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Lakshmeesha K.

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Physics (Newtonian Mechanics)

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Question:

The moon moves around the earth in a circular orbit with a period of 27 days. The radius of the earth $$R$$ is $$6.4 \times 10^6 m$$ and the acceleration due to gravity on the earth surface is $$9.8 m/s^2$$. If $$D$$ is the distance of the moon from the center of the earth, find the value of $$D/R$$ using the given data.

Lakshmeesha K.

Answer:

The problem is based on circular motion. The moon moves in the circular orbit and the force causing it is gravitational in nature. Since the orbit is circular we can equate this force to centripetal force. i.e., $$m \omega ^2 D= {GmM \over D^2}$$ where $$\omega$$ is the angular velocity of moon and $$m$$ and $$M$$ correspond to mass of the moon and the earth. Writing the angular velocity in terms of the time period of revolution, we have $$\omega ={2 \pi \over T}$$ thus we get $$D^3= {GMT^2 \over 4 \pi^2}$$ dividing both the sides with $$R^3$$ we get $$[{D \over R}]^3= {GM\over R^2}{T^2 \over 4 \pi^2 R}$$ but the acceleration due to gravity is given by $$g={GM \over R^2}$$ thus, $$[{D\over R}] = [\frac{gT^2}{4 \pi^2 R}]^{1\over3}= [\frac{9.8 \times (27*24*60*60)^2}{4 \pi^2 6.4*10^6}]^{1 \over 3}= 59.5$$ Notice that the answer has no units since it is a ratio of two dimensionally identical quantities.

Nuclear Physics

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Question:

A Particle $$A$$ of mass $$m$$ and initial velocity $$v$$ collides with a particle $$B$$ of mass $$m/2$$ which is at rest. The collision is head on and elastic. What is the ratio of de-Broglie wavelength $$\lambda_A$$ to $$\lambda_B$$ after the collision?

Lakshmeesha K.

Answer:

The question requires that we apply classical mechanics as far as collisions are concerned and bring in de-broglie hypothesis at the end. So we know from the question that the collision is elastic, thus the co-efficient of restitution is 1. Additionally the relations between the masses is given. Applying law of conservation of momentum, i.e., total momentum before and after collision are same, mass of A $$\times$$ initial velocity of A + mass of B $$\times$$ the initial velocity of B= mass of A $$\times$$ final velocity of A + mass of B $$\times$$ the final velocity of B i.e., $$m \times v+ \frac{m}{2} \times 0= m \times v_A+ \frac{m}{2} \times v_B$$ i.e., $$v=v_A+ \frac{v_B}{2}$$ $$(1)$$ thus the co-efficient of restitution formula gives; $$ e=\frac{v_B-v_A}{u_A-v_B}$$ $$\implies$$ $$1=\frac{v_B-v_A}{v}$$ $$\implies$$ $$v= v_B-v_A$$ $$(2)$$ from equations $$(1)$$ and $$(2)$$ we get $$v_A=\frac{v}{3}$$ and $$v_B=\frac{4v}{3}$$ Now invoking de-Broglie hypothesis, $$\lambda = h\p=h\mv$$ for $$A$$: $$\lambda_A= \frac{h}{mv_A} =\frac{3h}{mv}$$ and for $$B$$: $$\lambda_B= \frac{h}{\frac{m}{2} v_B}= \frac{6h}{4mv}$$ thus the ratio $$\frac{\lambda_A}{\lambda_B} =3 \times {4\over 6}=2$$

Physics

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Question:

A cyclist, weighing a total of $$80kg$$ with the bicycle, pedals at a speed of $$10 m/s$$. She stops pedalling at an instant which is taken to be $$t = 0$$ . Due to the velocity dependent frictional force, her velocity is found to vary as $$ v(t)= \frac{300}{30+t} m/s $$ , where t is measured in seconds. When the velocity drops to $$8 m/s$$ , she starts pedalling again to maintain a constant speed. What is the energy expended by her in $$ 1 $$ minute at this (new) speed?

Lakshmeesha K.

Answer:

The question above, asks for the energy expended in a certain time, and velocity is a variable that is given. So we need to look at the relation between velocity and energy. Certainly velocity and kinetic energy are related by $$K.E=\frac{1}{2}mv^2$$ but here we are looking at velocity which is not constant and asked for work done in a certain time. So we need to look at it other way. From Newton's 2nd law we know, $$F=ma$$ where $$a$$ is acceleration which is time rate of change of velocity. i.e., $$a=\frac{dv}{dt}$$. So multiplying the mass of the cyclist, which is given we arrive at the force. Now since work done is force $$\times$$ displacement, we are left to find out the displacement which is straight forward given the relation between time and velocity. Now lets look at the algebra. $$ v(t)= \frac{300}{30+t} $$ $$\implies$$ $$\frac{dv}{dt}=a= \frac {-300}{(30+t)^2}$$ $$\implies$$ $$F=ma=80\times \frac {-300}{(30+t)^2}$$. Since the frictional force is mentioned as the only force acting, the net force is same as the frictional force. If we multiply and divide the above equation by 300 we can relate force to velocity as $$F= \frac{-80}{300} \times [\frac{300}{30+t}]^2 = \frac{-4}{15} [v(t)]^2$$. So when the cyclist moves at a constant speed of $$8 m/s$$ $$F=\frac{-4}{15}8^2$$ Since distance= time $$\times$$ velocity, in one minute(60 seconds) the displacement is $$d= 8\times60$$ thus the work done is $$work= Force \times displacement= \frac{-4}{15} \times 64 \times 8 \times 60 =-8192=-8.192 KJ$$. Thus since work done and energy are equivalent, to maintain a constant speed she has to expend $$8.192 KJ$ of energy in one minute.

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