Block A, with a mass of 2.0 kg, moves along the x axis with a velocity of 5.0 m/s in the positive x direction.It suffers an elastic collision with block B, initially at rest, and the blocks leave the collision along the x axis.If B is much more massive than A, the velocity of A after the collision is:
A)0 m/s
B)+5.0 m/s
C)-5.0 m/s
D)+10 m/s
E)-10 m/s

Question

Quizplus · Accepted Answer

In an elastic collision, the total momentum and total kinetic energy of the system are conserved. Since block B is initially at rest, the total momentum before the collision is simply the momentum of block A (p = mv). After the collision, the two blocks move together, so the total momentum is (M+m)V, where M is the mass of block B and V is the velocity of the two blocks moving together.

Therefore, we have:

total momentum before = total momentum after
mv = (M+m)V

Solving for V, we get:

V = (m/m+M)v

Since M is much greater than m (according to the question), V is very close to zero. Therefore, the velocity of block A after the collision is approximately:

Velocity of A = V = (m/m+M)v ≈ 0 - 5.0 = -5.0 m/s

So, the best choice is C.

Block A, with a Mass of 2