Question

Block 1, of mass m1 = 9.10 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mass m2 = 13.0 kg , which was initially at rest. The blocks stick together after the collision.

Before collision: m2 After collision:

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

Express your answer numerically in joules.

Answer 1

Mass of block 1 = m₁ = 9.1 kg

Mass of block 2 = m₂ = 13 kg

Speed of block 1 before the collision = V₁ = 27 m/s

Speed of block 2 before the collision = V₂ = 0 m/s (At rest)

Speed of the blocks after the collision = V₃

By conservation of linear momentum,

m₁V₁ + m₂V₂ = (m₁ + m₂)V₃

(9.1)(27) + (13)(0) = (9.1 + 13)V₃

V₃ = 11.117 m/s

Initial kinetic energy of the system = KE₁

KE₁ = m₁V₁²/2 + m₂V₂²/2

KE₁ = (9.1)(27)²/2 + (13)(0)²/2

KE₁ = 3316.95 J

Final kinetic energy of the system = KE₂

KE₂ = (m₁ + m₂)V₃²/2

KE₂ = (9.1 + 13)(11.117)²/2

KE₂ = 1365.64 J

Change in kinetic energy of the system = $\Delta$KE

$\Delta$KE = KE₂ - KE₁

$\Delta$KE = 1365.64 - 3316.95

$\Delta$KE = -1951.31 J

Negative as energy is lost in the collision.

Change in the kinetic energy of the system due to the collision = -1951.31 J