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Quiz Description :
Name: Physics Momentum online quiz
Subject: Physics
Topic: Momentum
Questions: 10 mcq type
Time Allowed: 10 minutes
Important for: Class 10th, 11th & 12th school students, Engineering and Medical Entrance Test.
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 Question 1 of 10
1. Question
1 pointsMomentum is closely related to
CorrectWe know that impulse is equal to the total change in momentum produced during the impact.
Therefore momentum is closely related to the impulse.IncorrectWe know that impulse is equal to the total change in momentum produced during the impact.
Therefore momentum is closely related to the impulse.UnattemptedWe know that impulse is equal to the total change in momentum produced during the impact.
Therefore momentum is closely related to the impulse.  Question 2 of 10
2. Question
1 pointsA body, whose momentum is constant, must have constant
CorrectWe know that momentum of a moving body = Mass × Velocity
Since mass of the body is constant, therefore for constant momentum, a body must have constant velocity.IncorrectWe know that momentum of a moving body = Mass × Velocity
Since mass of the body is constant, therefore for constant momentum, a body must have constant velocity.UnattemptedWe know that momentum of a moving body = Mass × Velocity
Since mass of the body is constant, therefore for constant momentum, a body must have constant velocity.  Question 3 of 10
3. Question
1 pointsA mass m falls freely form rest. The linear momentum, after it has fallen through height h, is
CorrectGiven: Mass = m; Initial velocity (u) = 0 (because it falls from rest) and height through which the mass has fallen = h.
v^{2} = u^{2} + 2gh = 0^{2} + 2gh = 2gh
v = √2gh
Therefore linear momentum = mv= m√2gh.IncorrectGiven: Mass = m; Initial velocity (u) = 0 (because it falls from rest) and height through which the mass has fallen = h.
v^{2} = u^{2} + 2gh = 0^{2} + 2gh = 2gh
v = √2gh
Therefore linear momentum = mv= m√2gh.UnattemptedGiven: Mass = m; Initial velocity (u) = 0 (because it falls from rest) and height through which the mass has fallen = h.
v^{2} = u^{2} + 2gh = 0^{2} + 2gh = 2gh
v = √2gh
Therefore linear momentum = mv= m√2gh.  Question 4 of 10
4. Question
1 pointsA molecule of mass m of an ideal gas collides with the wall of a vessel with a velocity v and returns back with the same velocity. The change in the momentum of molecule is
CorrectGiven: Mass of molecule = m and initial velocity of molecule = v.
We know that initial momentum of molecule (p_{1}) = mv.
And final momentum (p_{2}) = m(v) = mv (Minus sign of v due to opposite direction).
Therefore change in linear momentum of molecule (dp) = p_{1} – p_{2} = mv = (mv) = 2 mv.IncorrectGiven: Mass of molecule = m and initial velocity of molecule = v.
We know that initial momentum of molecule (p_{1}) = mv.
And final momentum (p_{2}) = m(v) = mv (Minus sign of v due to opposite direction).
Therefore change in linear momentum of molecule (dp) = p_{1} – p_{2} = mv = (mv) = 2 mv.UnattemptedGiven: Mass of molecule = m and initial velocity of molecule = v.
We know that initial momentum of molecule (p_{1}) = mv.
And final momentum (p_{2}) = m(v) = mv (Minus sign of v due to opposite direction).
Therefore change in linear momentum of molecule (dp) = p_{1} – p_{2} = mv = (mv) = 2 mv.  Question 5 of 10
5. Question
1 pointsTwo balls each of mass 60g moving in opposite directions with speeds 4 ms^{1} collide and rebound with the same speed. The change in momentum will be
CorrectGiven: Mass of each ball (m) = 60g and initial speed of balls (v) = 4 ms^{1}.
We know that initial momentum of each ball (p_{1}) = mv = 0.06 × 4 = 0.24 kgms^{1}.
And final momentum of each ball (p_{2}) = mv = 0.06 × ( 4) = 0.24 kgms^{1}. (Minus sign of v due of opposite direction)
Therefore change in momentum (dp) = p_{1} – p_{2} = 0.24 – ( 0.24) = 0.48 kgms^{1}.IncorrectGiven: Mass of each ball (m) = 60g and initial speed of balls (v) = 4 ms^{1}.
