1. Let the initial momentum of the body be $p$ and mass be $m$.

2. The kinetic energy in terms of momentum is $K = \frac{p^2}{2m}.$

3. The momentum is increased by $50\%$, so the new momentum is $p' = 1.5p.$

4. The new kinetic energy becomes $K' = \frac{(p')^2}{2m} = \frac{(1.5p)^2}{2m} = 2.25\frac{p^2}{2m} = 2.25K.$

5. Hence, the kinetic energy becomes $2.25$ times the original. So the increase in kinetic energy is $K' - K = 2.25K - K = 1.25K.$

6. Therefore, the percentage increase is $\frac{1.25K}{K}\times 100 = 125\%.$

So, the required answer is $125$.

Comparison with stored answer:

• Stored correct answer: $125$
• Derived answer: $125$
• They match.