NEET Test Series from KOTA - 10 Papers In MS WORD
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LAWS OF MOTION (ADDITIONAL)
371700
The distance covered by a body of mass \(5 \mathrm{~g}\) having linear momentum \(0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) in \(5 \mathrm{~s}\) is:
1 \(300 \mathrm{~m}\)
2 \(30 \mathrm{~m}\)
3 \(3 \mathrm{~m}\)
4 \(0.3 \mathrm{~m}\) [SRM JEE-2014]
Explanation:
A Given data, \(\mathrm{m}=5 \mathrm{~g}\) Linear momentum, \(\mathrm{p}=0.3 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}\) \(\mathrm{t}=5 \mathrm{sec}\) Distance \((\mathrm{d})=\) ? We know that, \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}} \Rightarrow \frac{0.3 \times 1000}{5}\) \(\mathrm{v}=60 \mathrm{~m} / \mathrm{sec}\) \(\because\) velocity, \(\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}\) So, \(\quad \mathrm{d}=\mathrm{vt}\) \(\mathrm{d}=60 \times 5\) \(\mathrm{d}=300 \mathrm{~m}\)
LAWS OF MOTION (ADDITIONAL)
371701
A bullet of mass \(20 \mathrm{~g}\) moving with a speed of \(100 \mathrm{~m} \mathrm{~s}^{-1}\) enters a heavy wooden block and stops after a distance of \(50 \mathrm{~cm}\). The average resistive force exerted by the block on the bullet is
1 \(100 \mathrm{~N}\)
2 \(10000 \mathrm{~N}\)
3 \(200 \mathrm{~N}\)
4 \(500 \mathrm{~N}\)
Explanation:
C Given that, \(\mathrm{u}=100 \mathrm{~ms}^{-1}, \mathrm{v}=0\) \(\mathrm{~m}=20 \mathrm{~g}=\frac{20}{1000} \mathrm{~kg}=0.02 \mathrm{~kg}\) \(\mathrm{~s}=50 \mathrm{~cm}=0.5 \mathrm{~m}\) Use the third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\therefore \quad(0)^{2}-(100)^{2}=2 \mathrm{a} \times 0.5\) \(\mathrm{a}=\frac{(100)^{2}}{2 \times 0.5}=-10000 \mathrm{~ms}^{-2}\) Negative sign shows the retardation. \(\therefore\) The average resistive force exerted by block on the bullet is \(\mathrm{F} =\mathrm{m} \times \mathrm{a}\) \(=0.02 \times 10,000\) \(=200 \mathrm{~N}\)
Assam CEE-2016
LAWS OF MOTION (ADDITIONAL)
371702
A gun of mass of \(10 \mathrm{~kg}\) fires 4 bullets per second. The mass of each bullet is \(20 \mathrm{~g}\) and the velocity of the bullet when it leaves the gun is \(300 \mathrm{~ms}^{-1}\). The force required to hold the gun while firing is:
1 \(6 \mathrm{~N}\)
2 \(8 \mathrm{~N}\)
3 \(24 \mathrm{~N}\)
4 \(240 \mathrm{~N}\)
Explanation:
C Given data, Mass of each bullet \((\mathrm{m})=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\) No. of bullet \((\mathrm{n})=4\) \(\text { Velocity }(\mathrm{v}) =300 \mathrm{~ms}^{-1}\) \(\text { Time }(\mathrm{t}) =1 \mathrm{sec}\) \(\text { Force } =?\) According to Newton's second law, the rate of change in momentum (p) of an object is equal to the force (F) applied. \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\mathrm{F}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}} \quad[\because \mathrm{p}=\mathrm{mv}]\) \(\mathrm{F}_{\text {single }}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\) \(\frac{\mathrm{dv}}{\mathrm{dt}} \times \mathrm{m}=\mathrm{ma} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]\) \(\mathrm{F}_{\text {single }}=\frac{300}{1} \times 20 \times 10^{-3} \Rightarrow 6 \mathrm{~N}\) So, \(\quad \mathrm{F}_{\text {Total }}=\mathrm{n} \times \mathrm{F}_{\text {single }}\) \(\mathrm{F}_{\text {Total }}=4 \times 6=24 \mathrm{~N}\)
