371631
Two masses of \(1 \mathrm{~g}\) and \(4 \mathrm{~g}\) are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is
1 \(4: 1\)
2 \(1: 2\)
3 \(\sqrt{2}: 1\)
4 \(1: 16\)
Explanation:
B Given that, Mass of bodies, \(\mathrm{m}_{1}=1 \mathrm{~kg}\) \(\mathrm{m}_{2}=4 \mathrm{~kg}\) Both kinetic energy are equal Kinetic energy \((K)=\frac{P^{2}}{2 m}\) \(P=\sqrt{2 m K}\) According to question kinetic energy is same. So, \(\mathrm{P} \propto \sqrt{\mathrm{m}}\) \(\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}\) \(=\sqrt{\frac{1}{4}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
TS EAMCET 28.09.2020
LAWS OF MOTION (ADDITIONAL)
371632
\(\quad\) Two bodies of masses \(\mathrm{m}\) and \(4 \mathrm{~m}\) have kinetic energies in the ratio \(1: 2\). Their momentum \(p_{1}\) and \(p_{2}\) are in the ratio
C Given that, Mass of the body \((\mathrm{m})=2 \mathrm{~kg}\) Acceleration of the body (a) \(=3 \mathrm{~m} / \mathrm{sec}^{2}\) By Newton's second law Rate of change of momentum is equal to applied force \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\int_{0}^{1} F d t=\int_{P_{1}}^{P_{2}} d p\) \(\mathrm{F} \int_{0}^{1} \mathrm{dt}=\int_{\mathrm{P}_{1}}^{\mathrm{P}_{2}} \mathrm{dp}\) \(\mathrm{F}[\mathrm{t}]_{0}^{1}=[\mathrm{p}]_{\mathrm{p}_{1}}^{\mathrm{p}_{2}}\) \(\mathrm{F}(1-0)=\mathrm{p}_{2}-\mathrm{p}_{1}\) \(\quad \mathrm{~F}=\Delta \mathrm{p}\) \(\because \quad \mathrm{F}=\mathrm{ma}\) \(\quad \mathrm{F}=2 \times 3\) \(\quad \mathrm{~F}=6 \mathrm{~N}\) \(\therefore \quad \Delta \mathrm{p}=6 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\) \(\text { So, change of momentum in one second is } 6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
AP EAMCET-06.09.2021
LAWS OF MOTION (ADDITIONAL)
371634
Area under the force-time graph gives the change in
1 velocity
2 acceleration
3 linear momentum
4 angular momentum
5 impulsive force
Explanation:
E If we calculate area of triangle \(\mathrm{OAB}\), Area of shaped portion, \(=\frac{1}{2} \times \text { Base } \times \text { Altitude }\) \(=\frac{1}{2} \times \text { time } \times F_{\text {peak }}\) The above term gives impulse or change in momentum of body. Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.
