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If KE increases 300 percent then momentum increased by how many percent?
Wiki User
∙ 14y ago
Assuming the mass isn't changing, 100%.
Let the initial velocity be v0 and the mass of the object be m. Then KE0= (1/2)mv02 and P=mv. The new kinetic energy is KE1=4*KE0. (A 300% increase means the change is 3 times the original, giving a new value of 3KE0+KE0=4KE0.) So 4*(1/2)mv02=(1/2)mv12. Simplifying by canceling 1/2 and m, we get 4*v02=v12. Taking the square root of each side: 2*v0=v1. So the new velocity is twice the original. Plugging this into the formula for momentum, we see that the new momentum, P1=mv1=m*2*v0. So the momentum is twice as large as it was initially. The change in momentum is P0, so the percent change is P0/P0, or 100%.
Wiki User
∙ 14y ago
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