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Two bodies having same mass 40 kg are moving in opposite directions, one with a velocity of 10 m/s and the other with 7 m/s. If they collide elastically and move as one body, the velocity of the combination is
- a)10 m/s
- b)7 m/s
- c)3 m/s
- d)1.5 m/s
Correct answer is option 'D'. Can you explain this answer?
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Given information:
- Two bodies of same mass (40 kg) are moving in opposite directions.
- One body has a velocity of 10 m/s and the other has a velocity of 7 m/s.
- The collision between the two bodies is elastic.
- The combined velocity of the two bodies after collision needs to be determined.
Solution:
1. Before collision:
- Let the two bodies be A and B.
- Mass of A = Mass of B = 40 kg.
- Velocity of A = 10 m/s.
- Velocity of B = -7 m/s (since it is moving in the opposite direction).
- Total momentum before collision = (Mass of A x Velocity of A) + (Mass of B x Velocity of B)
= (40 kg x 10 m/s) + (40 kg x -7 m/s)
= 400 kg m/s - 280 kg m/s
= 120 kg m/s in the direction of A.
2. During collision:
- Since the collision is elastic, the total momentum after collision should also be 120 kg m/s in the direction of A.
- Let the combined velocity of A and B after collision be V.
- Total mass after collision = Mass of A + Mass of B = 80 kg.
- Therefore, total momentum after collision = (Total mass after collision x V) = 80V.
3. After collision:
- Equating the total momentum before and after collision, we get:
Total momentum before collision = Total momentum after collision
=> 120 kg m/s = 80V
=> V = 1.5 m/s.
Therefore, the combined velocity of the two bodies after collision is 1.5 m/s.
Answer: Option D (1.5 m/s).
- Two bodies of same mass (40 kg) are moving in opposite directions.
- One body has a velocity of 10 m/s and the other has a velocity of 7 m/s.
- The collision between the two bodies is elastic.
- The combined velocity of the two bodies after collision needs to be determined.
Solution:
1. Before collision:
- Let the two bodies be A and B.
- Mass of A = Mass of B = 40 kg.
- Velocity of A = 10 m/s.
- Velocity of B = -7 m/s (since it is moving in the opposite direction).
- Total momentum before collision = (Mass of A x Velocity of A) + (Mass of B x Velocity of B)
= (40 kg x 10 m/s) + (40 kg x -7 m/s)
= 400 kg m/s - 280 kg m/s
= 120 kg m/s in the direction of A.
2. During collision:
- Since the collision is elastic, the total momentum after collision should also be 120 kg m/s in the direction of A.
- Let the combined velocity of A and B after collision be V.
- Total mass after collision = Mass of A + Mass of B = 80 kg.
- Therefore, total momentum after collision = (Total mass after collision x V) = 80V.
3. After collision:
- Equating the total momentum before and after collision, we get:
Total momentum before collision = Total momentum after collision
=> 120 kg m/s = 80V
=> V = 1.5 m/s.
Therefore, the combined velocity of the two bodies after collision is 1.5 m/s.
Answer: Option D (1.5 m/s).
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Two bodies having same mass 40 kg are moving in opposite directions, one with a velocity of 10 m/s and the other with 7 m/s. If they collide elastically and move as one body, the velocity of the combination isa)10 m/sb)7 m/sc)3 m/sd)1.5 m/sCorrect answer is option 'D'. Can you explain this answer?
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