Posted by David E. on Wednesday, April 25, 2012 at 11:54pm.
The law of conservation of linear momentum
0 = m1•v1 – m2•v2,
v2 = m1•v1/m2 = 1•1.2/3 = 0.4.
The law of conservation of energy
PE = KE1 +KE2 = m1•v1^2/2 + m2•v2^2/2 =
= 1•1.2^2/2 + 3• (0.4)^2/2 =0.96 J.
I got it already
Thanks so much
Related Questions
Physics - A block of mass m = 2.00 kg rests on the left edge of a block of mass ...
physics - A block of mass m = 2.00 kg slides down a 30.0° incline which is 3...
Physics - In Case A the mass of each block is 4.8 kg. In Case B the mass of ...
Physics - A block of mass m1 = 4.1 kg rests on a frictionless horizontal surface...
physics - In the Figure the pulley has negligible mass, and both it and the ...
physics - Block A has a mass of 4.1 kg, and it is on a frictionless surface. A ...
physics - block A (mass 2.04kg) rests on a tabletop. It is connected by a ...
Physics - Two blocks move along a linear path on a nearly frictionless air ...
physics - In the Figure the pulley has negligible mass, and both it and the ...
physics - A 3.50 g bullet is fired horizontally at two blocks at rest on a ...
For Further Reading