The object moves on the -axis under conservative force in such a way that its ''speed'' and position satisfy where and are positive constants. [2007]

Question

	Column I		Column II
(A)	The object moves on the $x$ -axis under conservative force in such a way that its ''speed'' and position satisfy $v = c_{1} \sqrt{c_{2} - x^{2}}$ where $c_{1}$ and $c_{2}$ are positive constants.	(p)	The object executes a simple harmonic motion.
(B)	The object moves on the $x$ -axis in such a way that its velocity and its displacement from the origin satisfy $v = - k x,$ where $k$ is a positive constant.	(q)	The object does not change its direction.
(C)	The object is attached to one end of a massless spring of given spring constant. The other end of the spring is attached to the ceiling of an elevator. Initially everything is at rest. The elevator starts going upwards with a constant acceleration $a$ . The motion of the object is observed from the elevator during the period it maintains this acceleration.	(r)	The kinetic energy of the object keeps on decreasing.
(D)	The object is projected from the earth's surface vertically upwards with a speed $2 \sqrt{\frac{G M_{e}}{R_{e}}},$ where $M_{e}$ is the mass of the earth and $R_{e}$ is the radius of the earth. Neglect forces from objects other than the earth.	(s)	The object can change its direction only once.

[2007]

Smart Achievers · Accepted Answer

(2)

(A): For a simple harmonic motion $v = ω \sqrt{a^{2} - x^{2}}$ . On comparing it with $v = c_{1} \sqrt{c_{2} - x^{2}}$ this equation is SHM with $ω = c_{1}$ and $a^{2} = c_{2}$

(B): $v = - k x$

when $x$ is positive; $v$ is $- v e$ , and as $x$ decreases, $v$ decreases. Therefore kinetic energy will decrease. When $x = 0, v = 0$ . Therefore the object does not change its direction.

When $2 \sqrt{\frac{G M_{e}}{R_{e}}},$ is negative, $v$ is positive. But as $x$ decreases in magnitude, $v$ also decreases. Therefore kinetic energy decreases. When $x = 0, v = 0$ . Therefore the object does not change its direction.

(C): When $a = 0$ , let the spring have an extension $x$ . Then $k x = m g$

When the elevator starts going upwards with a constant acceleration, as seen by the observer in the elevator, the object is at rest.

$∴$ $m a + m g = k x'$

$\Rightarrow m a = k (x' - x)$ (Since $a$ is constant)

(D): The object is projected with a speed is $\sqrt{2}$ times the escape speed $V_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$ . Therefore the object will leave the earth. It will therefore not change the direction and keeps on moving with decreasing speed.

Solution

1 $A \to q, r; B \to p; C \to p; D \to q, r$

2 $A \to p; B \to q, r; C \to p; D \to q, r$

3 $A \to q, r; B \to p; C \to q, r; D \to p$

4 $A \to p; B \to q, r; C \to q, r; D \to p$

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Solution

1 A→q,r; B→p; C→p; D→q,r

2 A→p; B→q,r; C→p; D→q,r

3 A→q,r; B→p; C→q,r; D→p

4 A→p; B→q,r; C→q,r; D→p

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1 $A \to q, r; B \to p; C \to p; D \to q, r$

2 $A \to p; B \to q, r; C \to p; D \to q, r$

3 $A \to q, r; B \to p; C \to q, r; D \to p$

4 $A \to p; B \to q, r; C \to q, r; D \to p$