Concept explainers
Block A is released from rest and slides down the frictionless surface of B until it hits a bumper on the right end of B. Block A has a mass of 10 kg and object B has a mass of 30 kg and B can roll freely on the ground. Determine the velocities of A and B immediately after impact when (a)
(a)
The velocities of A and B just after the impact if
Answer to Problem 13.182P
velocities of A and B just after the impact if :
Explanation of Solution
Given information:
Mass of A is
Mass of B is
Conclusion:
The total linear momentum of two particles is conserved. Therefore:
The co-efficient of restitution is defined as:
The principle of conservation of energy is defined as.
When a particle moves under the action of conservation of forces. the sum of kinetic energy and potential energy of that particle remains constant.
Calculation:
Let
Apply conservation of linear momentum while block is sliding down.
Therefore.
Apply conservation of linear momentum at impact.
Therefore.
Assume.
According to conservation of energy.
The initial potential energy of A is equal to:
For B its zero.
Initial kinetic energy is zero.
And just before the impact.
Therefore, according to:
Substitute.
Therefore.
Therefore. the velocities just before the impact.
Apply co-efficient of restitution equation.
Substitute.
Therefore:
But we know that:
As a result of it:
Conclusion:
The velocities of A and B just after the impact. if
(b)
The velocities of A and B just after the impact if
Answer to Problem 13.182P
Explanation of Solution
Given information:
Mass of A is
Mass of B is
Conclusion:
According to sub part ‘a’ we have found.
Where.
Calculation:
Find the exact value of
For that.
Then.
But according to co-efficient of restitution equation.
Therefore.
Then.
Conclusion:
The velocities of A and B just after the impact. if
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Chapter 13 Solutions
Vector Mechanics For Engineers
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