Q. A ball of mass $m$ is attached to the lower end of light vertical spring of force constant $k$. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstretched) length and comes to rest again after descending through a distance $x$.

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A

$x=m g / k$

B

$x=2 m g / k$

C

The ball will have no acceleration at the position where it has descended through $x / 2$.

D

The ball will have an upward acceleration equal to $g$ at its lowermost position

Solution:

Apply conservation of energy:
loss in PE of ball = gain in PE of spring
$m g x=\frac{1}{2} k x^{2}$
$\Rightarrow x=2 m g / k$
At position $x / 2$ :

$k x / 2=m g$. So net force is zero. Hence, no acceleration at lowest position: $k x-m g=m a$
$\Rightarrow a=g$

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Q. A ball of mass $m$ is attached to the lower end of light vertical spring of force constant $k$. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstretched) length and comes to rest again after descending through a distance $x$.

$x=m g / k$

$x=2 m g / k$

The ball will have no acceleration at the position where it has descended through $x / 2$.

The ball will have an upward acceleration equal to $g$ at its lowermost position

Apply conservation of energy: loss in PE of ball = gain in PE of spring $m g x=\frac{1}{2} k x^{2}$ $\Rightarrow x=2 m g / k$ At position $x / 2$ : $k x / 2=m g$. So net force is zero. Hence, no acceleration at lowest position: $k x-m g=m a$ $\Rightarrow a=g$

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1. The work done by a particle moving with a velocity of $0.7\, c$ (where c is the velocity of light) in empty space free of electromagnetic field and far away from all matter is:

2. The machine gun fires $240 \, $ bullets per minute. If the mass of each bullet is $10 \, g$ and the velocity of the bullets is $600 \, m \, s^{- 1}$ ,the power (in $kW$ )of the gun is

4. A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_{c}$ is varying with time $t$ as $a_{c}=k^{2} \, rt^{2}$ , where $k$ is a constant. The power delivered to the particle by the force acting on it is

5. A ball of mass $1\, kg$ collides with a wall with speed $8\, ms ^{-1}$ and rebounds on the same line with the same speed. If mass of the wall is taken as infinite, the work done by the ball on the wall is

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5. A barometer is constructed using a liquid (density $= 760 \,kg/m^{3})$ What would be the height of the liquid column, when a mercury barometer reads $76\, cm$ ? (density of mercury $= 13600 kg/m^{3})$

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Apply conservation of energy:
loss in PE of ball = gain in PE of spring
$m g x=\frac{1}{2} k x^{2}$
$\Rightarrow x=2 m g / k$
At position $x / 2$ :

$k x / 2=m g$. So net force is zero. Hence, no acceleration at lowest position: $k x-m g=m a$
$\Rightarrow a=g$

5. A barometer is constructed using a liquid (density $= 760 \,kg/m^{3})$ What would be the height of the liquid column, when a mercury barometer reads $76\, cm$ ?
(density of mercury $= 13600 kg/m^{3})$