Question

Vertical distance covered by an object of mass 1kg in one second is:

Answer 1

Answer:

K/25=2J

Explanation:

Consider object A attains a height 'x' just before collision and object B descends a height 'y' just before the collision. The collision takes place at time 't' after the two objects start their flight.(Assume 'g' = 10m/s2)

Then for object 'B':

y=0.5gt²=5m.

For object 'A':

x=20t−0.5gt²

or

x=20t−y

or x + y = 20t = 20m;

Hence t = 1 sec.

and x = 15m.

At this time the velocities of objects 'A' and 'B' can be calculated by eqn of motion

v² − u²=2aS

For object B: 

v2=2×10×5=100.

or v = 10m/s;

Likewise for object A:

by eqn of motion: v = u + at;

wehavev=20−10×1=10m/s.

Now by conservation of momentum at the time of collision:

(mA vA  - mB vB ) = ( mA + mB )v;

hence

v = mAvA − mBvB/( mA + mB)

or v = -2 x10/4 = -5m/s  ie 5m/s in the direction of vB ;

The KE after collision will be 

KE = 0.5mAv² + 0.5mBv² = 0.5(mA + mB)v²

or KE = 0.5×4×5² =50J

K/25=2J

Answer 2

Concept:

The three fundamental laws of Newtonian mechanics called Newton's laws of motion describe how an object's motion and also the forces working on it interact.

Given:

The mass of the object is $1kg$ in one second.

Find:

The vertical distance is covered by the object.

Solution:

Mass of the object, $m=1kg$.

The vertical distance i.e. on the y-axis. Therefore, the vertical distance is the height.

The object is in free fall.

So, the initial velocity is zero.

The acceleration of the object will be acceleration due to gravity, $g=9.8m/s^2$.

The time, $t=1sec$.

Now, according to the second equation of motion.

$s=ut+\frac{1}{2}gt^2$

$s=\frac{1}{2}(9.8)(1)^2$

$s=4.9m$

The vertical distance covered by an object of mass $1kg$ in one second is $4.9m$.

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