AP EAMCET(Medical)-2000
LAWS OF MOTION (ADDITIONAL)
371703
A force of \(100 \mathrm{~N}\) acts on a body of mass \(2 \mathrm{~kg}\) for \(10 \mathrm{~s}\). The change in momentum of the body is:
1 \(100 \mathrm{Ns}\)
2 \(250 \mathrm{Ns}\)
3 \(500 \mathrm{Ns}\)
4 \(1000 \mathrm{Ns}\)
Explanation:
D Given data, Mass of body \((\mathrm{m})=2 \mathrm{~kg}\) Applied force \((\mathrm{F})=100 \mathrm{~N}\) Time \((t)=10\) sec By Newton's second law, The rate of change of momentum is directly proportional to the force. \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}} \propto \mathrm{F}\) \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\mathrm{F}\) \(\Delta \mathrm{t}\) \(\Delta \mathrm{p}=\mathrm{F} . \Delta \mathrm{t}\) \(\Delta \mathrm{p}=100 \mathrm{~N} \times 10 \mathrm{~s}=1000 \mathrm{Ns}\)
371700
The distance covered by a body of mass \(5 \mathrm{~g}\) having linear momentum \(0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) in \(5 \mathrm{~s}\) is:
1 \(300 \mathrm{~m}\)
2 \(30 \mathrm{~m}\)
3 \(3 \mathrm{~m}\)
4 \(0.3 \mathrm{~m}\) [SRM JEE-2014]
Explanation:
A Given data, \(\mathrm{m}=5 \mathrm{~g}\) Linear momentum, \(\mathrm{p}=0.3 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}\) \(\mathrm{t}=5 \mathrm{sec}\) Distance \((\mathrm{d})=\) ? We know that, \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}} \Rightarrow \frac{0.3 \times 1000}{5}\) \(\mathrm{v}=60 \mathrm{~m} / \mathrm{sec}\) \(\because\) velocity, \(\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}\) So, \(\quad \mathrm{d}=\mathrm{vt}\) \(\mathrm{d}=60 \times 5\) \(\mathrm{d}=300 \mathrm{~m}\)
LAWS OF MOTION (ADDITIONAL)
371701
A bullet of mass \(20 \mathrm{~g}\) moving with a speed of \(100 \mathrm{~m} \mathrm{~s}^{-1}\) enters a heavy wooden block and stops after a distance of \(50 \mathrm{~cm}\). The average resistive force exerted by the block on the bullet is
1 \(100 \mathrm{~N}\)
2 \(10000 \mathrm{~N}\)
3 \(200 \mathrm{~N}\)
4 \(500 \mathrm{~N}\)
Explanation:
C Given that, \(\mathrm{u}=100 \mathrm{~ms}^{-1}, \mathrm{v}=0\) \(\mathrm{~m}=20 \mathrm{~g}=\frac{20}{1000} \mathrm{~kg}=0.02 \mathrm{~kg}\) \(\mathrm{~s}=50 \mathrm{~cm}=0.5 \mathrm{~m}\) Use the third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\therefore \quad(0)^{2}-(100)^{2}=2 \mathrm{a} \times 0.5\) \(\mathrm{a}=\frac{(100)^{2}}{2 \times 0.5}=-10000 \mathrm{~ms}^{-2}\) Negative sign shows the retardation. \(\therefore\) The average resistive force exerted by block on the bullet is \(\mathrm{F} =\mathrm{m} \times \mathrm{a}\) \(=0.02 \times 10,000\) \(=200 \mathrm{~N}\)
Assam CEE-2016
LAWS OF MOTION (ADDITIONAL)
371702
A gun of mass of \(10 \mathrm{~kg}\) fires 4 bullets per second. The mass of each bullet is \(20 \mathrm{~g}\) and the velocity of the bullet when it leaves the gun is \(300 \mathrm{~ms}^{-1}\). The force required to hold the gun while firing is:
1 \(6 \mathrm{~N}\)
2 \(8 \mathrm{~N}\)
3 \(24 \mathrm{~N}\)
4 \(240 \mathrm{~N}\)
Explanation:
