A particle A, of mass 2 kg, collides with
a particle B, of mass 2 kg. The velocity
of particle A before the collision was (5i
+4j) ms-1 and the velocity of particle B
before the collision was 7i m s-1. Given
the velocity of particle B after the collision
was (6i+ 4j) ms-1, whatwas the velocity
of particle A after the collision?
Answers
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Explanation:
Given that,
Mass m
1
=2kg
Mass m
2
=3kg
Initial velocity u
1
=5
i
^
m/s
Initial velocity u
2
=−2
i
^
m/s
Final velocity v
1
=−1.6m/s
Now, by conservation of momentum
m
1
u
1
+m
2
u
2
=m
1
v
1
+m
2
v
2
2×5
i
^
+3×−2
i
^
=2×−1.6
i
^
+3v
2
v
2
=2.4
i
^
m/s
Velocity of center of mass after collision
=
m
1
+m
2
m
1
×v
1
+m
2
×v
2
=
5
2×−1.6
i
^
+3×2.4
i
^
=
5
4
i
^
Coefficient of restitution is
v
1
−v
2
=−e×(u
1
−u
2
)
−e=
u
1
−u
2
v
1
−v
2
e=
5+2
1.6+2.4
e=
7
4
e=0.57
Hence, the velocity after collision is
5
4
i
^
and coefficient of restitution is 0.57
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