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**Question 1: A certain computer hard disk drive is rated to withstand an acceleration of 100g without damage. Assuming the drive decelerates through a distance of 2 mm when it hits the ground, from how high can you drop the drive without running it?****Concept:****The height say y from which the disk falls freely, can be related to its final speed before collision say v _{y} and the initial speed v_{0y }as:**

From the equation of kinematics, we have

y = v

v

Where v

From the second equation, the time t is given as: t = (v

Substitute the value in first equation

y = v

= [(v

= [(v

= [(v

So,

2ay = (v

= (v

Since the disk accelerates under the action of gravity and we take the convention of downward motion to be positive, we can equate a = g in the equation above as:

(v

Substitute the given value of v0y to be zero, the above equation can be written as:

(v

(v

Assume that the distance travelled by the disk during collision is y', the speed of the disk just before the contact is v'

(v'

Since the disk will come to rest after colliding, hits final speed v'

Given that the disk crushes to a depth of 2 mm, the height y' can be equated to it.

Substitute the appropriate values in the equation above,

(0 m/s)

Substitute the value of (v

-2gy = 2a (2mm)

Given that the limit to the deceleration that the disk can withstand during the collision is 100 g, we can equate a = 100 g as,

-2gy = 2(-100 g)(2mm)

y = 100 (2 mm) (10

So, y = 0.2 m

Therefore if the disk is thrown from height 0.2 m the disk will start to damage, therefore the person can throw the disk from any height which is less than 0.2 m and avoid damaging it.

Inclination of slope, ? = 4.3°.

Altitude of plane, h = 35 m.

Speed of air, v

Let us assume that the length of slope to the point where the plane meets the slope is given by length l. Also, the speed of the plane is equal in magnitude with that of the speed of air, therefore the speed of the plane (say v

The figure below shows the motion of the plane and various other variables

From the figure one can see that the altitude (say d) of the slope is given in terms of length l of slope and the angle ? as:

d = l sin ?

If the plane were to collide with the slope then the following condition should be fulfilled:

h - d = 0

d = h

Substitute the value of d and h to have,

l sin ? = 35 m

l = 35 ml sin (4.3°)

So, l = 466.7 m

Therefore the length of the slope after which the plane collides is 466.7 m.

Now, to calculate the horizontal distance (say x) that the plane will travel if it collides with the slope, we calculate the base of triangle ABC as

From triangle ABC, we have

x = l cos ?

Substitute the value of l and ?, to have

x = (466.7 m) [cos (4.3°)]

= 465.3 m

Therefore the horizontal distance at which the plane will collide with the slope is 465.3 m.

The time (say t) that the plane had before it run into the inclined ground ahead is given as:

t = x/v

Substituting the value of x and v

t = (465.3 m)/(1300 km/h)

= (465.3 m)/[(1300 km/h) (10

=(465.3 m)/(361.1 m/s)

= 1.28 s

Rounding off to two significant figures,

t = 1.3 s

Therefore the plane has 1.3 s before it will run into the ground, and avoid the collision.

Given that the ball rise 2 m above the hand, the distance y travelled by the ball when he descends from the maximum height to reach the juggler hand is 2 m. Also the free fall acceleration is 9.81 m/s

Substitute the given values in equation (1),

y = ½ gt

2 m = ½ (9.81 m/s

t = √4 m/(9.81 m/s

= 0.63 s

The time (t') for which a ball stays in the air is given by multiplying t by 2 as:

t ' = 2t

= 2 (0.63 s)

= 1.27 s

The juggler has to juggles each ball once before time t ' so that he gets ready to juggle the ball reaching him again.

Therefore if two hands toss 5 times in 1.27 seconds, then the number of times (n) each ball is tossed in 1 min is given using equation (2) as:

n = [(5 times/1.27 s) (1 min)]/2

= [(5 times/1.27 s) (60 s)]/2

= 234.9 times/2

= 117.4 times.

Round off to three significant figures,

n = 117 times.

Therefore the juggler juggles at a rate of 117 toss per minute with each hand.

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