C Given data, Mass of each bullet \((\mathrm{m})=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\) No. of bullet \((\mathrm{n})=4\) \(\text { Velocity }(\mathrm{v}) =300 \mathrm{~ms}^{-1}\) \(\text { Time }(\mathrm{t}) =1 \mathrm{sec}\) \(\text { Force } =?\) According to Newton's second law, the rate of change in momentum (p) of an object is equal to the force (F) applied. \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\mathrm{F}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}} \quad[\because \mathrm{p}=\mathrm{mv}]\) \(\mathrm{F}_{\text {single }}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\) \(\frac{\mathrm{dv}}{\mathrm{dt}} \times \mathrm{m}=\mathrm{ma} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]\) \(\mathrm{F}_{\text {single }}=\frac{300}{1} \times 20 \times 10^{-3} \Rightarrow 6 \mathrm{~N}\) So, \(\quad \mathrm{F}_{\text {Total }}=\mathrm{n} \times \mathrm{F}_{\text {single }}\) \(\mathrm{F}_{\text {Total }}=4 \times 6=24 \mathrm{~N}\)
AP EAMCET(Medical)-2000
LAWS OF MOTION (ADDITIONAL)
371703
A force of \(100 \mathrm{~N}\) acts on a body of mass \(2 \mathrm{~kg}\) for \(10 \mathrm{~s}\). The change in momentum of the body is:
1 \(100 \mathrm{Ns}\)
2 \(250 \mathrm{Ns}\)
3 \(500 \mathrm{Ns}\)
4 \(1000 \mathrm{Ns}\)
Explanation:
D Given data, Mass of body \((\mathrm{m})=2 \mathrm{~kg}\) Applied force \((\mathrm{F})=100 \mathrm{~N}\) Time \((t)=10\) sec By Newton's second law, The rate of change of momentum is directly proportional to the force. \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}} \propto \mathrm{F}\) \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\mathrm{F}\) \(\Delta \mathrm{t}\) \(\Delta \mathrm{p}=\mathrm{F} . \Delta \mathrm{t}\) \(\Delta \mathrm{p}=100 \mathrm{~N} \times 10 \mathrm{~s}=1000 \mathrm{Ns}\)
371700
The distance covered by a body of mass \(5 \mathrm{~g}\) having linear momentum \(0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) in \(5 \mathrm{~s}\) is:
1 \(300 \mathrm{~m}\)
2 \(30 \mathrm{~m}\)
3 \(3 \mathrm{~m}\)
4 \(0.3 \mathrm{~m}\) [SRM JEE-2014]
Explanation:
A Given data, \(\mathrm{m}=5 \mathrm{~g}\) Linear momentum, \(\mathrm{p}=0.3 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}\) \(\mathrm{t}=5 \mathrm{sec}\) Distance \((\mathrm{d})=\) ? We know that, \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}} \Rightarrow \frac{0.3 \times 1000}{5}\) \(\mathrm{v}=60 \mathrm{~m} / \mathrm{sec}\) \(\because\) velocity, \(\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}\) So, \(\quad \mathrm{d}=\mathrm{vt}\) \(\mathrm{d}=60 \times 5\) \(\mathrm{d}=300 \mathrm{~m}\)
LAWS OF MOTION (ADDITIONAL)
371701
A bullet of mass \(20 \mathrm{~g}\) moving with a speed of \(100 \mathrm{~m} \mathrm{~s}^{-1}\) enters a heavy wooden block and stops after a distance of \(50 \mathrm{~cm}\). The average resistive force exerted by the block on the bullet is
1 \(100 \mathrm{~N}\)
2 \(10000 \mathrm{~N}\)
3 \(200 \mathrm{~N}\)
4 \(500 \mathrm{~N}\)
Explanation:
C Given that, \(\mathrm{u}=100 \mathrm{~ms}^{-1}, \mathrm{v}=0\) \(\mathrm{~m}=20 \mathrm{~g}=\frac{20}{1000} \mathrm{~kg}=0.02 \mathrm{~kg}\) \(\mathrm{~s}=50 \mathrm{~cm}=0.5 \mathrm{~m}\) Use the third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\therefore \quad(0)^{2}-(100)^{2}=2 \mathrm{a} \times 0.5\) \(\mathrm{a}=\frac{(100)^{2}}{2 \times 0.5}=-10000 \mathrm{~ms}^{-2}\) Negative sign shows the retardation. \(\therefore\) The average resistive force exerted by block on the bullet is \(\mathrm{F} =\mathrm{m} \times \mathrm{a}\) \(=0.02 \times 10,000\) \(=200 \mathrm{~N}\)