We know that initial momentum of each ball (p_{1}) = mv = 0.06 × 4 = 0.24 kgms^{1}.
And final momentum of each ball (p_{2}) = mv = 0.06 × ( 4) = 0.24 kgms^{1}. (Minus sign of v due of opposite direction)
Therefore change in momentum (dp) = p_{1} – p_{2} = 0.24 – ( 0.24) = 0.48 kgms^{1}.UnattemptedGiven: Mass of each ball (m) = 60g and initial speed of balls (v) = 4 ms^{1}.
We know that initial momentum of each ball (p_{1}) = mv = 0.06 × 4 = 0.24 kgms^{1}.
And final momentum of each ball (p_{2}) = mv = 0.06 × ( 4) = 0.24 kgms^{1}. (Minus sign of v due of opposite direction)
Therefore change in momentum (dp) = p_{1} – p_{2} = 0.24 – ( 0.24) = 0.48 kgms^{1}.  Question 6 of 10
6. Question
1 pointsIf a force of 10N acts on a body of mass 20kg for 10s, then change in its momentum is
CorrectGiven: Force (F) = 10N; Mass of body (m) = 20 kg; Initial velocity of the body (u) = 0 (because it is at rest) time (t) = 10s.
We know that acceleration of the body (a) = F/m = 10/20 = 0.5 ms^{2}
And final velocity of the body (v) = u + at = 0 + at = 0 + (0.5 10) = 5 ms^{1}
Therefore change in momentum = mv – mu = m(v u) = 20(5 – 0) = 100 kgms^{1}.IncorrectGiven: Force (F) = 10N; Mass of body (m) = 20 kg; Initial velocity of the body (u) = 0 (because it is at rest) time (t) = 10s.
We know that acceleration of the body (a) = F/m = 10/20 = 0.5 ms^{2}
And final velocity of the body (v) = u + at = 0 + at = 0 + (0.5 10) = 5 ms^{1}
Therefore change in momentum = mv – mu = m(v u) = 20(5 – 0) = 100 kgms^{1}.UnattemptedGiven: Force (F) = 10N; Mass of body (m) = 20 kg; Initial velocity of the body (u) = 0 (because it is at rest) time (t) = 10s.
We know that acceleration of the body (a) = F/m = 10/20 = 0.5 ms^{2}
And final velocity of the body (v) = u + at = 0 + at = 0 + (0.5 10) = 5 ms^{1}
Therefore change in momentum = mv – mu = m(v u) = 20(5 – 0) = 100 kgms^{1}.  Question 7 of 10
7. Question
1 pointsIf a gun is firing n bullets per second, each of mass m with velocity v, then force to be exerted by the shoulder so as to stop it, is
CorrectGiven: Mass of each bullet = m; No. of bullets = n; Velocity of bullet = v and time interval (dt) = 1s.
We know that change in momentum per bullet = mv
Therefore total change in momentum (dp) = nmv
We also know that force exerted by the shoulder to stop the gun (F) = dp/dt = nmv/1 = nmvIncorrectGiven: Mass of each bullet = m; No. of bullets = n; Velocity of bullet = v and time interval (dt) = 1s.
We know that change in momentum per bullet = mv
Therefore total change in momentum (dp) = nmv
We also know that force exerted by the shoulder to stop the gun (F) = dp/dt = nmv/1 = nmvUnattemptedGiven: Mass of each bullet = m; No. of bullets = n; Velocity of bullet = v and time interval (dt) = 1s.
We know that change in momentum per bullet = mv
Therefore total change in momentum (dp) = nmv
We also know that force exerted by the shoulder to stop the gun (F) = dp/dt = nmv/1 = nmv  Question 8 of 10
8. Question
1 pointsIf an object of mass 2kg is thrown vertically upwards, then the rate change of its momentum is (Take g=10 ms^{2})
CorrectGiven: Mass of object (m) = 2 kg and acceleration due to gravity (g) = 10 Ns^{2}.
