Momentum is the quantity of motion possessed by a moving object, defined as the product of its mass and velocity. It is a vector quantity, having both magnitude and direction. Momentum plays a crucial role in collision analysis and follows the law of conservation in isolated systems.
Chapter 5 of Class 11 Physics explains work energy and power. It discusses how work is calculated energy conservation and transformations and different forms of energy like kinetic and potential. The chapter also defines power as the rate at which work is done.
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We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula.
p = mv, whereas K.E. = (1/2) mv².
Express the Kinetic Energy as a Function of Momentum:
Let’s convert Kinetic energy as a function of momentum; since momentum was multiplied by constant so was velocity by same amount:
K.E. = (1/2) mv = (1/2)(p/v).
From p = mv, we have:
v = p/m
Substituting v in the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Calculate the initial and final momentum
Let the initial momentum be p. If momentum increases by 20%, the new momentum (p’) is:
p’ = p + 0.2p = 1.2p
Step 4: Calculate initial and final kinetic energy
Initial K.E. (K.E.₁):
K.E.₁ = (1/2) (p²/m)
Final K.E. (K.E.₂):
K.E.₂ = (1/2) (p’²/m)
K.E.₂ = (1/2) [(1.2p)²/m]
K.E.₂ = (1/2) [1.44p²/m]
Step 5: Find the increase in kinetic energy
Now, we calculate the increase in kinetic energy:
Increase in K.E. = K.E.₂ – K.E.₁
Increase in K.E. = (1/2) [1.44p²/m] – (1/2) [p²/m]
Increase in K.E. = (1/2m) [1.44p² – p²]
Increase in K.E. = (1/2m) [0.44p²]
Percentage increase in K.E. is given by:
Percentage increase = (Increase in K.E. / K.E.₁) × 100
Percentage increase = [(0.44p²/2m) / (p²/2m)] × 100
Percentage increase = 0.44 × 100
Percentage increase = 44%
Final Answer:
If the momentum is increased by 20%, then it increases the Kinetic energy with 44%.