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
Answers
Answered by
0
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
Answered by
0
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 .
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