A 1kg ball moving at 12ms collides headon with a 2kg ball moving in the opposite
direction at 24 m/s. If the coefficient of restitution is 2/3, then the energy lost in the
collision is
1) 60J
2) 120J
3) 240J
4) 480 J
Answers
Answer:
Given : Coefficient of restitution e=
3
2
Velocity of mass 1 kg after the collision is given by
V
1
=
m
1
+m
2
(m
1
−em
2
)u
1
+(1+e)m
2
u
2
Or V
1
=
1+2
(1−
3
2
×2)×12+(1+
3
2
)×2×(−24)
=−28 m/s
Velocity of mass 2 kg after the collision is given by
V
2
=
m
1
+m
2
(m
−
em
1
)u
2
+(1+e)m
1
u
1
Or V
2
=
1+2
(2−
3
2
×1)×(−24)+(1+
3
2
)×1×(12)
=−4 m/s
Explanation:
Answer: -28 m/s and -4 m/s
Explanation:
From law of conservation of momentum, when there is no net external force is acting on the system, the momentum is conserved.
This means, Momentum before the collision = Momentum after the collision
⇒m₁u₁+m₂u₂=m₁v₁+m₂v₂
where, m₁ and m₂ are the masses of the colliding objects, u₁ and u₂ are their initial velocities and v₁ and v₂ are their final velocities.
It is given that:
The coefficient of restitution,
It is given that, coefficient of restitution, e = 2/3
⇒2/3 (12-(-24))=v₂-v₁
⇒24 =v₂-v₁
⇒v₂=24+v₁ ... (1)
From law of conservation of momentum,
1 kg × 12 m/s + 2 kg × -24 m/s = 1 kg v₁+ 2 kg v₂
⇒12-48 =v₁+2v₂ ⇒ v₁+2v₂= -36... (2)
Substitute (1) into (2)
⇒v₁+2×(24+v₁)=-36
⇒3v₁+48=-36
⇒3v₁=-84 ⇒v₁=-28 m/s
Substitute this value in (1)
v₂= 24+(-28) =-4 m/s
Hence, the velocity of ball of mass 1 kg after collision is -28 m/s and ball with mass 2 kg, - 4m/s. This means, both the balls move in same direction after collision.