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A 1 kg ball moving at 12 m/s collides head-on with a 2 kg ball moving in the opposite direction at 24 m/s. If the coefficient of restitution is 2/3 then the final speeds of the two balls are respectively  
  • a)
    28 m/s, 4 m/s 
  • b)
    228/9  m/s, 154/9 m/s  
  • c)
    28 m/s,  154/9 m/s 
  • d)
    228/9 m/s, 4 m/s  
Correct answer is option 'A'. Can you explain this answer?
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A 1 kg ball moving at 12 m/s collides head-on with a 2 kg ball moving ...
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A 1 kg ball moving at 12 m/s collides head-on with a 2 kg ball moving ...
Given:
Mass of ball 1 (m1) = 1 kg
Initial velocity of ball 1 (u1) = 12 m/s
Mass of ball 2 (m2) = 2 kg
Initial velocity of ball 2 (u2) = -24 m/s (opposite direction)

Coefficient of restitution (e) = 2/3

To find:
Final speeds of the two balls after collision.

Solution:
1. Conservation of momentum:
Before the collision, the total momentum of the system is given by:
Initial momentum = m1 * u1 + m2 * u2 (1)

After the collision, the total momentum of the system is given by:
Final momentum = m1 * v1 + m2 * v2 (2)

According to the principle of conservation of momentum, the initial momentum and final momentum of the system should be equal.

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2 (3)

2. Coefficient of restitution (e):
The coefficient of restitution (e) is defined as the ratio of relative velocity after collision to the relative velocity before collision.

e = (v2 - v1) / (u1 - u2) (4)

3. Solving equations (3) and (4) simultaneously:
Substituting the values of m1, u1, m2, and u2 into equation (3):
1 * 12 + 2 * (-24) = 1 * v1 + 2 * v2
12 - 48 = v1 + 2v2
-36 = v1 + 2v2 (5)

Substituting the values of e, u1, and u2 into equation (4):
2/3 = (v2 - v1) / (12 - (-24))
2/3 = (v2 - v1) / 36
36 * 2/3 = v2 - v1
24 = v2 - v1 (6)

4. Solving equations (5) and (6) simultaneously:
We have two equations (5) and (6) with two variables v1 and v2. Solving these equations simultaneously will give us the values of v1 and v2.

From equation (6), we have:
v2 - v1 = 24
v2 = v1 + 24 (7)

Substituting the value of v2 from equation (7) into equation (5):
-36 = v1 + 2(v1 + 24)
-36 = v1 + 2v1 + 48
-36 - 48 = 3v1
-84 = 3v1
v1 = -84 / 3
v1 = -28 m/s

Substituting the value of v1 into equation (7):
v2 = -28 + 24
v2 = -4 m/s

5. Final speeds:
The final speeds of the two balls are:
v1 = -28 m/s (ball 1)
v2 = -4 m/s (ball 2)

Since the magnitude of velocity represents speed, we take the absolute values of v1 and v2:
Final speed of ball 1 = |v1| = |-28| = 28 m/s
Final speed of ball 2 =
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A 1 kg ball moving at 12 m/s collides head-on with a 2 kg ball moving in the opposite direction at 24 m/s. If the coefficient of restitution is 2/3 then the final speeds of the two balls are respectively a)28 m/s, 4 m/sb)228/9m/s, 154/9m/s c)28 m/s, 154/9m/sd)228/9m/s, 4 m/s Correct answer is option 'A'. Can you explain this answer?
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