We know that force applied on the object (F) = mg = 2 × 10 = 20N.
Since the rate of change of momentum is equal to force applied on the object, therefore rate of change of momentum (dp/dt) = F = 20N.IncorrectGiven: Mass of object (m) = 2 kg and acceleration due to gravity (g) = 10 Ns^{2}.
We know that force applied on the object (F) = mg = 2 × 10 = 20N.
Since the rate of change of momentum is equal to force applied on the object, therefore rate of change of momentum (dp/dt) = F = 20N.UnattemptedGiven: Mass of object (m) = 2 kg and acceleration due to gravity (g) = 10 Ns^{2}.
We know that force applied on the object (F) = mg = 2 × 10 = 20N.
Since the rate of change of momentum is equal to force applied on the object, therefore rate of change of momentum (dp/dt) = F = 20N.  Question 9 of 10
9. Question
1 pointsA machine gun fires a bullet of mass 40g with a velocity 1200 ms1. The man holding it can exert a maximum force of 144N on the gun. How many bullets can he fire per second at the most?
CorrectGiven: Mass of bullet (m) = 40g = 0.04 kg; Velocity of bullet (v) = 1200 ms1 and maximum force exerted by man on the gun (F) = 144N.
We know that force exerted on man’s hand when one bullet is fired (F1) = Change in momentum for one bullet fired per second = mv = 0.04 × 1200 = 48N.
Therefore number of bullets that can be fired per second (n) = F/F1 = 144/48 = 3.IncorrectGiven: Mass of bullet (m) = 40g = 0.04 kg; Velocity of bullet (v) = 1200 ms1 and maximum force exerted by man on the gun (F) = 144N.
We know that force exerted on man’s hand when one bullet is fired (F1) = Change in momentum for one bullet fired per second = mv = 0.04 × 1200 = 48N.
Therefore number of bullets that can be fired per second (n) = F/F1 = 144/48 = 3.UnattemptedGiven: Mass of bullet (m) = 40g = 0.04 kg; Velocity of bullet (v) = 1200 ms1 and maximum force exerted by man on the gun (F) = 144N.
We know that force exerted on man’s hand when one bullet is fired (F1) = Change in momentum for one bullet fired per second = mv = 0.04 × 1200 = 48N.
Therefore number of bullets that can be fired per second (n) = F/F1 = 144/48 = 3.  Question 10 of 10
10. Question
1 pointsA gun fires the bullets each of mass 10g with a velocity of 50 ms^{1}. If 1000 bullets are fired per second, then thrust on the shoulder will be
CorrectGiven: Mass of each bullet (m) = 10g = 10×10^{3} kg; Velocity of bullet (v) = 50 ms^{1}; No. of bullets (n) = 1000 and time interval (dt) = 1s.
We know that change in momentum per bullet = mv = (10 × 10^{3}) × 50 = 0.5 kgms^{1}.
Therefore total change in momentum (dp) = nmv = 1000 × 0.5 = 500 kgms^{1}.
We also know that thrust on the shoulder (F) = dp/dt = 500/1 = 500 N.IncorrectGiven: Mass of each bullet (m) = 10g = 10×10^{3} kg; Velocity of bullet (v) = 50 ms^{1}; No. of bullets (n) = 1000 and time interval (dt) = 1s.
We know that change in momentum per bullet = mv = (10 × 10^{3}) × 50 = 0.5 kgms^{1}.
Therefore total change in momentum (dp) = nmv = 1000 × 0.5 = 500 kgms^{1}.
We also know that thrust on the shoulder (F) = dp/dt = 500/1 = 500 N.UnattemptedGiven: Mass of each bullet (m) = 10g = 10×10^{3} kg; Velocity of bullet (v) = 50 ms^{1}; No. of bullets (n) = 1000 and time interval (dt) = 1s.
We know that change in momentum per bullet = mv = (10 × 10^{3}) × 50 = 0.5 kgms^{1}.
Therefore total change in momentum (dp) = nmv = 1000 × 0.5 = 500 kgms^{1}.
We also know that thrust on the shoulder (F) = dp/dt = 500/1 = 500 N.