A projectile of mass${\mathbf{m}}_{\mathbf{1}}$moving with speed ${\mathit{v}}_{\mathbf{1}}$in the +xdirection strikes a stationary target of mass${\mathbf{m}}_{\mathbf{2}}$head-on. The collision is elastic. Use the Momentum Principle and the Energy Principle to determine the final velocities of the projectile and target, making no approximations concerning the masses. After obtaining your results, see what your equations would predict if${\mathbf{m}}_{\mathbf{1}}\mathbf{\gg }{\mathbf{m}}_{\mathbf{2}}$, or if${\mathbf{m}}_{\mathbf{2}}\mathbf{\gg }{\mathbf{m}}_{\mathbf{1}}$. Verify that these predictions are in agreement with the analysis in this chapter of the Ping-Pong ball hitting the bowling ball, and of the bowling ball hitting the Ping-Pong ball.

The given data is listed below as,

• The mass of the projectile is, $m_1$.
• The initial velocity of the projectile is, .
• The direction in which the projectile is moving is, direction.
• The mass of the stationary target that the projectile strikes is, .
• The initial velocity of the target is, .

The law of conservation of momentum states that the momentum of a system before and after collision with another system remains constant if no external force acts on it.

Let the initial velocity of the projectile be $v_{1i}$and the initial velocity of the target is $v_{2i}$that is 0 as it was at rest and the final velocity of the projectile is $v_{1f}$and the final velocity of the target is $v_{2f}$.

From the law of conservation of momentum, the equation of the momentum of the projectile and the target can be expressed as,

role="math" localid="1657858281462" $(m_1-v_{1i})+(m_2-v_{2i})=(m_1-v_{1f})+(m_2-v_{2f})$

$\mathrm{For}{v}_{2i}=0$

$$\begin{gathered} (m_1-v_{1i})+(m_2-0)=(m_1-v_{1f})+(m_2-v_{2f}) \\ (m_1-v_{1i})=(m_1-v_{1f})+(m_2-v_{2f}) \end{gathered}$$...........(1)

The expression for the velocity of the projectile and the target is as follows,

$v_{1i}+v_{1f}=v_{2i}+v_{2f}$

For $v_{2i}=0$,

$v_{1i}+v_{1f}=v_{2f}$ …(2)

For $v_{1j}+v_{1f}=v_{2f}$in the equation (1),

$$\begin{gathered} (m_1-v_{1i})=(m_1-v_{1f})+(m_2-(v_{1i}+v_{1f}\left)\right) \\ (m_1-v_{1i})=(m_1-v_{1f})+(m_2-v_{1i}+m_2-v_{1f}) \end{gathered}$$

Hence, from the above equation, the value of the final velocity of the projectile becomes,

role="math" localid="1657859475686" $$\begin{gathered} (m_1-v_{1i})=(m_1-v_{1f})+(m_2-v_{1i}+m_2-v_{1f}) \\ {v}_{1f}(m_1+m_1)=m_1{v}_{1f}-m_2{v}_{1f} \\ {v}_{1f}=\frac{v_{1f}(m_1-m_2)}{m_1+m_2} \end{gathered}$$

From the equation (2), the value of the final velocity of the projectile can be expressed as,

$v_{1f}=v_{2f}-v_{1i}$

For $v_{1f}=v_{2f}-v_{1i}$in the equation (1),

$$\begin{gathered} (m_1-v_{1i})=(m_1.(v_{2f}-v_{1f}\left)\right)+(m_2-v_{2f}) \\ (m_1-m_1)=m_1{v}_{2f}-m_2{v}_{1i}+m_2{v}_{2f} \\ {v}_{1f}\left(2m_1\right)=v_{2f}(m_1+m_2) \\ {v}_{1f}=\frac{2m{v}_{1f}}{(m_1+m_2)} \end{gathered}$$

From the gathered value of the final velocity of the projectile, it is evident that the value of $m_1$is larger than the value of$m_2$, otherwise the velocity will be negative which cannot be possible.

Thus, the value of the final velocity of the projectile is $\frac{v_{1j}(m_1-m_2)}{m_1+m_2}$, the final velocity of the target is $\frac{2m_1{v}_{1j}}{m_1+m_2}$, and the final velocity of the projectile shows that $m_1\gg m_2$.