a 75-kg swimmer dives horizontally off a 200-kg raft initially at rest. if the diver's speed immediately after leaving the raft is 2 m/s, what is the corresponding raft speed?

A 75-kg swimmer dives horizontally off a 200-kg raft initially at rest. If the diver's speed immediately after leaving the raft is 2 m/s, The corresponding raft speed will be  0.75 m/s . The law of conservation of momentum states that in an isolated system the total momentum of two or more bodies acting upon each other remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.using momentum conservationm1v1 + m2v2 = m1u1 + m2u2 since , diver and raft , both are initially at rest that mean , initial velocity of diver and raft will be zero m1v1 + m2v2 = 075*2 + 200 * v2 = 0v2 = 0.75 m/sTo learn more about conservation of momentum  here :https://brainly.com/question/24989124#SPJ4