371649
\(\quad\) A batsman hits a ball of mass \(0.2 \mathrm{~kg}\) straight towards the bowler without changing its initial speed of \(6 \mathrm{~m} / \mathrm{s}\). What is the impulse imparted to the ball?
1 \(3.2 \mathrm{Ns}\)
2 \(2.4 \mathrm{Ns}\)
3 \(1.6 \mathrm{Ns}\)
4 \(4 \mathrm{Ns}\)
Explanation:
B We know that change in momentum equal to impulse \(\mathrm{p}_{\mathrm{i}}=\mathrm{mv}\) \(\mathrm{p}_{\mathrm{f}}=-\mathrm{mv}\) \(|\Delta \mathrm{p}|=\mathrm{mv}-(-\mathrm{mv})=2 \mathrm{mv}\) \(=2 \times 0.2 \times 6=2.4 \mathrm{Ns}\)
MHT-CET 2020
LAWS OF MOTION (ADDITIONAL)
371650
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
AP EAMCET (18.09.2020) Shift-I
LAWS OF MOTION (ADDITIONAL)
371651
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 (ADDITIONAL)
371652
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 \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
371649
\(\quad\) A batsman hits a ball of mass \(0.2 \mathrm{~kg}\) straight towards the bowler without changing its initial speed of \(6 \mathrm{~m} / \mathrm{s}\). What is the impulse imparted to the ball?
1 \(3.2 \mathrm{Ns}\)
2 \(2.4 \mathrm{Ns}\)
3 \(1.6 \mathrm{Ns}\)
4 \(4 \mathrm{Ns}\)
Explanation:
B We know that change in momentum equal to impulse \(\mathrm{p}_{\mathrm{i}}=\mathrm{mv}\) \(\mathrm{p}_{\mathrm{f}}=-\mathrm{mv}\) \(|\Delta \mathrm{p}|=\mathrm{mv}-(-\mathrm{mv})=2 \mathrm{mv}\) \(=2 \times 0.2 \times 6=2.4 \mathrm{Ns}\)
MHT-CET 2020
LAWS OF MOTION (ADDITIONAL)
371650
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
AP EAMCET (18.09.2020) Shift-I
LAWS OF MOTION (ADDITIONAL)
371651
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 (ADDITIONAL)
371652
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 \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
371649
\(\quad\) A batsman hits a ball of mass \(0.2 \mathrm{~kg}\) straight towards the bowler without changing its initial speed of \(6 \mathrm{~m} / \mathrm{s}\). What is the impulse imparted to the ball?
1 \(3.2 \mathrm{Ns}\)
2 \(2.4 \mathrm{Ns}\)
3 \(1.6 \mathrm{Ns}\)
4 \(4 \mathrm{Ns}\)
Explanation:
B We know that change in momentum equal to impulse \(\mathrm{p}_{\mathrm{i}}=\mathrm{mv}\) \(\mathrm{p}_{\mathrm{f}}=-\mathrm{mv}\) \(|\Delta \mathrm{p}|=\mathrm{mv}-(-\mathrm{mv})=2 \mathrm{mv}\) \(=2 \times 0.2 \times 6=2.4 \mathrm{Ns}\)
MHT-CET 2020
LAWS OF MOTION (ADDITIONAL)
371650
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
AP EAMCET (18.09.2020) Shift-I
LAWS OF MOTION (ADDITIONAL)
371651
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 (ADDITIONAL)
371652
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 \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
371649
\(\quad\) A batsman hits a ball of mass \(0.2 \mathrm{~kg}\) straight towards the bowler without changing its initial speed of \(6 \mathrm{~m} / \mathrm{s}\). What is the impulse imparted to the ball?
1 \(3.2 \mathrm{Ns}\)
2 \(2.4 \mathrm{Ns}\)
3 \(1.6 \mathrm{Ns}\)
4 \(4 \mathrm{Ns}\)
Explanation:
B We know that change in momentum equal to impulse \(\mathrm{p}_{\mathrm{i}}=\mathrm{mv}\) \(\mathrm{p}_{\mathrm{f}}=-\mathrm{mv}\) \(|\Delta \mathrm{p}|=\mathrm{mv}-(-\mathrm{mv})=2 \mathrm{mv}\) \(=2 \times 0.2 \times 6=2.4 \mathrm{Ns}\)
MHT-CET 2020
LAWS OF MOTION (ADDITIONAL)
371650
Find the apparent weight of a body of mass, 1.0 \(\mathrm{kg}\) falling with an acceleration of \(10 \mathrm{~ms}^{-2}\). \(\left(\mathrm{g} \approx \mathbf{1 0} \mathrm{ms}^{-2}\right)\)
1 \(1 \mathrm{~kg}\)-wt
2 \(2 \mathrm{~kg}-\mathrm{wt}\)
3 0
4 \(0.5 \mathrm{~kg}-\mathrm{wt}\)
Explanation:
C Given, mass \((\mathrm{m})=1 \mathrm{~kg}\), acceleration \((\mathrm{a})=10\) \(\mathrm{m} / \mathrm{s}^{2}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) As the body is falling so using second law of motion \(\mathrm{mg}-\mathrm{N}=\mathrm{ma}\) \(\mathrm{N}=\mathrm{m}(\mathrm{g}-\mathrm{a})\) \(\mathrm{N}=1(10-10)\) \(\mathrm{N}=0\)
AP EAMCET (18.09.2020) Shift-I
LAWS OF MOTION (ADDITIONAL)
371651
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 (ADDITIONAL)
371652
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 \(\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{15}{0.5}=30 \mathrm{~N}\)
