A 1kg ball moving at 12m/s collides head on with 2kg ball moving in opposite direction at 24m/s.The velocity of each ball after impact, if the coefficient of restitution is 2/3is
Answers
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:
m₁ = 1 kg
m₂ = 2 kg
u₁ = 12 m/s
u₂ = -24 m/s
The coefficient of restitution,
⇒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.
m1 u1 + m2 u2 = m1 v1 + m2 v2 …............. [1]
where:
m1 = 1 kg
m2 = 2 kg
u1 = 12 m/s
u2 = -24 m/s
and the coefficient of restitution, e, is given by:
e = (v2 – v1)/(u1 – u2) …......................... [2]
a) e = 2/3
[1] …............... 1 * 12 + 2 * (-24) = 1 * v1 + 2 * v2
v1 + 2 v2 = -36
v1 = -2 v2 - 36
[2] …................ 2/3 = (v2 – v1)/36
2/3 = (v2 – (-2 v2 – 36))/36
72/3 = 3 v2 + 36
► v2 = -12/3 = -4 m/s
► v1 = -2 v2 – 36 = -28 m/s
b)
[1] …............... m1 u1 + m2 u2 = (m1 + m2) v
1 * 12 + 2 * (-24) = (1 + 2) v
3 v = -36
v = -12 m/s
c) e = 1
[1] …............... 1 * 12 + 2 * (-24) = 1 * v1 + 2 * v2
v1 + 2 v2 = -36
v1 = -2 v2 - 36
[2] …................ 1 = (v2 – v1)/36
1 = (v2 – (-2 v2 – 36))/36
36 = 3 v2 + 36
► v2 = 0
► v1 = -2 v2 – 36 = -36 m/s
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