Calculate the energy loss in a perfectly inelastic collision if the mass of the object is 40kg with velocity 4m/s hits the object of mass 60kg with velocity 2m/s.

The correct answer is 48J.

To find the energy loss during a perfectly inelastic collision, we first calculate the total initial kinetic energy of the system, determine the final common velocity of the combined mass after the collision using the principle of conservation of linear momentum, calculate the final kinetic energy, and then find the difference between the initial and final kinetic energy.

Step 1: Identify the given values
Mass of the first object ($m_1$) = 40 kg
Initial velocity of the first object ($u_1$) = 4 m/s
Mass of the second object ($m_2$) = 60 kg
Initial velocity of the second object ($u_2$) = 2 m/s

Step 2: Calculate the total initial kinetic energy ($K_i$)
The total initial kinetic energy is the sum of the kinetic energies of both objects before the collision:

$K_i=\frac{1}{2}{m}_1{u_1}^2+\frac{1}{2}{m}_2{u_2}^2$

Substituting the given values:

$K_i=\frac{1}{2}\left(40\right){\left(4\right)}^2+\frac{1}{2}\left(60\right){\left(2\right)}^2$

$K_i=20\left(16\right)+30\left(4\right)$

$K_i=320+120=440\text{J}$

Step 3: Calculate the final velocity ($v$) after the collision
In a perfectly inelastic collision, the two objects stick together and move with a common final velocity ($v$). According to the law of conservation of linear momentum:

$m_1{u}_1+m_2{u}_2=(m_1+m_2)v$

Substitute the values to find $v$:

$\left(40\right)\left(4\right)+\left(60\right)\left(2\right)=(40+60)v$

$160+120=100v$

$280=100v$

$v=2.8\text{m/s}$

Step 4: Calculate the total final kinetic energy ($K_f$)
The final kinetic energy of the combined mass is:

$K_f=\frac{1}{2}(m_1+m_2)v^2$

$K_f=\frac{1}{2}\left(100\right){\left(2.8\right)}^2$

$K_f=50\left(7.84\right)$

$K_f=392\text{J}$

Step 5: Calculate the energy loss ($\Delta K$)
The energy loss during the collision is the difference between the initial kinetic energy and the final kinetic energy:

$\Delta K=K_i-K_f$

$\Delta K=440-392=48\text{J}$

Thus, the energy loss in the perfectly inelastic collision is 48J.