Work, Energy, and Power

This is a short answer type question as classified in NCERT Exemplar

mass of the rain drop = 1g=1 $*{10}^{-3}kg$

Height of falling h= 1km = 10³m and g = 10m/s² and sped of drop =50m/s

(a) Loss of PE of the drop =mgh= 1 $*{10}^{-3}*10*{10}^3=10J$

(b) Gain in KE of the drop = 1/2mv²= ½ $*1*{10}^{-3}*{50}^2$ =1.250J

(c) No gain in KE is not equal to the loss in its Pe, because a part of PE is utilised in doing work against the viscous drag in air.

This is a short answer type question as classified in NCERT Exemplar

When the ball A reaches bottom point its velocity in horizontal direction

(a) two balls have same mass and the collision between them is elastic therefore ball A transfers its entire linear momentum to ball B. hence ball A will come to rest after collision and does not rise at all.

(b) speed at B = speed with which A hits the ball B

= $\sqrt[]{2gh}$

$=\sqrt[]{2*9.8*1}$

$=\sqrt[]{19.6}=4.42m/s$

This is a short answer type question as classified in NCERT Exemplar

TE=KE+PE

E=V+K

For region A given V>E 

so 'K=E-V

V>E

 So E-V <0

Hence K<0 this is not possible.

For region B given V

So E-V>0

This is only possible because total energy is greater than PE

For region C given K>E

 So K-E>0

So PE =V= E-K <0

Which is possible because PE can be negative.

For region D given V>K

This is possible because for the system PE may be greater than KE

This is a short answer type question as classified in NCERT Exemplar

Let v₁ and v₂ are velocities of two balls after collision

According to conservation of momentum

2mv_o = mv₁+ mv₂

2v_o= v₁+v₂

and e= v₂-v₁/2v_o

v₂=v₁+2v_oe

2v₁=2v_o-2ev_o

V₁=v_o (1-e) since e<1 so ball will move after collision.

b)by principle of conservation of linear momentum

P=P₁+P₂

For inelastic collision some KE is lost hence $\frac{p^2}{2m}$ > $\frac{p_1^2}{2m}\frac{p_2^2}{2m}$

P²>p₁²+p₂²

Thus P, P₁ and P₂ are related as shown in fig

P²>p₁²+p₂² this condition only holds when angle is 90.

This is a short answer type question as classified in NCERT Exemplar

mechanical energy =KE+PE

E_o = KE + V (x)

KE= E₀ -V (x)

At A x=0 v (x)=E_o

KE= E_o-E_o =0

atB, V (x)o

so KE>0

at C and D, V (x)= 0

KE is maximum at FV (x)= E_o

Hence KE= 0

As KE= 1/2mv²

Therefore at A and F where KE =0, v=0

At C and D KE is maximum therefore v is maximum.

At B  KE is positive but not maximum but it has some value.

This is a short answer type question as classified in NCERT Exemplar

At B the velocity of B is vertically downward, therefore when string is cut at B then it fall downwards.

At C velocity along horizontally right, so when we cut it at C then it will move to right. But under the action of gravity its path becomes parabola.

At X when we cut it moves tangentially in forward direction. So under of the action of gravity it also follows parabola path.

This is a short answer type question as classified in NCERT Exemplar

According to work energy theorem = change in kinetic energy = work done

Kinetic energy of the body = force (displacement)

As kinetic energy is same so both will move with same velocities.

This is a short answer type question as classified in NCERT Exemplar

When charge particle moves in a uniform magnetic field the path of particle is circular . so when it moves in circular path its radius Is also constant . so its kinetic energy also constant.

This is a short answer type question as classified in NCERT Exemplar

Average work done by a human heart per beat =0.5J

Total work done during 72 beats= 72 $*0.5$ =36J

Power = work done/time

=36/60=0.6W

This is a short answer type question as classified in NCERT Exemplar

Total linear momentum of the system of two balls is always conserved. While balls are in contact, there may be deformation which means elastic PE which came from part of KE .But they are in contact while collision so according to above case kinetic energy will not conserved.