371657
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 ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
VITEEE - 2017
LAWS OF MOTION (ADDITIONAL)
371658
The momentum of the particle at any instant is given by \(3 \cos 4 t \hat{i}+3 \sin 4 t \hat{j}\). What is the angle between momentum and force acting on it?
371659
A reference frame attached to earth cannot be an inertial frame because
1 Earth is revolving around the sun
2 Earth is rotating about its axis
3 Newton's laws are applicable in this frame
4 both (a) and (b)
Explanation:
D A reference frame in which the observers are not subjected to any accelerating force is know inertial frame. Force will act on earth whether it is revolving about its own axis or around sun. This force will cause acceleration on earth. Hence, Earth is not an inertial frame of reference.
JCECE-2007
LAWS OF MOTION (ADDITIONAL)
371660
A machine gun of mass \(10 \mathrm{~kg}\) fires \(30 \mathrm{~g}\) bullets at the rate of \(6 \mathrm{bullet} / \mathrm{s}\) with a speed of \(400 \mathrm{~m} / \mathrm{s}\). The force required to keep the gun in position will be
1 \(30 \mathrm{~N}\)
2 \(40 \mathrm{~N}\)
3 \(72 \mathrm{~N}\)
4 \(400 \mathrm{~N}\)
Explanation:
C Given that, \(10 \mathrm{~kg} \& 30 \times 10^{-3} \mathrm{~kg}\) mass of machine gun and bullet respectively \(\& 6 \mathrm{bullet} / \mathrm{s}\) fired Let \(\mathrm{v}\) be the velocity of each unit, According to conservation of momentum \(\mathrm{m}_{1} \times \mathrm{v}=\mathrm{m}_{2} \times 400\) \(10 \times \mathrm{v}=6 \times 30 \times 10^{-3} \times 400\) \(\mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\) \(=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}} \quad[\because \Delta \mathrm{t}=1 \mathrm{sec}]\) \(=\frac{\mathrm{m}_{1} \mathrm{v}}{1}\) \(=\frac{10 \times 7.2}{1}=72 \mathrm{~N}\) \(=72 \mathrm{~N}\) \(\therefore \quad \mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\)
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LAWS OF MOTION (ADDITIONAL)
371657
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 ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
VITEEE - 2017
LAWS OF MOTION (ADDITIONAL)
371658
The momentum of the particle at any instant is given by \(3 \cos 4 t \hat{i}+3 \sin 4 t \hat{j}\). What is the angle between momentum and force acting on it?
371659
A reference frame attached to earth cannot be an inertial frame because
1 Earth is revolving around the sun
2 Earth is rotating about its axis
3 Newton's laws are applicable in this frame
4 both (a) and (b)
Explanation:
D A reference frame in which the observers are not subjected to any accelerating force is know inertial frame. Force will act on earth whether it is revolving about its own axis or around sun. This force will cause acceleration on earth. Hence, Earth is not an inertial frame of reference.
JCECE-2007
LAWS OF MOTION (ADDITIONAL)
371660
A machine gun of mass \(10 \mathrm{~kg}\) fires \(30 \mathrm{~g}\) bullets at the rate of \(6 \mathrm{bullet} / \mathrm{s}\) with a speed of \(400 \mathrm{~m} / \mathrm{s}\). The force required to keep the gun in position will be
1 \(30 \mathrm{~N}\)
2 \(40 \mathrm{~N}\)
3 \(72 \mathrm{~N}\)
4 \(400 \mathrm{~N}\)
Explanation:
C Given that, \(10 \mathrm{~kg} \& 30 \times 10^{-3} \mathrm{~kg}\) mass of machine gun and bullet respectively \(\& 6 \mathrm{bullet} / \mathrm{s}\) fired Let \(\mathrm{v}\) be the velocity of each unit, According to conservation of momentum \(\mathrm{m}_{1} \times \mathrm{v}=\mathrm{m}_{2} \times 400\) \(10 \times \mathrm{v}=6 \times 30 \times 10^{-3} \times 400\) \(\mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\) \(=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}} \quad[\because \Delta \mathrm{t}=1 \mathrm{sec}]\) \(=\frac{\mathrm{m}_{1} \mathrm{v}}{1}\) \(=\frac{10 \times 7.2}{1}=72 \mathrm{~N}\) \(=72 \mathrm{~N}\) \(\therefore \quad \mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\)
371657
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 ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
VITEEE - 2017
LAWS OF MOTION (ADDITIONAL)
371658
The momentum of the particle at any instant is given by \(3 \cos 4 t \hat{i}+3 \sin 4 t \hat{j}\). What is the angle between momentum and force acting on it?
