Two small glass spheres of masses 10 g and
20 g are moving in a straight line in the same
direction with velocities of 3 m s-1 and 2 m s-1
respectively. They collide with each other and
after collision, glass sphere of mass 10 g moves
with a velocity of 2.5 m s-1. Find the velocity
of the second ball after collision.
I AM GETTING THE ANSWER AS 2.25. BUT IF THE VELOCITY OF THE SECOND BALL IS LESSER THAN THE FIRST THEN THE FIRST SHOULD BE PUSHING THE SECOND BALL ALONG WITH IT.
PLEASE GIVE ME A PROPER SOLUTION (that befits 25 points ) OR YOUR ANSWER WILL BE REPORTED.
Answers
Here, m
1
=10g=
1000
10
=0.01kg
m
2
=20g=0.02kg
u
1
=3ms
−1
;u
2
=2ms
−1
v
1
=2.5ms
−1
;v
2
=?
Total momentum of both the spheres before collision =m
1
u
1
+m
2
u
2
=0.01×3+0.02×2=0.07kgms
−1
Total momentum of both the spheres after collision
=m
1
v
1
+m
2
v
2
=0.01×2.5+0.02v
2
=0.025+0.02v
2
Now, according to the law of conservation of momentum,
Total momentum after collision =Total momentum before collision
∴0.025+0.02v
2
=0.07
or 0.02v
2
=0.07−0.025=0.045
or v
2
=
0.02
0.045
=2.25ms
−1
Answer:
Here, m
1
=10g=
1000
10
=0.01kg
m
2
=20g=0.02kg
u
1
=3ms
−1
;u
2
=2ms
−1
v
1
=2.5ms
−1
;v
2
=?
Total momentum of both the spheres before collision =m
1
u
1
+m
2
u
2
=0.01×3+0.02×2=0.07kgms
−1
Total momentum of both the spheres after collision
=m
1
v
1
+m
2
v
2
=0.01×2.5+0.02v
2
=0.025+0.02v
2
Now, according to the law of conservation of momentum,
Total momentum after collision =Total momentum before collision
∴0.025+0.02v
2
=0.07
or 0.02v
2
=0.07−0.025=0.045
or v
2
=
0.02
0.045
=2.25ms
−1