The original momentum is mv
the final momentum is m/2 * v1' + m/2*v2'
set the equal.
2v=v1' + v2'
a) v1'=2v-v2'
Now, energy.
1/2 m v^2+16.7=1/4 m v1'^2 + 1/4 m v2'^2
b) 2 v^2+ 16.7=v1'^2 + v2'^2
put the expression a) into b). multiply it out, gather terms, and solve the quadratic.
A block of mass 7.33 kg in outerspace is moving at 1.72 m/s with no external forces acting on it. After an explosion, the block is split into two parts both having mass equal to half the mass of the original block. The explosion supplies the two masses with an additional 16.7 J of kinetic energy. Neither mass leaves the line of original motion. Calculate the magnitude of the velocity of the mass that is moving at a greater velocity.
4 answers
bob- thanks, this helped alot. Just one thing though- when you're elminating the mass and the fractions in the Kinetic energy equation, you forgot about the normal number(energy added). It should be 4 * (16.7) / m after doing the elimination. The rest was perfect though, thanks!
wow thanks
I don't understand why 16.7 J of kinetic energy has to be added to the initial kinetic energy of the mass instead of the final kinetic energy of the two split masses. Anyone have a good explanation ?