Question

Block 1, of mass m1 = 1.70 kg , moves along a frictionless air track with speed v1 = 29.0 m/s . It collides with block 2, of mass m2 = 59.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1)

Find the magnitude pi of the total initial momentum of the two-block system.

Find vf, the magnitude of the final velocity of the two-block system.

What is the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision?

Answer 1

Mass of block 1 = m₁ = 1.7 kg

Mass of block 2 = m₂ = 59 kg

Velocity of block 1 before the collision = V₁ = 29 m/s

Velocity of block 2 before the collision = V₂ = 0 m/s

Velocity of the blocks after the collision = V_f

Initial momentum of the two block system = P_i

P_i = m₁V₁ + m₂V₂

P_i = (1.7)(29) + (59)(0)

P_i = 49.3 kg.m/s

By conservation of linear momentum,

m₁V₁ + m₂V₂ = (m₁ + m₂)V_f

(1.7)(29) + (59)(0) = (1.7 + 59)V_f

V_f = 0.812 m/s

Initial kinetic energy = KE_i

KE_i = m₁V₁²/2 + m₂V₂²/2

KE_i = (1.7)(29)²/2 + (59)(0)²/2

KE_i = 714.85 J

Final kinetic energy = KE_f

KE_f = (m₁ + m₂)V_f²/2

KE_f = (1.7 + 59)(0.812)²/2

KE_f = 20.01 J

Change in kinetic energy = KE

KE = KE_f - KE_i

KE = 20.01 - 714.85

KE = -694.84 J

Total initial momentum of the two block system = 49.3 kg.m/s

Final velocity of the two block system = 0.812 m/s

Change in kinetic energy due to the collision = - 694.84 J