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Two equal masses m1 and m2 moving along the same straight line with velocities + 3 m/s and – 5m/s respectively, collide elastically.
Their velocities after the collision will be respectively
- a)–3 m/s & +5 m/s
- b)+ 4 m/s for both
- c)–4 m/s & +4 m/s
- d)–5m/s & +3 m/s
Correct answer is option 'D'. Can you explain this answer?
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Two equal masses m1 and m2 moving along the same straight line with v...
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity. Therefore, the total momentum before the collision is equal to m1*v1 + m2*v2, where m1 and m2 are the masses of the objects, and v1 and v2 are their velocities.
Since the masses are equal (m1 = m2), we can simplify the total momentum to 2m*v1 + 2m*v2 = 2m(v1 + v2).
After the collision, the masses will still be equal and their velocities will be denoted as v'1 and v'2. Therefore, the total momentum after the collision is equal to 2m(v'1 + v'2).
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we have:
2m(v1 + v2) = 2m(v'1 + v'2)
Since the masses are equal (m1 = m2), we can divide both sides of the equation by 2m to get:
v1 + v2 = v'1 + v'2
Now, we can substitute the given values into the equation. v1 = 3 m/s and v2 = -5 m/s, so we have:
3 m/s + (-5 m/s) = v'1 + v'2
-2 m/s = v'1 + v'2
Since the masses are equal, the velocities after the collision must also be equal. Therefore, we can denote them as v'1 = v'2 = v'.
Substituting this into the equation, we have:
-2 m/s = 2v'
Dividing both sides of the equation by 2, we get:
v' = -1 m/s
Therefore, the velocities after the collision will be -1 m/s for both masses.
The momentum of an object is given by the product of its mass and velocity. Therefore, the total momentum before the collision is equal to m1*v1 + m2*v2, where m1 and m2 are the masses of the objects, and v1 and v2 are their velocities.
Since the masses are equal (m1 = m2), we can simplify the total momentum to 2m*v1 + 2m*v2 = 2m(v1 + v2).
After the collision, the masses will still be equal and their velocities will be denoted as v'1 and v'2. Therefore, the total momentum after the collision is equal to 2m(v'1 + v'2).
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we have:
2m(v1 + v2) = 2m(v'1 + v'2)
Since the masses are equal (m1 = m2), we can divide both sides of the equation by 2m to get:
v1 + v2 = v'1 + v'2
Now, we can substitute the given values into the equation. v1 = 3 m/s and v2 = -5 m/s, so we have:
3 m/s + (-5 m/s) = v'1 + v'2
-2 m/s = v'1 + v'2
Since the masses are equal, the velocities after the collision must also be equal. Therefore, we can denote them as v'1 = v'2 = v'.
Substituting this into the equation, we have:
-2 m/s = 2v'
Dividing both sides of the equation by 2, we get:
v' = -1 m/s
Therefore, the velocities after the collision will be -1 m/s for both masses.
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Two equal masses m1 and m2 moving along the same straight line with v...
In elastic collision, the velocities get inter changed if the colliding objects have equal masses.
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Two equal masses m1 and m2 moving along the same straight line with velocities + 3 m/s and – 5m/s respectively, collide elastically.Their velocities after the collision will be respectivelya)–3 m/s & +5 m/sb)+ 4 m/s for bothc)–4 m/s & +4 m/sd)–5m/s & +3 m/sCorrect answer is option 'D'. Can you explain this answer?
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