Assam CEE-2016
LAWS OF MOTION (ADDITIONAL)
371702
A gun of mass of \(10 \mathrm{~kg}\) fires 4 bullets per second. The mass of each bullet is \(20 \mathrm{~g}\) and the velocity of the bullet when it leaves the gun is \(300 \mathrm{~ms}^{-1}\). The force required to hold the gun while firing is:
1 \(6 \mathrm{~N}\)
2 \(8 \mathrm{~N}\)
3 \(24 \mathrm{~N}\)
4 \(240 \mathrm{~N}\)
Explanation:
C Given data, Mass of each bullet \((\mathrm{m})=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\) No. of bullet \((\mathrm{n})=4\) \(\text { Velocity }(\mathrm{v}) =300 \mathrm{~ms}^{-1}\) \(\text { Time }(\mathrm{t}) =1 \mathrm{sec}\) \(\text { Force } =?\) According to Newton's second law, the rate of change in momentum (p) of an object is equal to the force (F) applied. \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\mathrm{F}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}} \quad[\because \mathrm{p}=\mathrm{mv}]\) \(\mathrm{F}_{\text {single }}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\) \(\frac{\mathrm{dv}}{\mathrm{dt}} \times \mathrm{m}=\mathrm{ma} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]\) \(\mathrm{F}_{\text {single }}=\frac{300}{1} \times 20 \times 10^{-3} \Rightarrow 6 \mathrm{~N}\) So, \(\quad \mathrm{F}_{\text {Total }}=\mathrm{n} \times \mathrm{F}_{\text {single }}\) \(\mathrm{F}_{\text {Total }}=4 \times 6=24 \mathrm{~N}\)
AP EAMCET(Medical)-2000
LAWS OF MOTION (ADDITIONAL)
371703
A force of \(100 \mathrm{~N}\) acts on a body of mass \(2 \mathrm{~kg}\) for \(10 \mathrm{~s}\). The change in momentum of the body is:
1 \(100 \mathrm{Ns}\)
2 \(250 \mathrm{Ns}\)
3 \(500 \mathrm{Ns}\)
4 \(1000 \mathrm{Ns}\)
Explanation:
D Given data, Mass of body \((\mathrm{m})=2 \mathrm{~kg}\) Applied force \((\mathrm{F})=100 \mathrm{~N}\) Time \((t)=10\) sec By Newton's second law, The rate of change of momentum is directly proportional to the force. \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}} \propto \mathrm{F}\) \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\mathrm{F}\) \(\Delta \mathrm{t}\) \(\Delta \mathrm{p}=\mathrm{F} . \Delta \mathrm{t}\) \(\Delta \mathrm{p}=100 \mathrm{~N} \times 10 \mathrm{~s}=1000 \mathrm{Ns}\)
371700
The distance covered by a body of mass \(5 \mathrm{~g}\) having linear momentum \(0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) in \(5 \mathrm{~s}\) is:
1 \(300 \mathrm{~m}\)
2 \(30 \mathrm{~m}\)
3 \(3 \mathrm{~m}\)
4 \(0.3 \mathrm{~m}\) [SRM JEE-2014]
Explanation:
A Given data, \(\mathrm{m}=5 \mathrm{~g}\) Linear momentum, \(\mathrm{p}=0.3 \mathrm{~kg}-\mathrm{m} / \mathrm{sec}\) \(\mathrm{t}=5 \mathrm{sec}\) Distance \((\mathrm{d})=\) ? We know that, \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}} \Rightarrow \frac{0.3 \times 1000}{5}\) \(\mathrm{v}=60 \mathrm{~m} / \mathrm{sec}\) \(\because\) velocity, \(\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}\) So, \(\quad \mathrm{d}=\mathrm{vt}\) \(\mathrm{d}=60 \times 5\) \(\mathrm{d}=300 \mathrm{~m}\)
