Kinetic Energy of 2 masses after a collision?

Question

Two masses are 2.9 kg each. The first mass is moving with a velocity v1 immediately before colliding with the second mass, which is suspended by a string of length 1m. The 2 masses are stuck together as a result of the collision. The compound system then swings to the right and rises to the horizontal level. The acceleration of gravity is 9.8m/s/s.

I drew a diagram for this problem, which can be found at http://img363.imageshack.us/img363/6962/physicssv4.jpg

Find the kinetic energy of the compound system immediately after the collision. Answer in units of J.

Any help is greatly appreciated! Thanks!

Also, what is v1, the speed of m1 immediately before the collision? Answer in units of m/s. Thanks!!!!

Answer 1

Since the masses stick together, the collision is not elastic but momentum must still be conserved so:
M = 2.9 kg ; h = 1 m
v (1)= velocity before collision
V = velocity immediately after the collision
Potential energy of the double mass at the height = 2Mgh

Mv(1) + M ( 0 ) = (2M) V; V= v (1) / 2

KE of the compound system = (1/2) (2M) {v(1)}^2}/4 = (1/4)M {v(1)}^2

Answer 2

i believe you try this utilising momentum. p=mv. My guess is that momentum is concept to be conserved for that reason for the reason that they supply you barely sufficient information to discern that out. (m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f) v2f = [(m1 * v1i) + (m2 * v2i) - (m1 * v1f)] / m2 m1 = 8 m2 = 2 v1i = 4 v2i = -3 v1f = 2 v2f = [(m1 * v1i) + (m2 * v2i) - (m1 * v1f)] / m2 v2f = [(32) + (-6) - (sixteen)] / 2 = 5 m/s complete KE = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 complete KE = 0.5*8*2^2 + 0.5*2*5^2 = sixteen + 25 complete KE = 40-one Joules