# A block of mass 2 kg moving at 2 m/s collides head-on with another block of equal mass kept at...

## Question:

A block of mass 2 kg moving at 2 m/s collides head-on with another block of equal mass kept at rest.

(A) Find the maximum possible loss in kinetic energy due to collision.

(B) If the actual loss in kinetic energy is half of this maximum value, find the value of e?

## Kinetic energy loss:

The total energy is a conserved quantity. It remains constant throughout the system. The total energy is a total of all the energies e.g. Kinetic energy, Potential energy, Thermal energy, etc. It transforms from one form to another. But in the kinetic energy, it remains throughout some cases like in the case of the elastic collision the conditions where kinetic energy is conserved. In some cases, the kinetic energy goes in loss due to thermal energy emission in the system. The total energy is conserved in this case but the kinetic energy is not.

Given:

• The masses of the blocks {eq}m_1=m_2=2 \ \text{kg} {/eq}
• The velocity of the first block {eq}u_1=2 \ \text{m/s} {/eq}
• The second block is initially at rest.

(a)

The maximum loss of kinetic energy is possible if both the objects move together after the collision.

From the momentum conservation:

{eq}mu_1+0=mv+mv {/eq}

Thus the velocity after the collision is:

{eq}\displaystyle{v=\frac{u}{2}=1 \ \text{m/s}} {/eq}

Now the kinetic energy loss in the collision is computed as:

{eq}\begin{align} \Delta \text{K}&=\frac{1}{2}(m+m)v^2-\frac{1}{2}mu_1^2\\ &=\frac{1}{2}(2+2)(1)^2-\frac{1}{2}(2)(2)^2\\ &=(2 \ \text{J})-(4 \ \text{J})\\ &=\color{blue}{-2 \ \text{J}} \end{align} {/eq}

50% of the kinetic energy is lost.

(b)

• The final kinetic energy is one half to the maximum kinetic energy loss means the final kinetic energy is three-fourth of the initial kinetic energy.

From the momentum conservation:

{eq}\Rightarrow mu_1+0=mv_1+mv_2 {/eq}

{eq}\Rightarrow v_1+v_2=u_1............(1) {/eq}

Now recall the formula for the coefficient of restitution:

{eq}\Rightarrow \displaystyle{e=\frac{v_2-v_1}{u_1-0}} {/eq}

{eq}\Rightarrow v_2-v_1=eu_1...........(2) {/eq}

From equations (1) & (2):

{eq}\begin{align} \Rightarrow v_2&=\frac{(1+e)}{2}u_1\\ \Rightarrow v_1&=\frac{(1-e)}{2}u_1\\ \end{align} {/eq}

Now the final kinetic energy is three fourth of the original kinetic energy thus:

{eq}\displaystyle{\Rightarrow \frac{3}{4}\left(\frac{1}{2}mu_1^2\right)=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2} {/eq}

{eq}\displaystyle{\Rightarrow \frac{3}{4}u_1^2=\left(\frac{1-e}{2}u_1\right)^2+\left(\frac{1+e}{2}u_1\right)^2} {/eq}

{eq}\displaystyle{\Rightarrow (1-e)^2+(1+e)^2=3} {/eq}

{eq}\displaystyle{\Rightarrow 2(1+e^2)=3} {/eq}

Thus the value of coefficient of restitution is:

{eq}\begin{align} \Rightarrow e&=\frac{2}{\sqrt{2}}\\ &=\color{blue}{0.707} \end{align} {/eq} 