LAWS OF MOTION (ADDITIONAL)
371701
A bullet of mass \(20 \mathrm{~g}\) moving with a speed of \(100 \mathrm{~m} \mathrm{~s}^{-1}\) enters a heavy wooden block and stops after a distance of \(50 \mathrm{~cm}\). The average resistive force exerted by the block on the bullet is
1 \(100 \mathrm{~N}\)
2 \(10000 \mathrm{~N}\)
3 \(200 \mathrm{~N}\)
4 \(500 \mathrm{~N}\)
Explanation:
C Given that, \(\mathrm{u}=100 \mathrm{~ms}^{-1}, \mathrm{v}=0\) \(\mathrm{~m}=20 \mathrm{~g}=\frac{20}{1000} \mathrm{~kg}=0.02 \mathrm{~kg}\) \(\mathrm{~s}=50 \mathrm{~cm}=0.5 \mathrm{~m}\) Use the third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\therefore \quad(0)^{2}-(100)^{2}=2 \mathrm{a} \times 0.5\) \(\mathrm{a}=\frac{(100)^{2}}{2 \times 0.5}=-10000 \mathrm{~ms}^{-2}\) Negative sign shows the retardation. \(\therefore\) The average resistive force exerted by block on the bullet is \(\mathrm{F} =\mathrm{m} \times \mathrm{a}\) \(=0.02 \times 10,000\) \(=200 \mathrm{~N}\)
Assam CEE-2016
LAWS OF MOTION (ADDITIONAL)
371702
A gun of mass of \(10 \mathrm{~kg}\) fires 4 bullets per second. The mass of each bullet is \(20 \mathrm{~g}\) and the velocity of the bullet when it leaves the gun is \(300 \mathrm{~ms}^{-1}\). The force required to hold the gun while firing is:
1 \(6 \mathrm{~N}\)
2 \(8 \mathrm{~N}\)
3 \(24 \mathrm{~N}\)
4 \(240 \mathrm{~N}\)
Explanation:
C Given data, Mass of each bullet \((\mathrm{m})=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}\) No. of bullet \((\mathrm{n})=4\) \(\text { Velocity }(\mathrm{v}) =300 \mathrm{~ms}^{-1}\) \(\text { Time }(\mathrm{t}) =1 \mathrm{sec}\) \(\text { Force } =?\) According to Newton's second law, the rate of change in momentum (p) of an object is equal to the force (F) applied. \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\mathrm{F}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}} \quad[\because \mathrm{p}=\mathrm{mv}]\) \(\mathrm{F}_{\text {single }}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}\) \(\frac{\mathrm{dv}}{\mathrm{dt}} \times \mathrm{m}=\mathrm{ma} \quad\left[\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\right]\) \(\mathrm{F}_{\text {single }}=\frac{300}{1} \times 20 \times 10^{-3} \Rightarrow 6 \mathrm{~N}\) So, \(\quad \mathrm{F}_{\text {Total }}=\mathrm{n} \times \mathrm{F}_{\text {single }}\) \(\mathrm{F}_{\text {Total }}=4 \times 6=24 \mathrm{~N}\)
AP EAMCET(Medical)-2000
LAWS OF MOTION (ADDITIONAL)
371703
A force of \(100 \mathrm{~N}\) acts on a body of mass \(2 \mathrm{~kg}\) for \(10 \mathrm{~s}\). The change in momentum of the body is:
1 \(100 \mathrm{Ns}\)
2 \(250 \mathrm{Ns}\)
3 \(500 \mathrm{Ns}\)
4 \(1000 \mathrm{Ns}\)
Explanation:
D Given data, Mass of body \((\mathrm{m})=2 \mathrm{~kg}\) Applied force \((\mathrm{F})=100 \mathrm{~N}\) Time \((t)=10\) sec By Newton's second law, The rate of change of momentum is directly proportional to the force. \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}} \propto \mathrm{F}\) \(\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\mathrm{F}\) \(\Delta \mathrm{t}\) \(\Delta \mathrm{p}=\mathrm{F} . \Delta \mathrm{t}\) \(\Delta \mathrm{p}=100 \mathrm{~N} \times 10 \mathrm{~s}=1000 \mathrm{Ns}\)