Kerala CEE 2021
LAWS OF MOTION (ADDITIONAL)
371635
A ball of mass \(0.15 \mathrm{~kg}\) is dropped from a height \(10 \mathrm{~m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 0
2 \(4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
3 \(2.1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
4 \(1.4 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
Explanation:
B Given, Mass of the ball \(=0.15 \mathrm{~kg}\) Height from which ball is dropped \(=10 \mathrm{~m}\) We know that, Impulse, \(\mathrm{I}=\) change in linear momentum \((\Delta \mathrm{p})\) Velocity of ball at ground, \(\mathrm{v}=\sqrt{2 \mathrm{gh}}\) \(\mathrm{v}=\sqrt{2 \times 10 \times 10}\) \(\mathrm{v}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\) \(\mathrm{I} =\Delta \mathrm{p}\) \(=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(=\mathrm{mv}-[\mathrm{m}(-\mathrm{v})]\) \(=2 \mathrm{mv}\) \(=2 \times 0.15 \times 10 \sqrt{2}\) \(=4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
371631
Two masses of \(1 \mathrm{~g}\) and \(4 \mathrm{~g}\) are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is
1 \(4: 1\)
2 \(1: 2\)
3 \(\sqrt{2}: 1\)
4 \(1: 16\)
Explanation:
B Given that, Mass of bodies, \(\mathrm{m}_{1}=1 \mathrm{~kg}\) \(\mathrm{m}_{2}=4 \mathrm{~kg}\) Both kinetic energy are equal Kinetic energy \((K)=\frac{P^{2}}{2 m}\) \(P=\sqrt{2 m K}\) According to question kinetic energy is same. So, \(\mathrm{P} \propto \sqrt{\mathrm{m}}\) \(\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}\) \(=\sqrt{\frac{1}{4}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
TS EAMCET 28.09.2020
LAWS OF MOTION (ADDITIONAL)
371632
\(\quad\) Two bodies of masses \(\mathrm{m}\) and \(4 \mathrm{~m}\) have kinetic energies in the ratio \(1: 2\). Their momentum \(p_{1}\) and \(p_{2}\) are in the ratio
C Given that, Mass of the body \((\mathrm{m})=2 \mathrm{~kg}\) Acceleration of the body (a) \(=3 \mathrm{~m} / \mathrm{sec}^{2}\) By Newton's second law Rate of change of momentum is equal to applied force \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\int_{0}^{1} F d t=\int_{P_{1}}^{P_{2}} d p\) \(\mathrm{F} \int_{0}^{1} \mathrm{dt}=\int_{\mathrm{P}_{1}}^{\mathrm{P}_{2}} \mathrm{dp}\) \(\mathrm{F}[\mathrm{t}]_{0}^{1}=[\mathrm{p}]_{\mathrm{p}_{1}}^{\mathrm{p}_{2}}\) \(\mathrm{F}(1-0)=\mathrm{p}_{2}-\mathrm{p}_{1}\) \(\quad \mathrm{~F}=\Delta \mathrm{p}\) \(\because \quad \mathrm{F}=\mathrm{ma}\) \(\quad \mathrm{F}=2 \times 3\) \(\quad \mathrm{~F}=6 \mathrm{~N}\) \(\therefore \quad \Delta \mathrm{p}=6 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\) \(\text { So, change of momentum in one second is } 6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
AP EAMCET-06.09.2021
LAWS OF MOTION (ADDITIONAL)
371634
Area under the force-time graph gives the change in
1 velocity
2 acceleration
3 linear momentum
4 angular momentum
5 impulsive force
Explanation:
E If we calculate area of triangle \(\mathrm{OAB}\), Area of shaped portion, \(=\frac{1}{2} \times \text { Base } \times \text { Altitude }\) \(=\frac{1}{2} \times \text { time } \times F_{\text {peak }}\) The above term gives impulse or change in momentum of body. Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.
Kerala CEE 2021
LAWS OF MOTION (ADDITIONAL)
371635
A ball of mass \(0.15 \mathrm{~kg}\) is dropped from a height \(10 \mathrm{~m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 0
2 \(4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
3 \(2.1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
