Acceleration \((\mathrm{a})=\frac{\mathrm{F}}{\mathrm{m}}=\frac{10}{0.5}=20 \mathrm{~m} / \mathrm{s}^{2}\) From the Newton's first law of motion \(\therefore \quad \mathrm{v} =\mathrm{u}+\mathrm{at}\) \(=0+20 \times 0.25\) \(=5 \mathrm{~m} / \mathrm{s}\) Then, from equation (i) \(\mathrm{p}=0.5 \times 5\) \(=2.5 \mathrm{~N}-\mathrm{s}\)
JIPMER-2018
Laws of Motion
145798
A man throws a ball of mass \(3.0 \mathrm{~kg}\) with a speed of \(5.0 \mathrm{~ms}^{-1}\). His hand is in contact with the ball for \(0.2 \mathrm{~s}\). If the throws 4 balls in 2 seconds, the average force exerted by him in 1 second is
1 \(15 \mathrm{~N}\)
2 \(30 \mathrm{~N}\)
3 \(150 \mathrm{~N}\)
4 \(75 \mathrm{~N}\)
Explanation:
B Given, Mass of the ball \((\mathrm{m})=3 \mathrm{~kg}\) Speed \((\mathrm{v})=5 \mathrm{~m} / \mathrm{s}\) Time taken by the man to throw a ball is, \(\Delta \mathrm{t}=\frac{2}{4}=0.5 \mathrm{~s}\) Change in momentum of the ball \(=\) Mass \(\times\) Speed \(=\mathrm{m} . \mathrm{v}\) \(=3 \times 5\) \(=15 \mathrm{Ns}\) Force \(=\) Rate of change in linear momentum \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
AMU-2018
Laws of Motion
145799
When a train takes a turn, the passengers are thrown outwards because of-
1 acceleration of motion
2 speed of motion
3 inertia of direction
4 Both (a) and (c)
Explanation:
C Law of inertia states that, everybody continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force. There are three types of inertia- 1.Inertia of rest 2.Inertia of motion 3.Inertia of direction In this question third type of inertia that causes the passengers to be thrown outwards when the driver makes sudden turn.
BCECE-2018
Laws of Motion
145803
A block \(P\) of mass \(M_{P}\) is in contact with another block \(Q\) of mass \(M_{Q}\) as shown in the figure and they are placed on a smooth floor. Force on block \(Q\) is
1 \(\frac{M_{P}}{M_{P}+M_{Q}}\)
2 \(\frac{M_{Q} F}{M_{P}+M_{Q}}\)
3 \(\frac{M_{P} F}{M_{Q}}\)
4 \(\frac{M_{Q} F}{M_{P}}\)
Explanation:
B Given, mass of block \(\mathrm{P}=\mathrm{M}_{\mathrm{P}}\), mass of block \(Q=M_{Q}\) Since, \(\quad \mathrm{F}_{\mathrm{ext}}=\mathrm{ma}\) \(\mathrm{F}=\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}\right) \mathrm{a}\) \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\) Now, \(\therefore \quad \mathrm{N}=\mathrm{M}_{\mathrm{Q}} \mathrm{a}\) Put the value of ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
145797
A force of \(10 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\) for \(0.25 \mathrm{~s}\) starting from rest. What is its momentum now?
1 \(0.25 \mathrm{~N}-\mathrm{s}\)
2 \(2.5 \mathrm{~N}-\mathrm{s}\)
3 \(0.5 \mathrm{~N}-\mathrm{s}\)
4 \(0.75 \mathrm{~N}-\mathrm{s}\)
Explanation:
B Given, \(\mathrm{F}=10 \mathrm{~N}, \mathrm{~m}=0.5 \mathrm{~kg}\) and \(\mathrm{t}=0.25 \mathrm{sec}\) We know that, Momentum (p) \(=\mathrm{mv}\) Then, \(\text { Force }(F)=m a\) Acceleration \((\mathrm{a})=\frac{\mathrm{F}}{\mathrm{m}}=\frac{10}{0.5}=20 \mathrm{~m} / \mathrm{s}^{2}\) From the Newton's first law of motion \(\therefore \quad \mathrm{v} =\mathrm{u}+\mathrm{at}\) \(=0+20 \times 0.25\) \(=5 \mathrm{~m} / \mathrm{s}\) Then, from equation (i) \(\mathrm{p}=0.5 \times 5\) \(=2.5 \mathrm{~N}-\mathrm{s}\)
JIPMER-2018
Laws of Motion
145798
A man throws a ball of mass \(3.0 \mathrm{~kg}\) with a speed of \(5.0 \mathrm{~ms}^{-1}\). His hand is in contact with the ball for \(0.2 \mathrm{~s}\). If the throws 4 balls in 2 seconds, the average force exerted by him in 1 second is
1 \(15 \mathrm{~N}\)
2 \(30 \mathrm{~N}\)
3 \(150 \mathrm{~N}\)
4 \(75 \mathrm{~N}\)
Explanation:
B Given, Mass of the ball \((\mathrm{m})=3 \mathrm{~kg}\) Speed \((\mathrm{v})=5 \mathrm{~m} / \mathrm{s}\) Time taken by the man to throw a ball is, \(\Delta \mathrm{t}=\frac{2}{4}=0.5 \mathrm{~s}\) Change in momentum of the ball \(=\) Mass \(\times\) Speed \(=\mathrm{m} . \mathrm{v}\) \(=3 \times 5\) \(=15 \mathrm{Ns}\) Force \(=\) Rate of change in linear momentum \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
AMU-2018
Laws of Motion
145799
When a train takes a turn, the passengers are thrown outwards because of-
1 acceleration of motion
2 speed of motion
3 inertia of direction
4 Both (a) and (c)
Explanation:
C Law of inertia states that, everybody continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force. There are three types of inertia- 1.Inertia of rest 2.Inertia of motion 3.Inertia of direction In this question third type of inertia that causes the passengers to be thrown outwards when the driver makes sudden turn.
