A metal ball of mass 2kg moving with a velocity of 36km / h has an head-on collision with a stationary ball of mass 3kg . If after the collision, the two balls move together, the loss in kinetic energy due to collision is

The correct option is 60 J.

Step-by-step Explanation:

First, let's identify the given values from the problem statement:
- Mass of the moving ball ($m_1$) = 2 kg
- Initial velocity of the moving ball ($u_1$) = 36 km/h
- Mass of the stationary ball ($m_2$) = 3 kg
- Initial velocity of the stationary ball ($u_2$) = 0 m/s

Step 1: Convert the velocity to SI units (meters per second)
To convert the velocity from km/h to m/s, we multiply by $\frac{5}{18}$:

$u_1=36\times \frac{5}{18}\text{m/s}=10\text{m/s}$

Step 2: Apply the Principle of Conservation of Linear Momentum
Since the two balls stick and move together after the collision, this is a completely inelastic collision. Let their common final velocity be $v$.
According to the conservation of momentum:

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

Substituting the given values:

$(2\times 10)+(3\times 0)=(2+3)v$

$20=5v$

$v=4\text{m/s}$

Step 3: Calculate the Initial Kinetic Energy ($K_i$)
Before the collision, only the first ball is in motion:

$K_i=\frac{1}{2}{m}_1{u}_1^2$

$K_i=\frac{1}{2}\times 2\times {10}^2=100\text{J}$

Step 4: Calculate the Final Kinetic Energy ($K_f$)
After the collision, both balls move together as a combined mass of $5\text{kg}$ with velocity $4\text{m/s}$:

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

$K_f=\frac{1}{2}\times 5\times 4^2$

$K_f=\frac{1}{2}\times 5\times 16=40\text{J}$

Step 5: Calculate the loss in Kinetic Energy ($\Delta K$)
The loss in kinetic energy is the difference between the initial and final kinetic energies:

$\Delta K=K_i-K_f$

$\Delta K=100\text{J}-40\text{J}=60\text{J}$

Thus, the loss in kinetic energy due to the collision is 60 J.