Answer to Question #220054 in Mechanics | Relativity for Alex

Question

a. The orbit of a satellite about the Earth is given by r= 10000 /(1+0.6cos theta)

What is the eccentricity and shape of the orbit? Also calculate the apogee and

perigee distances.

b. A steel ball A collides elastically with another steel ball B (at rest initially) and the

ball B is observed to move off at an angle theta with the initial direction of motion of A.

The mass of B is five times that of A. Determine the direction in which A moves

after collision and the speeds of the two balls.

c. The displacement of an object executing simple harmonic oscillations is given by:

x = 0.02 sin 2pie (t+0.01) m

Determine (a) amplitude of the oscillatory motion, (b) time-period of oscillation,

(c) maximum velocity, (d) maximum acceleration, and (e) initial displacement of

the object.

Answer 1

Solution.

a. $r=\dfrac{10000}{1+0.6cos\theta}$ ;

$r=\dfrac{p}{1+ecos\theta};$

$e=0.6 -$ the orbit is elliptical;

$r_p=\dfrac{p}{1+e}; r_a=\dfrac{p}{1-e};$

$r_p=\dfrac{10000}{1+0.6}=6250; r_a=\dfrac{10000}{1-0.6}=25000;$

b. $m_b=5m_a;$

$\theta_b=\theta;$

$m_av_0+0=m_av_acos\theta_a+m_bv_bcos\theta_b;$

$m_av_0=m_av_acos\theta_a+5m_av_bcos\theta;$

$v_o=v_acos\theta_a+5v_bcos\theta;$ $cos\theta_a=\dfrac{v_0-5v_bcos\theta}{v_a};$

$\dfrac{m_av_0^2}{2}=\dfrac{m_av_a^2}{2}+\dfrac{5m_av_b^2}{2};$

$v_0^2=v_a^2+5v_b^2;$

c. $x=0.02sin2\pi(t+0.01);$

$x=0.02sin(2\pi t+0.02\pi);$

$A=0.02m;$

$T=\dfrac{2\pi}{2\pi}=1s;$

$v_{max}=\omega A;$

$v_{max}=2\pi\sdot0.02=0.04\sdot3.14=0.1256m/s;$

$a_{max}=\omega^2A;$

$a_{max}=(2\pi)^2\sdot0.02=0.788m/s^2;$

$x_o=0.02sin0.02\pi=0.0012m;$

Answer:$a. e=0.6;r_p=6250; r_a=25000;$

$b.$ $cos\theta_a=\dfrac{v_0-5v_bcos\theta}{v_a};$

$c. A=0.02m; T=1s; v_{max}=0.1256m/s; a_{max}=0.788m/s^2; x_0=0.0012m.$