BCECE-2018
Laws of Motion
145803
A block \(P\) of mass \(M_{P}\) is in contact with another block \(Q\) of mass \(M_{Q}\) as shown in the figure and they are placed on a smooth floor. Force on block \(Q\) is
1 \(\frac{M_{P}}{M_{P}+M_{Q}}\)
2 \(\frac{M_{Q} F}{M_{P}+M_{Q}}\)
3 \(\frac{M_{P} F}{M_{Q}}\)
4 \(\frac{M_{Q} F}{M_{P}}\)
Explanation:
B Given, mass of block \(\mathrm{P}=\mathrm{M}_{\mathrm{P}}\), mass of block \(Q=M_{Q}\) Since, \(\quad \mathrm{F}_{\mathrm{ext}}=\mathrm{ma}\) \(\mathrm{F}=\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}\right) \mathrm{a}\) \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\) Now, \(\therefore \quad \mathrm{N}=\mathrm{M}_{\mathrm{Q}} \mathrm{a}\) Put the value of ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
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Laws of Motion
145797
A force of \(10 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\) for \(0.25 \mathrm{~s}\) starting from rest. What is its momentum now?
1 \(0.25 \mathrm{~N}-\mathrm{s}\)
2 \(2.5 \mathrm{~N}-\mathrm{s}\)
3 \(0.5 \mathrm{~N}-\mathrm{s}\)
4 \(0.75 \mathrm{~N}-\mathrm{s}\)
Explanation:
B Given, \(\mathrm{F}=10 \mathrm{~N}, \mathrm{~m}=0.5 \mathrm{~kg}\) and \(\mathrm{t}=0.25 \mathrm{sec}\) We know that, Momentum (p) \(=\mathrm{mv}\) Then, \(\text { Force }(F)=m a\) Acceleration \((\mathrm{a})=\frac{\mathrm{F}}{\mathrm{m}}=\frac{10}{0.5}=20 \mathrm{~m} / \mathrm{s}^{2}\) From the Newton's first law of motion \(\therefore \quad \mathrm{v} =\mathrm{u}+\mathrm{at}\) \(=0+20 \times 0.25\) \(=5 \mathrm{~m} / \mathrm{s}\) Then, from equation (i) \(\mathrm{p}=0.5 \times 5\) \(=2.5 \mathrm{~N}-\mathrm{s}\)
JIPMER-2018
Laws of Motion
145798
A man throws a ball of mass \(3.0 \mathrm{~kg}\) with a speed of \(5.0 \mathrm{~ms}^{-1}\). His hand is in contact with the ball for \(0.2 \mathrm{~s}\). If the throws 4 balls in 2 seconds, the average force exerted by him in 1 second is
1 \(15 \mathrm{~N}\)
2 \(30 \mathrm{~N}\)
3 \(150 \mathrm{~N}\)
4 \(75 \mathrm{~N}\)
Explanation:
B Given, Mass of the ball \((\mathrm{m})=3 \mathrm{~kg}\) Speed \((\mathrm{v})=5 \mathrm{~m} / \mathrm{s}\) Time taken by the man to throw a ball is, \(\Delta \mathrm{t}=\frac{2}{4}=0.5 \mathrm{~s}\) Change in momentum of the ball \(=\) Mass \(\times\) Speed \(=\mathrm{m} . \mathrm{v}\) \(=3 \times 5\) \(=15 \mathrm{Ns}\) Force \(=\) Rate of change in linear momentum \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
AMU-2018
Laws of Motion
145799
When a train takes a turn, the passengers are thrown outwards because of-
1 acceleration of motion
2 speed of motion
3 inertia of direction
4 Both (a) and (c)
Explanation:
C Law of inertia states that, everybody continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force. There are three types of inertia- 1.Inertia of rest 2.Inertia of motion 3.Inertia of direction In this question third type of inertia that causes the passengers to be thrown outwards when the driver makes sudden turn.
BCECE-2018
Laws of Motion
145803
A block \(P\) of mass \(M_{P}\) is in contact with another block \(Q\) of mass \(M_{Q}\) as shown in the figure and they are placed on a smooth floor. Force on block \(Q\) is
1 \(\frac{M_{P}}{M_{P}+M_{Q}}\)
2 \(\frac{M_{Q} F}{M_{P}+M_{Q}}\)
3 \(\frac{M_{P} F}{M_{Q}}\)
4 \(\frac{M_{Q} F}{M_{P}}\)
Explanation:
B Given, mass of block \(\mathrm{P}=\mathrm{M}_{\mathrm{P}}\), mass of block \(Q=M_{Q}\) Since, \(\quad \mathrm{F}_{\mathrm{ext}}=\mathrm{ma}\) \(\mathrm{F}=\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}\right) \mathrm{a}\) \(\mathrm{a}=\frac{\mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\) Now, \(\therefore \quad \mathrm{N}=\mathrm{M}_{\mathrm{Q}} \mathrm{a}\) Put the value of ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
