A ball collide directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by impact, the value of cofficient of restitution is nearly
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Answer:
Let u
1
and v
1
be the initial and final velocities of ball 1 and u
2
and v
2
be the similar quantities for ball 2. Here,m u
2
=0 and v
1
=0.
∴ initial KE,K
i
=
2
1
mu
1
2
+
2
1
mu
2
2
=
2
1
mu
1
2
and final KE,K
f
=
2
1
mv
1
2
+
2
1
mv
2
2
=
2
1
mv
2
2
Loss of KE,△K
i
−K
f
=
2
1
mu
1
2
−
2
1
mv
2
2
According to question,
2
1
(
2
1
mv
1
2
)=
2
1
mu
1
2
−
2
1
mv
2
2
(∵ half of its KE is lost by impact)
or u
1
2
=2v
2
2
or v
2
=
2
u
1
∴ Coefficient of restitution,
e=
∣
∣
∣
∣
∣
u
1
−u
2
v
2
−v
1
∣
∣
∣
∣
∣
=
u
1
v
2
=
2
1
.
Explanation:
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