Question

Block 1, of mass m1 = 1.30 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mass m2 = 23.0 kg , which was initially at rest. The blocks stick together after the collision
Part A. Find the magnitude pi of the total initial momentum of the two-block system.
Part B. Find vf, the magnitude of the final velocity of the two-block system.
Part C. What is the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?

Answer 1

(A) The total initial momentum of the system is

(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s

(B) Momentum is conserved, so that the total momentum of the system after the collision is

35.1 kg•m/s = (1.30 kg + 23.0 kg) v

where v is the speed of the combined blocks. Solving for v gives

v = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s

(C) The kinetic energy of the system after the collision is

1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J

and before the collision, it is

1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J

so that the change in kinetic energy is

∆K = 25.4 J - 474 J ≈ -449 J