6.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system
6.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system
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1 Answer
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6.7 (a) False. The total momentum and energy is conserved, not the individual.
(b) False. External forces can change the energy of a body.
(c) False. Only work done by conservative force over a closed loop is zero.
(d) True. In an inelastic collision, the final velocity reduces, resulting in loss of the initial kinetic energy.
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Using Newton’s formula,
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(b) conserving energy between “O” ans ”A”

Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1= ………….2
From eqn 1 and 2
So v1 = v’/ =v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path
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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.
Hence total work done =-mgL + mgL=0
As the point of application of the contact forces does not move hence work done by reaction forces will be zero.
This is a multiple choice answer as classified in NCERT Exemplar
(c) m =150g =3/20kg
Time of contact =0.001s
U=126km/h=
V= -35m/s
Change in momentum of the ball = m (v-u)=
=21/2
F= dp/dt=- = - 1.05
Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.
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