In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)?
(a) Kinetic energy.
(b) Total linear momentum.
Give reason for your answer in each case.
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 Therefore, KE may not be conserved.
Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.
$$\begin{aligned} \text { mass } & =m=100 \mathrm{~kg} \\ \text { height } & =h=10 \mathrm{~m} \text { time duration } t=20 \mathrm{~s} \\ \text { power } & =\text { Rate of work done } \\ & =\frac{\text { change of } \mathrm{PE}}{\text { time }}=\frac{m g h}{t} \\ & =\frac{100 \times 9.8 \times 10}{20} \\ & =5 \times 98=490 \mathrm{~W} \end{aligned}$$
The average work done by a human heart while it beats once is 0.5 J . Calculate the power used by heart if it beats 72 times in a minute.
Given, average work done by a human heart per beat $$=0.5 \mathrm{~J}$$
$$\begin{aligned} &\text { Total work done during } 72 \text { beats }\\ &\begin{aligned} & =72 \times 0.5 \mathrm{~J}=36 \mathrm{~J} \\ \text { Power } & =\frac{\text { Work done }}{\text { Time }}=\frac{36 \mathrm{~J}}{60 \mathrm{~s}}=0.6 \mathrm{~W} \end{aligned} \end{aligned}$$
Give example of a situation in which an applied force does not result in a change in kinetic energy.
When a charged particle moves in a uniform normal magnetic field, the path of the particle is circular, as given field is uniform hence, radius of the circular path is also constant.
As the force is central and movement is tangential work done by the force is zero. As speed is also constant we can say that $$\Delta K=0$$.
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
According to work-energy theorem,
Change in $$\mathrm{KE}=$$ Work done by the retarding force
KE of the body $$=$$ Retarding force $\times$ Displacement
As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.