Since kinetic energy is given by K.E.=21mv2
Momentum is given by, P=mv
⟹v=P/m
Then, I can write it as,
K.E.=21mv2=21m(p/m)2=2mp2
Given, kinetic energy, K.E.=21mv2=275J
Momentum, P=mv=25.0kgm/s
Dividing both, 2mvmv2=25275
v=2×11=22m/s
Using the relationship which we have derived,
K.E.=2mp2
275=2m252
m=1.14Kg
Comments