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Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.
- a)Same as that of B [2012]
- b)Opposite to that of B
- c)θ = tan–1 (1/2) to the x-axis
- d)θ = tan–1 (–1/2) to the x-axis
Correct answer is option 'C'. Can you explain this answer?
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Two spheres A and B of masses m1 and m2 respectively collide. A is at ...
Perpendicular to the original direction of B
d)Cannot be determined
Answer:
c) Perpendicular to the original direction of B
Explanation:
We can solve this problem using conservation of momentum and conservation of kinetic energy.
Before the collision, the total momentum of the system is:
p = m2v
Since sphere A is at rest, its momentum is zero.
The total kinetic energy of the system before the collision is:
K = (1/2)m2v^2
After the collision, the spheres move in different directions. Let the velocity of sphere A be u and the velocity of sphere B be w. Then, the total momentum of the system after the collision is:
p' = m1u + m2w
Since sphere B moves perpendicular to its original direction, we can write:
w = 2v
Using conservation of momentum, we have:
m2v = m1u + m2(2v)
Simplifying, we get:
u = (m2 - 2m1)v / m1
Now, using conservation of kinetic energy, we have:
(1/2)m1u^2 + (1/2)m2w^2 = (1/2)m2v^2
Substituting the values of u and w, we get:
(m2 - 2m1)v^2 = m1u^2
Simplifying, we get:
u = sqrt((m2 - 2m1)/m1) v
Since m1 and m2 are positive, (m2 - 2m1)/m1 is negative. Therefore, u is imaginary, which means that sphere A moves in a direction perpendicular to the original direction of sphere B. Hence, the answer is option c) Perpendicular to the original direction of B.
d)Cannot be determined
Answer:
c) Perpendicular to the original direction of B
Explanation:
We can solve this problem using conservation of momentum and conservation of kinetic energy.
Before the collision, the total momentum of the system is:
p = m2v
Since sphere A is at rest, its momentum is zero.
The total kinetic energy of the system before the collision is:
K = (1/2)m2v^2
After the collision, the spheres move in different directions. Let the velocity of sphere A be u and the velocity of sphere B be w. Then, the total momentum of the system after the collision is:
p' = m1u + m2w
Since sphere B moves perpendicular to its original direction, we can write:
w = 2v
Using conservation of momentum, we have:
m2v = m1u + m2(2v)
Simplifying, we get:
u = (m2 - 2m1)v / m1
Now, using conservation of kinetic energy, we have:
(1/2)m1u^2 + (1/2)m2w^2 = (1/2)m2v^2
Substituting the values of u and w, we get:
(m2 - 2m1)v^2 = m1u^2
Simplifying, we get:
u = sqrt((m2 - 2m1)/m1) v
Since m1 and m2 are positive, (m2 - 2m1)/m1 is negative. Therefore, u is imaginary, which means that sphere A moves in a direction perpendicular to the original direction of sphere B. Hence, the answer is option c) Perpendicular to the original direction of B.
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Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v/2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.a)Same as that of B [2012]b)Opposite to that of Bc)θ = tan–1 (1/2) to the x-axisd)θ = tan–1 (–1/2) to the x-axisCorrect answer is option 'C'. Can you explain this answer?
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