Could someone please help me figure out this physics linear momentum problem?

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/2 of its original speed. (a) What is the mass of the other body?





I don't understand how you can solve this without the final velocity of the 2nd body..





Thanks.Could someone please help me figure out this physics linear momentum problem?
If we assume that by ';elastic'; collision, they mean ';perfectly elastic';, then we have enough information. In a perfectly elastic collision, kinetic energy is conserved. Kinetic energy is T = (1/2)mv^2. If you cut the speed in half, you cut the kinetic energy to 1/4 its original value. So we know that the kinetic energy of the second body after the collision is 3/4 the kinetic energy of the first body before the collision.





Because momentum is directly proportional to velocity, we know that the momentum of the first body after the collision is 1/2 what it was before the collision. So, after the collision, the second body must have half of the momentum and three-fourths of the kinetic energy of the first body before the collision. Put another way, after the collision the second body has the same momentum as the first body and three times its kinetic energy.





Let M be the mass of the second body and V be its speed after the collision. For algebraic convenience, let m bet the mass of the first body (2.7 kg) and v be its speed before the collision. We know its speed after the collision is v/2. So,





mv/2 = MV





and





3(1/2)m(v/2)^2 = (1/2)MV^2 (because we've figured out that the second body after the collision has three times the kinetic energy of the first after the collision).





Let's work with the second equation. The (1/2)'s cancel out. Squaring the (v/2), we end up with





(3/4)mv^2 = MV^2





From the first one, V = mv/(2M), so





(3/4)mv^2 = M(mv/(2M))^2





3mv^2/4 = m^2v^2/(4M)





The v^2 cancel out, as do the 4's in the denominator and one of the m's, leaving





3 = m/M





M = m/3 = 2.7 kg/3 = 0.9 kg





Check it: We've figured that the mass of the second object is 1/3 that of the first. If they have the same momentum after the collision, then the speed of the second body after the collision must be three times as great as that of the first body after the collision. Let v' be the speed of the first body after the collision, then V = 3v'





T (first body, after the collision) = (1/2)m(v')^2





T (second body after the collision) = (1/2)MV^2 = (1/2)(m/3)(3v')^2 = 3(1/2)m(v')^2 = three times the kinetic energy of the first object after the collision, which we know to be the case. It checks..