Question

A metal ball of mass $2kg$ moving with a velocity of $36km/h$ has a 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

• A. $40J$
• B. $60J$
• C. $100J$
• D. $140J$

Answer 1

The correct option is B

$60J$

Step 1. Given data:

Mass of the stationary ball, $m_2=3kg$

Speed of stationary ball, $v_2=0$

Moving metal ball mass, $m_1=2kg$

Speed of metal ball, $v_1=36km/h$

$=10m/s$

Step 2. Applying law of conservation of momentum:

Energy before collision $=$ Energy after collision

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

$\Rightarrow v=\frac{[m_1{v}_1+m_2{v}_2]}{(m_1+m_2)}$

$\Rightarrow v=\frac{\left[\right(2x10)+(3x0\left)\right]}{\left(2+3\right)}$

$\Rightarrow v=\frac{20}{5}$

$\therefore$ Final velocity, $v=4m/s$.

Step 3. Calculating Initial and final Kinetic energy:

Initial Kinetic Energy, ${(K.E)}_1=\frac{1}{2}\times m_1\times {\left(v_1\right)}^2$

$$\begin{gathered} \Rightarrow {(K.E)}_1=\frac{1}{2}\times 2\times {\left(10\right)}^2 \\ \therefore {(K.E)}_1=100J \end{gathered}$$
Final Kinetic Energy, ${(K.E)}_2=\frac{1}{2}\times \left(m_1+m_2\right)\times {\left(v\right)}^2$

$$\begin{gathered} \Rightarrow {(K.E)}_2=\frac{1}{2}\times \left(2+3\right)\times {\left(4\right)}^2 \\ \Rightarrow {(K.E)}_2=\frac{1}{2}\times 5\times 16 \\ \therefore {(K.E)}_2=40J \end{gathered}$$

Step 4. Calculating loss in kinetic energy:

Loss of energy, $∆(K.E)=$Initial Kinetic Energy - Final Kinetic Energy

$$\begin{gathered} {=(K.E)}_1-{(K.E)}_2 \\ =100-40 \\ =60J \end{gathered}$$

Hence, the correct option is (B).