4 \(1.4 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
Explanation:
B Given, Mass of the ball \(=0.15 \mathrm{~kg}\) Height from which ball is dropped \(=10 \mathrm{~m}\) We know that, Impulse, \(\mathrm{I}=\) change in linear momentum \((\Delta \mathrm{p})\) Velocity of ball at ground, \(\mathrm{v}=\sqrt{2 \mathrm{gh}}\) \(\mathrm{v}=\sqrt{2 \times 10 \times 10}\) \(\mathrm{v}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\) \(\mathrm{I} =\Delta \mathrm{p}\) \(=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(=\mathrm{mv}-[\mathrm{m}(-\mathrm{v})]\) \(=2 \mathrm{mv}\) \(=2 \times 0.15 \times 10 \sqrt{2}\) \(=4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
371631
Two masses of \(1 \mathrm{~g}\) and \(4 \mathrm{~g}\) are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is
1 \(4: 1\)
2 \(1: 2\)
3 \(\sqrt{2}: 1\)
4 \(1: 16\)
Explanation:
B Given that, Mass of bodies, \(\mathrm{m}_{1}=1 \mathrm{~kg}\) \(\mathrm{m}_{2}=4 \mathrm{~kg}\) Both kinetic energy are equal Kinetic energy \((K)=\frac{P^{2}}{2 m}\) \(P=\sqrt{2 m K}\) According to question kinetic energy is same. So, \(\mathrm{P} \propto \sqrt{\mathrm{m}}\) \(\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}\) \(=\sqrt{\frac{1}{4}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
TS EAMCET 28.09.2020
LAWS OF MOTION (ADDITIONAL)
371632
\(\quad\) Two bodies of masses \(\mathrm{m}\) and \(4 \mathrm{~m}\) have kinetic energies in the ratio \(1: 2\). Their momentum \(p_{1}\) and \(p_{2}\) are in the ratio
C Given that, Mass of the body \((\mathrm{m})=2 \mathrm{~kg}\) Acceleration of the body (a) \(=3 \mathrm{~m} / \mathrm{sec}^{2}\) By Newton's second law Rate of change of momentum is equal to applied force \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\int_{0}^{1} F d t=\int_{P_{1}}^{P_{2}} d p\) \(\mathrm{F} \int_{0}^{1} \mathrm{dt}=\int_{\mathrm{P}_{1}}^{\mathrm{P}_{2}} \mathrm{dp}\) \(\mathrm{F}[\mathrm{t}]_{0}^{1}=[\mathrm{p}]_{\mathrm{p}_{1}}^{\mathrm{p}_{2}}\) \(\mathrm{F}(1-0)=\mathrm{p}_{2}-\mathrm{p}_{1}\) \(\quad \mathrm{~F}=\Delta \mathrm{p}\) \(\because \quad \mathrm{F}=\mathrm{ma}\) \(\quad \mathrm{F}=2 \times 3\) \(\quad \mathrm{~F}=6 \mathrm{~N}\) \(\therefore \quad \Delta \mathrm{p}=6 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\) \(\text { So, change of momentum in one second is } 6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
AP EAMCET-06.09.2021
LAWS OF MOTION (ADDITIONAL)
371634
Area under the force-time graph gives the change in
1 velocity
2 acceleration
3 linear momentum
4 angular momentum
5 impulsive force
Explanation:
E If we calculate area of triangle \(\mathrm{OAB}\), Area of shaped portion, \(=\frac{1}{2} \times \text { Base } \times \text { Altitude }\) \(=\frac{1}{2} \times \text { time } \times F_{\text {peak }}\) The above term gives impulse or change in momentum of body. Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.
Kerala CEE 2021
LAWS OF MOTION (ADDITIONAL)
371635
A ball of mass \(0.15 \mathrm{~kg}\) is dropped from a height \(10 \mathrm{~m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 0
2 \(4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
3 \(2.1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
4 \(1.4 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
Explanation:
