Question:
A body of mass $M$ hits normally a rigid wall with velocity $V$ and bounces back with the same velocity. The impulse experienced by the body is
A body of mass $M$ hits normally a rigid wall with velocity $V$ and bounces back with the same velocity. The impulse experienced by the body is
Updated On: Jul 28, 2024
- MV
- 1.5 MV
- 2 MV
- zero
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The Correct Option is C
Solution and Explanation
A Body of mass $=m$
velocity $=V(\hat{i})$
Knock out (Bounce velocity) $=V(-\hat{i})$
Change in momentum $=m(v)-(m)(-v)$
$=2 M V$
Rate of change of momentum is Force
Thus $\frac{d \vec{p}}{d t}=$ force.
force $\times$ time $=$ Impulse
Thus Impulse Experience by body $=2 m v .$
velocity $=V(\hat{i})$
Knock out (Bounce velocity) $=V(-\hat{i})$
Change in momentum $=m(v)-(m)(-v)$
$=2 M V$
Rate of change of momentum is Force
Thus $\frac{d \vec{p}}{d t}=$ force.
force $\times$ time $=$ Impulse
Thus Impulse Experience by body $=2 m v .$
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Concepts Used:
Newtons Laws of Motion
Newton’s First Law of Motion:
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion:
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:
Newton’s Third Law of Motion:
Newton’s 3rd law states that there is an equal and opposite reaction for every action.