Question

A ball is dropped from a height of 5.0 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Initial velocity(u) of the ball is 0 m/s.

Let the final velocity of the ball when it just strikes the sand is v m/s.

Applying Newton's laws,we get,

v^2-u^2=2gs

v^2-0=2(10)(5)

v^2=100

v=10 m/s.

Now, initial velocity(v) when the ball strikes the sand is 10 m/s

and the final velocity v' when the ball stops inside the sand is 0 m/s.

Here,displacement inside the sand is 10 cm or 0.1 m

and the accn. of the ball inside the sand is a'm/s^2.

again,applying Newton's laws,we get,

v'^2-v^2=2a's

0-100=2(-a)(0.1).........(here the negative sign indicates retardation of the ball,i.e.negative acc. of the ball.)

-100=-0.2a

a=500m/s^2

So,the retardation of the ball is 500m/s^2.

Answer 2

Solution:
Initial velocity=u=0 m/s
a=9.8 m/s2
Let us consider the motion of ball from Point A to B. B is just above the sand
s=5m
From second equation of motion :
s=ut +1/2at²
5=0+1/2 x9.8 (t²)
t²=5/4.9 =1.02 sec
t=1.01 sec
Velocity at point B ,
V=u+at
=9.8 x1.01=9.89 m/s
For motion of ball in sand : u1 =9.89 m/s  , v1=0
s1=10cm=0.1m
v1²-u1²=2as2
s2= v1²-u1²/2a=0-(9.8)²/2x0.1
= -490 m/s²

Hence retardation in sand is 490m/s2