B Given, Mass of the ball \(=0.15 \mathrm{~kg}\) Height from which ball is dropped \(=10 \mathrm{~m}\) We know that, Impulse, \(\mathrm{I}=\) change in linear momentum \((\Delta \mathrm{p})\) Velocity of ball at ground, \(\mathrm{v}=\sqrt{2 \mathrm{gh}}\) \(\mathrm{v}=\sqrt{2 \times 10 \times 10}\) \(\mathrm{v}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\) \(\mathrm{I} =\Delta \mathrm{p}\) \(=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(=\mathrm{mv}-[\mathrm{m}(-\mathrm{v})]\) \(=2 \mathrm{mv}\) \(=2 \times 0.15 \times 10 \sqrt{2}\) \(=4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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LAWS OF MOTION (ADDITIONAL)
371631
Two masses of \(1 \mathrm{~g}\) and \(4 \mathrm{~g}\) are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is
1 \(4: 1\)
2 \(1: 2\)
3 \(\sqrt{2}: 1\)
4 \(1: 16\)
Explanation:
B Given that, Mass of bodies, \(\mathrm{m}_{1}=1 \mathrm{~kg}\) \(\mathrm{m}_{2}=4 \mathrm{~kg}\) Both kinetic energy are equal Kinetic energy \((K)=\frac{P^{2}}{2 m}\) \(P=\sqrt{2 m K}\) According to question kinetic energy is same. So, \(\mathrm{P} \propto \sqrt{\mathrm{m}}\) \(\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}\) \(=\sqrt{\frac{1}{4}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
TS EAMCET 28.09.2020
LAWS OF MOTION (ADDITIONAL)
371632
\(\quad\) Two bodies of masses \(\mathrm{m}\) and \(4 \mathrm{~m}\) have kinetic energies in the ratio \(1: 2\). Their momentum \(p_{1}\) and \(p_{2}\) are in the ratio
C Given that, Mass of the body \((\mathrm{m})=2 \mathrm{~kg}\) Acceleration of the body (a) \(=3 \mathrm{~m} / \mathrm{sec}^{2}\) By Newton's second law Rate of change of momentum is equal to applied force \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\int_{0}^{1} F d t=\int_{P_{1}}^{P_{2}} d p\) \(\mathrm{F} \int_{0}^{1} \mathrm{dt}=\int_{\mathrm{P}_{1}}^{\mathrm{P}_{2}} \mathrm{dp}\) \(\mathrm{F}[\mathrm{t}]_{0}^{1}=[\mathrm{p}]_{\mathrm{p}_{1}}^{\mathrm{p}_{2}}\) \(\mathrm{F}(1-0)=\mathrm{p}_{2}-\mathrm{p}_{1}\) \(\quad \mathrm{~F}=\Delta \mathrm{p}\) \(\because \quad \mathrm{F}=\mathrm{ma}\) \(\quad \mathrm{F}=2 \times 3\) \(\quad \mathrm{~F}=6 \mathrm{~N}\) \(\therefore \quad \Delta \mathrm{p}=6 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\) \(\text { So, change of momentum in one second is } 6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
AP EAMCET-06.09.2021
LAWS OF MOTION (ADDITIONAL)
371634
Area under the force-time graph gives the change in
1 velocity
2 acceleration
3 linear momentum
4 angular momentum
5 impulsive force
Explanation:
E If we calculate area of triangle \(\mathrm{OAB}\), Area of shaped portion, \(=\frac{1}{2} \times \text { Base } \times \text { Altitude }\) \(=\frac{1}{2} \times \text { time } \times F_{\text {peak }}\) The above term gives impulse or change in momentum of body. Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.
Kerala CEE 2021
LAWS OF MOTION (ADDITIONAL)
371635
A ball of mass \(0.15 \mathrm{~kg}\) is dropped from a height \(10 \mathrm{~m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 0
2 \(4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
3 \(2.1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
4 \(1.4 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
Explanation:
B Given, Mass of the ball \(=0.15 \mathrm{~kg}\) Height from which ball is dropped \(=10 \mathrm{~m}\) We know that, Impulse, \(\mathrm{I}=\) change in linear momentum \((\Delta \mathrm{p})\) Velocity of ball at ground, \(\mathrm{v}=\sqrt{2 \mathrm{gh}}\) \(\mathrm{v}=\sqrt{2 \times 10 \times 10}\) \(\mathrm{v}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\) \(\mathrm{I} =\Delta \mathrm{p}\) \(=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(=\mathrm{mv}-[\mathrm{m}(-\mathrm{v})]\) \(=2 \mathrm{mv}\) \(=2 \times 0.15 \times 10 \sqrt{2}\) \(=4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
