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A ball impinges directly on another ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by the impact, the value of coefficient of restitution is
- a)2√3
- b)√3
- c)√2
- d)2
Correct answer is option 'C'. Can you explain this answer?
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The coefficient of restitution (e) is defined as the ratio of relative velocity of separation to the relative velocity of approach between two colliding objects.
Let us consider the two balls before and after the impact.
Before the impact:
Let the mass of the first ball be m1 and its initial velocity be u1.
Let the mass of the second ball be m2 and its initial velocity be 0 (as it is at rest).
The total initial kinetic energy of the system is given by:
KEi = 1/2 m1u1^2 + 1/2 m2(0)^2 = 1/2 m1u1^2
After the impact:
Let the velocity of the first ball after the impact be v1 and the velocity of the second ball be v2.
As the first ball is brought to rest by the impact, we have v1 = 0.
The total final kinetic energy of the system is given by:
KEf = 1/2 m1(0)^2 + 1/2 m2v2^2 = 1/2 m2v2^2
From the law of conservation of momentum, we have:
m1u1 = m1v1 + m2v2
Since v1 = 0, we have:
m1u1 = m2v2
Solving for v2, we get:
v2 = (m1/m2)u1
The relative velocity of approach between the two balls is u1 - 0 = u1.
The relative velocity of separation between the two balls is 0 - v2 = -v2.
Therefore, the coefficient of restitution is given by:
e = -(-v2/u1) = v2/u1 = (m1/m2)u1/u1 = m1/m2
As half of the kinetic energy is lost by the impact, we have:
KEf = 1/2 KEi
1/2 m2v2^2 = 1/2 (1/2 m1u1^2)
m2v2^2 = 1/4 m1u1^2
Substituting the value of v2, we get:
m2((m1/m2)u1)^2 = 1/4 m1u1^2
m1^2u1^2 = 1/4 m1m2u1^2
4m2 = m1
m2/m1 = 1/4
Therefore, the coefficient of restitution is:
e = m1/m2 = 4
Hence, the value of coefficient of restitution is 4.
Let us consider the two balls before and after the impact.
Before the impact:
Let the mass of the first ball be m1 and its initial velocity be u1.
Let the mass of the second ball be m2 and its initial velocity be 0 (as it is at rest).
The total initial kinetic energy of the system is given by:
KEi = 1/2 m1u1^2 + 1/2 m2(0)^2 = 1/2 m1u1^2
After the impact:
Let the velocity of the first ball after the impact be v1 and the velocity of the second ball be v2.
As the first ball is brought to rest by the impact, we have v1 = 0.
The total final kinetic energy of the system is given by:
KEf = 1/2 m1(0)^2 + 1/2 m2v2^2 = 1/2 m2v2^2
From the law of conservation of momentum, we have:
m1u1 = m1v1 + m2v2
Since v1 = 0, we have:
m1u1 = m2v2
Solving for v2, we get:
v2 = (m1/m2)u1
The relative velocity of approach between the two balls is u1 - 0 = u1.
The relative velocity of separation between the two balls is 0 - v2 = -v2.
Therefore, the coefficient of restitution is given by:
e = -(-v2/u1) = v2/u1 = (m1/m2)u1/u1 = m1/m2
As half of the kinetic energy is lost by the impact, we have:
KEf = 1/2 KEi
1/2 m2v2^2 = 1/2 (1/2 m1u1^2)
m2v2^2 = 1/4 m1u1^2
Substituting the value of v2, we get:
m2((m1/m2)u1)^2 = 1/4 m1u1^2
m1^2u1^2 = 1/4 m1m2u1^2
4m2 = m1
m2/m1 = 1/4
Therefore, the coefficient of restitution is:
e = m1/m2 = 4
Hence, the value of coefficient of restitution is 4.
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A ball impinges directly on another ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by the impact, the value of coefficient of restitution isa)2√3b)√3c)√2d)2Correct answer is option 'C'. Can you explain this answer?
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