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.

Answer 2

The decrease in the total kinetic energy of the given blocks after the collision is 13.50 J.

From the conservation of motion,  

$p_i = p_f$  

 

$p_i$ - total initial momentum

$p_f$ - total final momentum

So,

$4.00\rm \ kg \times 2.00 \rm \ m/s + 6.00 \rm \ kg \ \times (-2,5 \rm \ m/s) = 4.00 \times \rm V_A\ kg \times \rm \ m/s + 6.00 \rm \ kg \rm \ (0.5 \ m/s)$  

Solve for $V_A$, we get,

$V_A$ = -2.50 m/s  

The decrease in total kinetic energy,

$\Delta K = K_f - K_i$

Where,

$K_i$ - initial kinetic energy

$K_f$- final kinetic energy

 

$Ki = \dfrac 12\times 4.00 \rm \ kg\times (2.00)^2(m/s)^2 +\dfrac 12 \times 6.00 \times (-2.50)^2(m/s)^2 =\bold {26.75\rm \ J}$  

$K_f = \dfrac 12\times 4.00 \rm \ kg\times (2.50)^2(m/s)^2 +\dfrac 12 \times 6.00 \times (-0.50)^2(m/s)^2 =\bold {13.57 \rm \ J}$

Put the values in the formula,  

$\Delta K = 13.25\rm \ J - 26.75 \rm \ J$

$\Delta K = -13.50\rm \ J$  

Therefore, the decrease in the total kinetic energy of the given blocks after the collision is 13.50 J.

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