Question

Two large blocks of wood are sliding toward each other on the frictionless surface of a frozen pond. Block A has mass 4.00 kg and is initially sliding east at 2.00 m/s. Block B has mass 6.00 kg and is initially sliding west at 2.50 m/s. The blocks collide head-on. After the collision block B is sliding east at 0.50 m/s. What is the decrease in the total kinetic energy of the two blocks as a result of the collision? Express your answer with the appropriate units.

Answer 1

Answer:

13.50 J

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, so we can write the following equation:

Δp = 0 ⇒ p₁ = p₀

Assuming that the mass moving to the east has positive speed, we can write:

p₀ (initial momentum) = 4.00 kg*2.00 m/s + 6.00kg*(-2,5 m/s) = -7.00 kg*m/s

p₁ (final momentum) = 4.00kg*vAm/s + 6.00kg*(0.5m/s) = -7.00 kg*m/s

Solving for vA, we have:

vA = -2.50 m/s

Now, we can find the initial and final kinetic energies, as follows:

Ki = 1/2*4.00kg*(2.00)²(m/s)² +1/2*6.00*(-2.50)²(m/s)² = 26.75 J

Kf=  1/2*4.00kg*(-2.50)²(m/s)² +1/2*6.00*(0.50)²(m/s)² = 13.25 J

⇒ ΔK = Kf-Ki = 13.25 J - 26.75 J = -13.50 J

So, the decrease in the total kinetic energy of the two blocks as a result of the collision is equal to 13.50 J.