371659
A reference frame attached to earth cannot be an inertial frame because
1 Earth is revolving around the sun
2 Earth is rotating about its axis
3 Newton's laws are applicable in this frame
4 both (a) and (b)
Explanation:
D A reference frame in which the observers are not subjected to any accelerating force is know inertial frame. Force will act on earth whether it is revolving about its own axis or around sun. This force will cause acceleration on earth. Hence, Earth is not an inertial frame of reference.
JCECE-2007
LAWS OF MOTION (ADDITIONAL)
371660
A machine gun of mass \(10 \mathrm{~kg}\) fires \(30 \mathrm{~g}\) bullets at the rate of \(6 \mathrm{bullet} / \mathrm{s}\) with a speed of \(400 \mathrm{~m} / \mathrm{s}\). The force required to keep the gun in position will be
1 \(30 \mathrm{~N}\)
2 \(40 \mathrm{~N}\)
3 \(72 \mathrm{~N}\)
4 \(400 \mathrm{~N}\)
Explanation:
C Given that, \(10 \mathrm{~kg} \& 30 \times 10^{-3} \mathrm{~kg}\) mass of machine gun and bullet respectively \(\& 6 \mathrm{bullet} / \mathrm{s}\) fired Let \(\mathrm{v}\) be the velocity of each unit, According to conservation of momentum \(\mathrm{m}_{1} \times \mathrm{v}=\mathrm{m}_{2} \times 400\) \(10 \times \mathrm{v}=6 \times 30 \times 10^{-3} \times 400\) \(\mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\) \(=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}} \quad[\because \Delta \mathrm{t}=1 \mathrm{sec}]\) \(=\frac{\mathrm{m}_{1} \mathrm{v}}{1}\) \(=\frac{10 \times 7.2}{1}=72 \mathrm{~N}\) \(=72 \mathrm{~N}\) \(\therefore \quad \mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\)
371657
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 ' \(a\) ' \(\mathrm{N}=\frac{\mathrm{M}_{\mathrm{Q}} \mathrm{F}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\)
VITEEE - 2017
LAWS OF MOTION (ADDITIONAL)
371658
The momentum of the particle at any instant is given by \(3 \cos 4 t \hat{i}+3 \sin 4 t \hat{j}\). What is the angle between momentum and force acting on it?
371659
A reference frame attached to earth cannot be an inertial frame because
1 Earth is revolving around the sun
2 Earth is rotating about its axis
3 Newton's laws are applicable in this frame
4 both (a) and (b)
Explanation:
D A reference frame in which the observers are not subjected to any accelerating force is know inertial frame. Force will act on earth whether it is revolving about its own axis or around sun. This force will cause acceleration on earth. Hence, Earth is not an inertial frame of reference.
JCECE-2007
LAWS OF MOTION (ADDITIONAL)
371660
A machine gun of mass \(10 \mathrm{~kg}\) fires \(30 \mathrm{~g}\) bullets at the rate of \(6 \mathrm{bullet} / \mathrm{s}\) with a speed of \(400 \mathrm{~m} / \mathrm{s}\). The force required to keep the gun in position will be
1 \(30 \mathrm{~N}\)
2 \(40 \mathrm{~N}\)
3 \(72 \mathrm{~N}\)
4 \(400 \mathrm{~N}\)
Explanation:
C Given that, \(10 \mathrm{~kg} \& 30 \times 10^{-3} \mathrm{~kg}\) mass of machine gun and bullet respectively \(\& 6 \mathrm{bullet} / \mathrm{s}\) fired Let \(\mathrm{v}\) be the velocity of each unit, According to conservation of momentum \(\mathrm{m}_{1} \times \mathrm{v}=\mathrm{m}_{2} \times 400\) \(10 \times \mathrm{v}=6 \times 30 \times 10^{-3} \times 400\) \(\mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\) \(=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}} \quad[\because \Delta \mathrm{t}=1 \mathrm{sec}]\) \(=\frac{\mathrm{m}_{1} \mathrm{v}}{1}\) \(=\frac{10 \times 7.2}{1}=72 \mathrm{~N}\) \(=72 \mathrm{~N}\) \(\therefore \quad \mathrm{v}=\frac{12 \times 6}{10}=7.2 \mathrm{~m} / \mathrm{s}\)
