Question

A point mass of 1 kg moving at constant speed of 5 m/s on an elliptical path experiences a centripetal force of 20N when at either end point of the major axis and a similar force of 2.5 N at each end of minor axis. Then :-

• A. length of semi major axis is 5 m
• B. length of semi major axis is 10 m
• C. length of semi minor axis is 2.5 m
• D. length of semi minor axis is 5 m

Answer 1

Answer: (A), (C)

a, c

Answer 2

The problem describes a point mass moving on an elliptical path at a constant speed. We are given the mass of the object, its speed, and the centripetal force it experiences at the endpoints of the major and minor axes.

Let the mass of the object be $m = 1$ kg.
Let the constant speed of the object be $v = 5$ m/s.
Let the length of the semi-major axis be $a$ and the length of the semi-minor axis be $b$.

The centripetal force $F_c$ experienced by an object moving in a curved path with radius of curvature $R$ and speed $v$ is given by $F_c = \frac{mv^2}{R}$.

At the endpoints of the major axis, the radius of curvature is $R_a = \frac{b^2}{a}$. The centripetal force at these points is given as $F_a = 20$ N.
So, $F_a = \frac{mv^2}{R_a} = \frac{mv^2}{b^2/a} = \frac{mv^2 a}{b^2}$.
Substituting the given values, we have $20 = \frac{(1)(5^2)a}{b^2} = \frac{25a}{b^2}$.
This gives the equation $20b^2 = 25a$, which simplifies to $4b^2 = 5a$ (Equation 1).

At the endpoints of the minor axis, the radius of curvature is $R_b = \frac{a^2}{b}$. The centripetal force at these points is given as $F_b = 2.5$ N.
So, $F_b = \frac{mv^2}{R_b} = \frac{mv^2}{a^2/b} = \frac{mv^2 b}{a^2}$.
Substituting the given values, we have $2.5 = \frac{(1)(5^2)b}{a^2} = \frac{25b}{a^2}$.
Multiplying by 4, we get $10 = \frac{100b}{a^2}$, or $10a^2 = 100b$, which simplifies to $a^2 = 10b$ (Equation 2).

Now we solve the system of equations:

1. $4b^2 = 5a$
2. $a^2 = 10b$

From Equation 2, we can express $b$ in terms of $a$: $b = \frac{a^2}{10}$.
Substitute this expression for $b$ into Equation 1:
$4 \left(\frac{a^2}{10}\right)^2 = 5a$
$4 \frac{a^4}{100} = 5a$
$\frac{a^4}{25} = 5a$
$a^4 = 125a$
$a^4 - 125a = 0$
$a(a^3 - 125) = 0$
Since $a$ is the length of the semi-major axis, $a > 0$. Thus, we can divide by $a$:
$a^3 - 125 = 0$
$a^3 = 125$
$a = \sqrt[3]{125} = 5$.

So, the length of the semi-major axis is $a = 5$ m.

Now substitute the value of $a$ back into the expression for $b$:
$b = \frac{a^2}{10} = \frac{5^2}{10} = \frac{25}{10} = 2.5$.

So, the length of the semi-minor axis is $b = 2.5$ m.

Let's check the given options:
a. length of semi major axis is 5 m (Matches our result)
b. length of semi major axis is 10 m (Does not match)
c. length of semi minor axis is 2.5 m (Matches our result)
d. length of semi minor axis is 5 m (Does not match)

Based on our calculations, both option a and option c are correct.