Question

two steel spheres of radii 2cm and 3cm move with a velocities of 24 cm /s in opposite direction and collide head on if collision is elastic, calculate velocities after collision​

Answer 1

Let us consider m  

1

​  

 & m  

2

​  

 are masses of the sphere and velocity is v of the sphere before collision 'v' is the velocity of other sphere and fist sphere at rest after collision.

From conservation of energy

2

1

​  

m  

1

​  

v  

1

2

​  

+  

2

1

​  

m  

2

​  

v  

2

2

​  

=  

2

1

​  

m  

1

​  

v  

1

2

​  

+  

2

1

​  

m  

2

​  

v  

2

2

​  

 

2

1

​  

m  

1

​  

v  

2

+  

2

1

​  

m  

2

​  

v  

2

=  

2

1

​  

m  

2

​  

v  

′2

      ....(1)

2

1

​  

m  

1

​  

v  

2

−  

2

1

​  

m  

2

​  

v  

2

=  

2

1

​  

m  

2

​  

v  

′2

 

v  

′

=  

m  

2

​  

 

n(m  

1

​  

−m  

2

​  

)

​  

 

Put the value of v  

′

 in eq (1)

2

1

​  

m  

1

​  

v  

2

+  

2

1

​  

m  

2

​  

v  

2

=  

2

1

​  

m  

2

​  

×(  

m  

2

​  

 

v(m  

1

​  

−m  

2

​  

)

​  

)  

2

​  

 

m  

1

​  

+m  

2

​  

=  

m  

2

​  

 

(m  

1

​  

−m  

2

​  

)  

2

 

​  

 

m  

1

​  

m  

2

​  

+m  

2

2

​  

=m  

1

2

​  

+m  

2

2

​  

−2m  

1

​  

m  

2

​  

 

m  

2

​  

=  

3

m  

1

​  

 

​  

       ...(2)

The volume of the fiist sphere is v and the volume of other is v/3

Now, the radius of the sphere is r and other sphere is r  

′

 

The volume of first sphere is

3

4

​  

πr  

′

3=  

3

v

​  

       ...(3)

the volume of radius other sphere is

3

4

​  

πr  

′

3=  

3

v

​  

       ....(4)

Divide eq. (4) by (3)

r  

′

=  

(3)  

3

1

​  

 

r

​

Answer 2

Answer:

As both balls of steel so same density and so their mass will be proportional to their volume and volume is proportional to cube of radius so their mass m

1

and m

2

will be in ratio of 2

3

:3

3

i.e. 8:27

so suppose m

1

=8m and m

2

=27m then use equation

m

1

u

1

+m

2

u

2

=m

1

v

1

+m

2

v

2

and Newton's law of collision as

v

2

−v

1

u

1

−u

2

=e ............................1

given e=1 and u

1

=5cm/s u

2

=0

From the equation-1 we get v

2

=v

1

+5 put it in the momentum conservation equation we will get v

1

=−2.7cm/s and v

2

=2.3cm/s

Note- Negative sign shows that the lighter mass bounced means went back with less speed .