To calculate the percentage change in momentum when the kinetic energy increases by 300%, let’s use the relationship between kinetic energy and momentum.

Step 1: Relationship between kinetic energy and momentum

Kinetic energy (K.E.) is given by:
K.E. = (1/2) m v²

Momentum (p) is given by:
p = m v

We know:
K.E. = p² / (2m)

Step 2: Express the relationship between K.E. and p

Rearranging:
p² = 2m K.E.

Step 3: Analyze the change in K.E.

If kinetic energy increases by 300%, then the new K.E. is:
New K.E. = Initial K.E. × (1 + 300/100)
New K.E. = 4 × Initial K.E.
.
Step 4: Calculate the change in momentum

From the relationship (p² ∝ K.E.):
If K.E. increases by a factor of 4, then:
New p = √(4 × Initial p²)
New p = 2 × Initial p
Thus, momentum increases by 100%.

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For a relation between the linear momenta of the two bodies moving with equal kinetic energies, we use the relation of kinetic energy with momentum.
Step 1: Kinetic energy and momentum relation
The kinetic energy K.E. is related to the momentum p by,
Kinetic energy = p2 / (2m)
Now, rearranging in terms of momentum we get,
p = √(2m × K.E.)
Step 2: Given
For the two bodies
Mass of the first body = m
Mass of the second body = 4m
– Both have the same kinetic energy.

Let the kinetic energy of both bodies be \\( K.E. \\).

Step 3: Calculate the momentum of each body

For the first body:
p₁ = √(2m × K.E.)

For the second body:
p₂ = √(2 × 4m × K.E.) = √(8m × K.E.)

Step 4: Find the ratio of momenta

The ratio of their momenta is:
p₁ : p₂ = √(2m × K.E.) : √(8m × K.E.)
p₁ : p₂ = √2 : √8
p₁ : p₂ = 1 : 2

The ratio of their linear momenta is: 1 : 2

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