371631
Two masses of \(1 \mathrm{~g}\) and \(4 \mathrm{~g}\) are moving with equal kinetic energy. The ratio of the magnitudes of their linear momentum is
1 \(4: 1\)
2 \(1: 2\)
3 \(\sqrt{2}: 1\)
4 \(1: 16\)
Explanation:
B Given that, Mass of bodies, \(\mathrm{m}_{1}=1 \mathrm{~kg}\) \(\mathrm{m}_{2}=4 \mathrm{~kg}\) Both kinetic energy are equal Kinetic energy \((K)=\frac{P^{2}}{2 m}\) \(P=\sqrt{2 m K}\) According to question kinetic energy is same. So, \(\mathrm{P} \propto \sqrt{\mathrm{m}}\) \(\therefore \quad \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}}\) \(=\sqrt{\frac{1}{4}}\) \(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{1}{2}\) \(\mathrm{P}_{1}: \mathrm{P}_{2}=1: 2\)
TS EAMCET 28.09.2020
LAWS OF MOTION (ADDITIONAL)
371632
\(\quad\) Two bodies of masses \(\mathrm{m}\) and \(4 \mathrm{~m}\) have kinetic energies in the ratio \(1: 2\). Their momentum \(p_{1}\) and \(p_{2}\) are in the ratio
C Given that, Mass of the body \((\mathrm{m})=2 \mathrm{~kg}\) Acceleration of the body (a) \(=3 \mathrm{~m} / \mathrm{sec}^{2}\) By Newton's second law Rate of change of momentum is equal to applied force \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(\int_{0}^{1} F d t=\int_{P_{1}}^{P_{2}} d p\) \(\mathrm{F} \int_{0}^{1} \mathrm{dt}=\int_{\mathrm{P}_{1}}^{\mathrm{P}_{2}} \mathrm{dp}\) \(\mathrm{F}[\mathrm{t}]_{0}^{1}=[\mathrm{p}]_{\mathrm{p}_{1}}^{\mathrm{p}_{2}}\) \(\mathrm{F}(1-0)=\mathrm{p}_{2}-\mathrm{p}_{1}\) \(\quad \mathrm{~F}=\Delta \mathrm{p}\) \(\because \quad \mathrm{F}=\mathrm{ma}\) \(\quad \mathrm{F}=2 \times 3\) \(\quad \mathrm{~F}=6 \mathrm{~N}\) \(\therefore \quad \Delta \mathrm{p}=6 \mathrm{~kg} \mathrm{~m} \mathrm{~s}\) \(\text { So, change of momentum in one second is } 6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
AP EAMCET-06.09.2021
LAWS OF MOTION (ADDITIONAL)
371634
Area under the force-time graph gives the change in
1 velocity
2 acceleration
3 linear momentum
4 angular momentum
5 impulsive force
Explanation:
E If we calculate area of triangle \(\mathrm{OAB}\), Area of shaped portion, \(=\frac{1}{2} \times \text { Base } \times \text { Altitude }\) \(=\frac{1}{2} \times \text { time } \times F_{\text {peak }}\) The above term gives impulse or change in momentum of body. Hence, Area under the force-time graph gives the impulse or change in momentum acting on the body.
Kerala CEE 2021
LAWS OF MOTION (ADDITIONAL)
371635
A ball of mass \(0.15 \mathrm{~kg}\) is dropped from a height \(10 \mathrm{~m}\), strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 0
2 \(4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
3 \(2.1 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
4 \(1.4 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
Explanation:
B Given, Mass of the ball \(=0.15 \mathrm{~kg}\) Height from which ball is dropped \(=10 \mathrm{~m}\) We know that, Impulse, \(\mathrm{I}=\) change in linear momentum \((\Delta \mathrm{p})\) Velocity of ball at ground, \(\mathrm{v}=\sqrt{2 \mathrm{gh}}\) \(\mathrm{v}=\sqrt{2 \times 10 \times 10}\) \(\mathrm{v}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\) \(\mathrm{I} =\Delta \mathrm{p}\) \(=\mathrm{mv}_{\mathrm{f}}-\mathrm{mv}_{\mathrm{i}}\) \(=\mathrm{mv}-[\mathrm{m}(-\mathrm{v})]\) \(=2 \mathrm{mv}\) \(=2 \times 0.15 \times 10 \sqrt{2}\) \(=4.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s}\)
